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The Universe of Discourse
https://blog.plover.com
The Universe of Discourse (Mark Dominus Blog)
en
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A puzzle about balancing test tubes in a centrifuge
https://blog.plover.com/2025/05/01#centrifuge
<p><img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5cdef%5cnk%231%232%7b%5cleft%5clangle%7b%231%20%5catop%20%232%7d%5cright%5crangle%7d%0a%5cdef%5cdd%231%7b%5cnk%7b12%7d%7b%231%7d%7d%0a%24"></p>
<p>Suppose a centrifuge has <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24n%24"> slots, arranged in a circle around the
center, and we have <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24k%24"> test tubes we wish to place into the slots.
If the tubes are not arranged symmetrically around the center, the
centrifuge will explode.</p>
<p>(By "arranged symmetrically around the center, I mean that if the
center is at <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%280%2c0%29%24">, then the sum of the positions of the tubes
must also be at <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%280%2c0%29%24">.)</p>
<p>Let's consider the example of <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24n%3d12%24">. Clearly we can arrange <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%242%24">,
<img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%243%24">, <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%244%24">, or <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%246%24"> tubes symmetrically:</p>
<div style="margin: auto;">
<img src="https://pic.blog.plover.com/math/centrifuge/n12k2.svg" style="width: 20%; margin: 2%;"
alt="twelve small circles arranged around a central point, like a
clock; circles 5 and 11 are filled in" />
<img src="https://pic.blog.plover.com/math/centrifuge/n12k3.svg" style="width: 20%; margin: 2%;"
alt="cirles 1, 5, and 9 are filled in" />
<img src="https://pic.blog.plover.com/math/centrifuge/n12k4.svg" style="width: 20%; margin: 2%;"
alt="circles 0, 3, 6, and 9 are filled" />
<img src="https://pic.blog.plover.com/math/centrifuge/n12k6.svg" style="width: 20%; margin: 2%;"
alt="circles 1, 3, 5, 7, 9, and 11 are filled" />
</div>
<p>Equally clearly
we can't arrange only <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%241%24">. Also it's easy to see we can do <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24k%24"> tubes if
and only if we can also do <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24n%2dk%24"> tubes, which rules out <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24n%3d12%2c%0ak%3d11%24">.</p>
<p>From now on I will write <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5cnk%20nk%24"> to mean the problem of balancing
<img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24k%24"> tubes in a centrifuge with <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24n%24"> slots. So <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5cdd%202%2c%20%5cdd%203%2c%20%5cdd%0a4%2c%20%24"> and <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5cdd%206%24"> are possible, and <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5cdd%201%24"> and <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5cdd%7b11%7d%24"> are
not. And <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5cnk%20nk%24"> is solvable if and only if <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5cnk%20n%7bn%2dk%7d%24"> is.</p>
<p>It's perhaps a little surprising that <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5cdd7%24"> is possible.
If you just ask this to someone out of nowhere they might
have a happy inspiration: “Oh, I'll just combine the solutions for
<img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5cdd3%24"> and <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5cdd4%24">, easy.” But that doesn't work because two groups
of the form <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%243i%2bj%24"> and <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%244i%2bj%24"> always overlap.</p>
<p>For example, if your group of <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%244%24"> is the
slots <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%240%2c%203%2c%206%2c%209%24"> then you can't also have your group of <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%243%24"> be
<img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%241%2c%205%2c%209%24">, because slot <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%249%24"> already has a tube in it. </p>
<div style="margin: auto;">
<img src="https://pic.blog.plover.com/math/centrifuge/n12k7x.svg" style="width: 30%;"
alt="The k=3 and k=4 diagrams from before, superimposed; there is a
small explosion symbol at 9 o'clock where they interfere">
</div>
<p>The
other balanced groups of <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%243%24"> are blocked in the same way. You
cannot solve the puzzle with <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%247%3d3%2b4%24">; you have to do <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%247%3d3%2b2%2b2%24"> as
below left.
The best way to approach this is to do <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5cdd5%24">, as below right.
This is easy,
since the triangle only blocks three of the six symmetric pairs.
Then you replace the holes with tubes and the tubes with holes to
turn <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5cdd5%24"> into <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5cdd%7b12%2d5%7d%3d%5cdd7%24">.</p>
<div style="margin: auto;">
<img src="https://pic.blog.plover.com/math/centrifuge/n12k7.svg" style="width: 25%; margin: 5%;"
alt="a triangle filling slots 1, 5, and 9, plus a pair at 0, 6 and
another pair at 2, 8" />
<img src="https://pic.blog.plover.com/math/centrifuge/n12k5.svg" style="width: 25%; margin: 5%;"
alt="the opposite of the previous diagram, consisting of a triangle
filling slots 3, 7, 11 and a pair at 4, 10" />
</div>
<p>Given <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24n%24"> and <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24k%24">, how can we decide whether the centrifuge can be
safely packed?</p>
<p>Clearly you can solve <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5cnk%20nk%24"> when <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24n%24"> is a multiple of <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24k%3e1%24">, but the example
of <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5cdd5%24"> (or <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5cdd7%24">) shows this isn't a necessary condition.</p>
<p>A generalization of this is that <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5cnk%20nk%24"> is always solvable
if <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5cgcd%28n%2ck%29%20%3e%201%24"> since you can easily
balance <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24g%20%3d%20%5cgcd%28n%2c%20k%29%24"> tubes at positions <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%240%2c%20%5cfrac%20ng%2c%20%5cfrac%7b2n%7dg%2c%20%5cdots%2c%0a%5cfrac%20%7b%28g%2d1%29n%7dg%24">, then do another <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24g%24"> tubes one position over, and
so on. For example, to do <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5cdd8%24"> you just put first four tubes
in slots <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%240%2c%203%2c%206%2c%209%24"> and the next four one position over, in slots
<img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%241%2c%204%2c%207%2c%2010%24">.</p>
<div style="margin: auto;">
<img src="https://pic.blog.plover.com/math/centrifuge/n12k8.svg" style="width: 30%;"
alt="Two squares this time, as described in the previous paragraph">
</div>
<p>An interesting counterexample is that the strategy for <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5cdd7%24">,
where we did <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%247%3d3%2b2%2b2%24">, cannot be extended to <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5cnk%7b14%7d9%24">. One
would want to do <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24k%3d7%2b2%24">, but there is no way to arrange the tubes
so that the group of <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%242%24"> doesn't conflict with the group of <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%247%24">,
which blocks one slot from every pair.</p>
<p>But we can see that this must be true without even considering the
geometry. <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5cnk%7b14%7d9%24"> is the reverse of <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5cnk%7b14%7d%7b14%2d9%7d%20%3d%20%5cnk%7b14%7d5%24">, which
impossible: the only nontrivial divisors of <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24n%3d14%24"> are <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%242%24"> and
<img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%247%24">, so <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24k%24"> must be a sum of <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%242%24">s and <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%247%24">s, and <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%245%24"> is not.</p>
<p>You can't fit <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24k%3d3%2b5%3d8%24"> tubes when <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24n%3d15%24">, but again the reason is
a bit tricky. When I looked at <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%248%24"> directly, I did a case analysis
to make sure that the <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%243%24">-group and the <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%245%24">-group would always
conflict. But again there was an easier was to see this: <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%248%3d15%2d7%24"> and
<img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%247%24"> clearly won't work, as <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%247%24"> is not a sum of <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%243%24">s and <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%245%24">s.
I wonder if there's an example where both <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24k%24"> and <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24n%2dk%24"> are not obvious?</p>
<p>For <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24n%3d20%24">, every <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24k%24"> works except <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24k%3d3%2c17%24"> and the always-impossible <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24k%3d1%2c19%24">.</p>
<p>What's the answer in general? I don't know.</p>
<h2>Addenda</h2>
<h3>20250502</h3>
<p>Now I am amusing myself thinking about the perversity of a centrifuge
with a prime number of slots, say <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%2413%24">. If you use it at all, you must
fill every slot. I hope you like explosions!</p>
<p>While I did not explode any centrifuges in university chemistry, I did
once explode an expensive <a href="https://en.wikipedia.org/wiki/Liebig_condenser">Liebig condenser</a>.</p>
<p><a href="https://pic.blog.plover.com/math/centrifuge/condenser.jpg"><img
src="https://pic.blog.plover.com/math/centrifuge/condenser-th.jpg" class="center" alt="A chemistry lab
apparatus. In the middle thje the Liebig condenser, a pair of long
concentric glass tubes, one inside the other, supported by a metal
ring stand. At left a heater heats a flash whose top is connected to
the upper end of the inner tube of the condenser. The condensate
collects in a flask at right. Two rubber tubes connect to the top and
bottom of the outer tube, carrying water through it."></a></p>
<p><span style="font-size: x-small">Condenser setup
by <a
href="https://commons.wikimedia.org/w/index.php?title=User:Mario_Link">Mario
Link</a> from <a
href="https://www.flickr.com/photos/24049265@N00/162821856">an
original image</a> by <a
href="https://www.flickr.com/photos/arlen/">Arlen on Flickr</a>.
Licensed <a
href="https://creativecommons.org/licenses/by/2.0/deed.en">cc-by-2.0</a>,
provided via <a
href="https://commons.wikimedia.org/wiki/File:Distillation_2-3.jpg">Wikimedia Commons</a>.</p>
<h3>20250503</h3>
<ul>
<li><p>Michael Lugo informs me that
<a href="https://mattbaker.blog/2018/06/25/the-balanced-centrifuge-problem/">a complete solution may be found on Matt Baker's math blog</a>. I
have not yet looked at this myself.</p></li>
<li><p><a href="https://mathstodon.xyz/@oantolin/114434195433254052">Omar Antolín points out an important consideration I missed</a>:
it may be necessary
to <em>subtract</em> polygons. Consider <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5cnk%7b30%7d6%24">. This is obviously
possible since <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%246%5cmid%2030%24">. But there is a more interesting
solution. We can add the pentagon <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5c%7b0%2c%206%2c%2012%2c%2018%2c%2024%5c%7d%24"> to the
digons <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5c%7b5%2c%2020%5c%7d%24"> and <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5c%7b10%2c%2025%5c%7d%24"> to obtain the solution
$${0,5,6,10,12,18, 20, 24, 25}.$$</p>
<p><img src="https://pic.blog.plover.com/math/centrifuge/n30k9.svg" style="width: 40%;" class="center"
alt=""/> Then from this we can <em>subtract</em> the triangle <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5c%7b0%2c%2010%2c%0a%20%2020%5c%7d%24"> to obtain $${5, 6, 12, 18, 24, 25},$$ a solution to
<img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5cnk%7b30%7d6%24"> which is <em>not</em> a sum of regular polygons:</p>
<p><img src="https://pic.blog.plover.com/math/centrifuge/n30k6b.svg" style="width: 40%;" class="center"
alt=""/></p></li>
<li><p>Thanks to Dave Long for pointing out a small but significant error,
which I have corrected.</p></li>
</ul>
<h3>20250505</h3>
<ul>
<li>Robin Houston points out this video,
<a href="https://www.youtube.com/watch?v=7DHE8RnsCQ8">The centrifuge Problem</a>
with Holly Krieger, on <a href="https://www.youtube.com/@numberphile">the Numberphile channel</a>.</li>
</ul>
-
Proof by insufficient information
https://blog.plover.com/2025/04/30#proof-by-symmetry
<p>Content warning: rambly</p>
<p>Given the coordinates of the three vertices of a triangle, can we find
the area? Yes. If by no other method, we can use the Pythagorean
theorem to find the lengths of the edges, and then
<a href="https://en.wikipedia.org/wiki/Heron%27s_formula">Heron's formula</a> to compute the area from
that.</p>
<p>Now, given the coordinates of the four vertices of a quadrilateral,
can we find the area? And the answer is, no, there is no method to do
that, <em>because there is not enough information</em>:</p>
<p><img class="center" border=0 src="https://pic.blog.plover.com/math/proof-by-symmetry/q1.svg" alt="three points
arranged in an irregular triangle, with one in the middle. Four of the six
possible edges are drawn in, definining a quadrilateral" />
<img class="center" border=0 src="https://pic.blog.plover.com/math/proof-by-symmetry/q2.svg" alt="the same
four points, but with three different edges drawn in to define a
different quadrilateral with the same vertices" />
<img class="center" border=0 src="https://pic.blog.plover.com/math/proof-by-symmetry/q3.svg" alt="a third
quadrilateral with the same vertices as the first two" /></p>
<p>These three quadrilaterals have the same vertices, but different
areas. Just knowing the vertices is not enough; you also need their order.</p>
<p>I suppose one could abstract this: Let <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24f%24"> be the function that maps
the set of vertices to the area of the quadrilateral. Can we
calculate values of <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24f%24">? No, because there is no such <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24f%24">, it is
not well-defined.</p>
<p>Put that way it seems less interesting. It's just another example of
the principle that, just because you put together a plausible sounding
description of some object, you cannot infer that such an object must
exist. One of the all-time pop hits here is:</p>
<blockquote>
<p>Let <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%ce%b5%24"> be the smallest [real / rational] number strictly greater than <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%240%24">…</p>
</blockquote>
<p>which appears on Math SE quite frequently. Another one I remember is
someone who asked about
<a href="https://math.stackexchange.com/q/838561/25554">the volume of a polyhedron with exactly five faces, all triangles</a>. This
is a fallacy at the ontological level, not the mathematical
level, so when it comes up I try to demonstrate it with a
nonmathematical counterexample, usually something like “the largest
purple hat in my closet” or perhaps “the current Crown Prince of the
Ottoman Empire”. The latter is less good because it relies on the
other person to know obscure stuff about the Ottoman Empire, whatever
that is.</p>
<p>This is also unfortunately also the error in Anselm's so-called
“ontological proof of God”. A philosophically-minded friend of mine
once remarked that being known for the discovery of the ontological
proof of God is like being known for the discovery that you can wipe
your ass with your hand.</p>
<p>Anyway, I'm digressing. The interesting part of the quadrilateral
thing, to me, is not so much that <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24f%24"> doesn't exist, but the specific
reasoning that demonstrates that it <em>can't</em> exist. I think there are
more examples of this proof strategy, where we prove nonexistence
by showing there is not enough information for the thing to exist, but
I haven't thought about it enough to come up with one.</p>
<p>There is a proof, the so-called
<a href="https://math.stackexchange.com/a/1313546/25554">“information-theoretic proof”</a>,
that a comparison sorting algorithm takes at least <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24O%28n%5clog%20n%29%24"> time, based
on comparing the amount of information gathered from the comparisons
(one bit each) with that required to distinguish all <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24n%21%20%24"> possible
permutations (<img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5clog_2%20n%21%20%5cge%20n%5clog_2%20n%24"> bits total). I'm not sure
that's what I'm looking for here. But I'm also not sure it isn't, or
why I feel it might be different.</p>
<h2>Addenda</h2>
<h3>20250430</h3>
<p>Carl Muckenhoupt suggests that logical independence proofs are of the
same sort. He says, for example:</p>
<blockquote>
<p>Is there a way to prove the parallel postulate from Euclid's other
axioms? No, there is not enough information. Here are two geometric
models that produce different results.</p>
</blockquote>
<p>This is just the sort of thing I was looking for.</p>
<h3>20250503</h3>
<p>Rik Signes has allowed me to reveal that he was the source of the
memorable disparagement of Anselm's dumbass argument.</p>
<!-- I had previously gotten blanket permission from him in connection
with still-unpublished article math/logic/properties.html -->
-
The mathematical past is a foreign country
https://blog.plover.com/2025/03/25#peano-starts-at-1
<p>A modern presentation of the <a href="https://en.wikipedia.org/wiki/Peano_axioms">Peano axioms</a> looks like
this:</p>
<ol>
<li><img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%240%24"> is a natural number</li>
<li>If <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24n%24"> is a natural number, then so is the result of appending an
<img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24S%24"> to the beginning of <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24n%24"></li>
<li>Nothing else is a natural number</li>
</ol>
<p>This baldly states that zero is a natural number.</p>
<p>I think this is a 20th-century development. In 1889, the natural
numbers started at <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%241%24">, not at <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%240%24">. Peano's
<a href="https://archive.org/details/arithmeticespri00peangoog/page/n6/mode/2up"><em>Arithmetices principia, nova methodo exposita</em></a>
(1889) is the source of the Peano axioms and in it Peano starts the
natural numbers at <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%241%24">, not at <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%240%24">:</p>
<p><a href="https://pic.blog.plover.com/math/peano-starts-at-1/Peano-1.png"><img src="https://pic.blog.plover.com/math/peano-starts-at-1/axiom-1.png"
class="center" alt="screencap of Peano's first axiom from his book.
It states “1∈N”."/></a></p>
<p>There's axiom 1: <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%241%5cin%5cBbb%20N%24">. No zero. I think starting at <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%200%24">
may be a Bourbakism.</p>
<p>In a modern presentation we define addition like this: </p>
<p>$$
\begin{array}{rrl}
(i) & a + 0 = & a \\
(ii) & a + Sb = & S(a+b)
\end{array}
$$</p>
<p>Peano doesn't have zero, so he doesn't need item <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%28i%29%24">. His definition
just has <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%28ii%29%24">.</p>
<p>But wait, doesn't his inductive definition need to have a base case? Maybe something like this?</p>
<p>\begin{array}{rrl}
(i') & a + 1 = & Sa \\
\end{array}</p>
<p>Nope, Peano has nothing like that. But surely the definition must
have a base case? How can Peano get around that?</p>
<p>Well, by modern standards, he cheats!</p>
<p>Peano doesn't have a special notation like <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24S%24"> for successor. Where
a modern presentation might write <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24Sa%24"> for the successor of the
number <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24a%24">, Peano writes “<img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24a%20%2b%201%24">”.</p>
<p>So his version of <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%28ii%29%24"> looks like this:</p>
<p>$$
a + (b + 1) = (a + b) + 1
$$</p>
<p>which is pretty much a symbol-for-symbol translation of <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%28ii%29%24">. But
if we try to translate <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%28i%27%29%24"> similarly, it looks like this:</p>
<p>$$
a + 1 = a + 1
$$</p>
<p>That's why Peano didn't include it: to him, it was tautological.</p>
<p>But to modern eyes that last formula is deceptive because it
equivocates between the "<img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%2b%201%24">" notation that is being used to
represent the successor operation (on the right) and the addition
operation that Peano is trying to define (on the left). In a modern
presentation, we are careful to distinguish between our formal symbol
for a successor, and our definition of the addition operation. </p>
<p>Peano, working pre-Frege and pre-Hilbert, doesn't have the same
concept of what this means. To Peano, constructing the successor of a
number, and adding a number to the constant <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%241%24">, are the <em>same</em>
operation: the successor operation <em>is</em> just adding <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%241%24">.</p>
<p>But to us, <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24Sa%24"> and <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24a%2bS0%24"> are different operations that happen to
yield the same value. To us, the successor operation is a purely
abstract or formal symbol manipulation (“stick an <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24S%24"> on the
front”). The fact that it also has an arithmetic interpretation,
related to addition, appears only once we contemplate the theorem
$$\forall a. a + S0 = Sa.$$ There is nothing like this in Peano.</p>
<p>It's things like this that make it tricky to read older mathematics
books. There are deep philosophical differences about what is being
done and why, and they are not usually explicit.</p>
<p>Another example: in the 19th century, the abstract presentation of
group theory had not yet been invented. The phrase “group” was
understood to be short for “group of permutations”, and the important
property was <em>closure</em>, specifically closure under composition of
permutations. In a 20th century abstract presentation, the closure
property is usually passed over without comment. In a modern view, the
notation <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24G_1%5ccup%20G_2%24"> is not even meaningful, because groups are
not sets and you cannot just mix together two sets of group elements
without also specifying how to extend the binary operation, perhaps
via a free product or something. In the 19th century, <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24G_1%5ccup%20G_2%24">
is perfectly ordinary, because <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24G_1%24"> and <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24G_2%24"> are just sets of
permutations. One can then ask whether that set is a group — that is,
whether it is closed under composition of permutations — and if not,
what is the smallest group that contains it.</p>
<p>It's something like a foreign language of a foreign
culture. You can try to translate the words, but the underlying ideas
may not be the same.</p>
<h3>Addendum 20250326</h3>
<p>Simon Tatham reminds me that Peano's equivocation has come up here
before.
<a href="https://blog.plover.com/math/se/2023-10.html#addition-with-successor">I previously discussed</a>
a Math SE post
<a href="https://math.stackexchange.com/q/4795154">in which OP was confused</a>
because Bertrand Russell's presentation of the Peano axioms similarly
used the notation “<img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%2b%201%24">” for the successor operation, and did not
understand why it was not tautological.</p>
-
Just ANSWER THE QUESTION
https://blog.plover.com/2025/02/03#just-answer-the-question
<!-- https://mathstodon.xyz/@mjd/113649962849017720 -->
<p>Here's a Math SE pathology that bugs me. OP will ask "I'm trying
to prove that groups <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24A%24"> and <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24B%24"> are isomorphic, I constructed this
bijection but I see that it's not a homomorphism. Is it sufficient,
or do I need to find a bijective homomorphism?"</p>
<p>And respondent <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24R%24"> will reply in the comments "How can a function which is
not an homomorphism prove that the groups are isomorphic?"</p>
<p>Which is literally the exact question that OP was asking! "Do I need
to find … a homomorphism?"</p>
<p>My preferred reply would be something like "Your function is not
enough. You are correct that it needs to be a homomorphism."</p>
<p>Because what problem did OP really have? Clearly, their problem is
that they are not sure what it means for two groups to be isomorphic.
For the respondent to ask "How can a function which is not an
homomorphism prove the the groups are isomorphic" is unhelpful because
they <em>know</em> that OP doesn't know the answer to that question.</p>
<p>OP knows too, that's exactly what their question was! They're trying
to find out the answer to that exact question! OP correctly
identified the gap in their own understanding. Then they formulated a
clear, direct question that would address the gap.</p>
<p>THEY ARE ASKING THE EXACT RIGHT QUESTION AND <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24R%24"> DID NOT ANSWER IT</p>
<p>My advice to people answering questions on MSE:</p>
<p><style type="text/css">
.maxim{ color: purple; font-size: 300%; text-align: center; }
.maxim-s{ color: purple; font-size: 200%; text-align: center; }
</style></p>
<div class="maxim">Just answer the question</div>
<p>It's all very well for <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24R%24"> to imagine that they are going to be
brilliant like Socrates, conducting a dialogue for that ages that draws from OP the
realization that the knowledge they sought was within them all along.
Except:</p>
<ol>
<li><img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24R%24"> is not Socrates</li>
<li>Nobody has time for this nonsense</li>
<li>The knowledge was <em>not</em> within them all along</li>
</ol>
<p>MSE is a site where people go to get answers to their questions. That
is its sole and stated purpose. If <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24R%24"> is not going to answer
questions, what are they even doing there? In my opinion, just
wasting everyone's time.</p>
<h3>Important pedagogical note</h3>
<p>It's sufficient to say "Your function is not enough", which answers
the question.</p>
<p>But it is <em>much better</em> to say "Your function is not enough. You are
correct that it needs to be a homomorphism". That acknowledges the
student's contribution. It tells them that their analysis of the
difficulty was correct!</p>
<p>They may not know what it means for two groups to be isomorphic, but
they <em>do</em> know one something almost as good: that they are unsure what
it means for two groups to be isomorphic. This is valuable knowledge.</p>
<p>This wise student recognises that they don't know. Socrates said that
he was the wisest of all men, because he at least “knew that he didn't
know”. If you want to take a lesson from Socrates, take that one, not
his stupid theory that all knowledge is already within us.</p>
<p>OP did what students are supposed to do: they reflected on their
knowledge, they realized it was inadequate, and they set about
rectifying it. This deserves positive reinforcement.</p>
<h2>Addenda</h2>
<ol>
<li><p>This is a real example. I have not altered it, because I am afraid
that if I did you would think I was exaggerating.</p></li>
<li><p>I have been banging this drum for decades, but I will cut the
scroll here. Expect a followup article.</p></li>
</ol>
<h3>20250206</h3>
<p><a href="https://blog.plover.com/misc/just-answer-the-question-2.html">The threatened followup article</a>, about the EFNet <code>#perl</code> channel
in the early 2000's.</p>
-
XKCD game theory question
https://blog.plover.com/2024/08/22#xkcd-game-theory
<p><img class="center"
src="https://pic.blog.plover.com/math/xkcd-game-theory/exam_numbers.png"
alt="Six-panel cartoon from XKCD.
Each panel gives a one-question mathematics ‘final exam’ from
a different level of education from ‘kindergarten’ to
‘postgraduate math’. This article concerns the fifth, which says
“Game Theory Final Exam: Q. Write down 10 more than the average of
the class’s answers. A. (blank).”" /></p>
<p><span style="font-size: x-small">(Source: <a href="https://xkcd.com/2966/">XKCD “Exam numbers”</a>.)</span></p>
<p>This post is about the bottom center panel, “Game Theory final exam”.</p>
<p>I don't know much about game theory and I haven't seen any other
discussion of this question. But I have a strategy I think is
plausible and I'm somewhat pleased with.</p>
<p>(I assume that answers to the exam question must be real numbers — not
<img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5cinfty%24"> — and
that “average” here is short for 'arithmetic mean'.)</p>
<p>First, I believe the other players and I must find a way to agree on
what the average will be, or else we are all doomed. We can't
communicate, so we should choose a Schelling point and hope that
everyone else chooses the same one. Fortunately, there is only one
distinguished choice: zero. So I will try to make the average zero
and I will hope that others are trying to do the same.</p>
<p>If we succeed in doing this, any winning entry will therefore be
<img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%2410%24">. Not all <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24n%24"> players can win because the average must be
<img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%240%24">. But <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24n%2d1%24"> can win, if the one other player writes
<img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%2d10%28n%2d1%29%24">. So my job is to decide whether I will be the loser. I
should select a random integer between <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%240%24"> and <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24n%2d1%24">. If it is
zero, I have drawn a short straw, and will write
<img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%2d10%28n%2d1%29%24">. otherwise I write <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%2410%24">.</p>
<p>(The straw-drawing analogy is perhaps misleading. Normally, exactly
one straw is short. Here, any or all of the straws might be short.)</p>
<p>If everyone follows this strategy, then I will win if exactly one
person draws a short straw and if that one person isn't me. The
former has a probability that rapidly approaches <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5cfrac1e%5capprox%0a36%2e8%5c%25%24"> as <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24n%24"> increases, and the latter is <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5cfrac%7bn%2d1%7dn%24">. In an <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24n%24">-person class,
the probability of my winning is $$\left(\frac{n-1}n\right)^n$$ which
is already better than <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5cfrac13%24"> when <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24n%3d%206%24">, and it increases slowly
toward <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%2436%2e8%5c%25%24"> after that.</p>
<p>Some miscellaneous thoughts:</p>
<ol>
<li><p>The whole thing depends on my idea that everyone will agree on
<img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%240%24"> as a Schelling point. Is that even how Schelling points
work? Maybe I don't understand Schelling points.</p></li>
<li><p>I like that the probability <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5cfrac1e%24"> appears. It's surprising
how often this comes up, often when multiple
agents try to coordinate without communicating. For example, in
<a href="https://en.wikipedia.org/wiki/ALOHAnet%23Slotted_ALOHA">ALOHAnet</a> a number of ground
stations independently try to send packets to a single satellite
transceiver, but if more than one tries to send a packet at a
particular time, the packets are garbled and must be retransmitted.
At most <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5cfrac1e%24"> of the available bandwidth can be used, the
rest being lost to packet collisions.</p></li>
<li><p>The first strategy I thought of was plausible but worse: flip a
coin, and write down <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%2410%24"> if it is heads and <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%2d10%24"> if it is
tails. With this strategy I win if exactly <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5cfrac%20n2%24"> of the
class flips heads and if I do too. The probability of this
happening is only $$\frac{n\choose n/2}{2^n}\cdot \frac12 \approx
\frac1{\sqrt{2\pi n}}.$$ Unlike the other strategy, this decreases to
zero as <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24n%24"> increases, and in no case is it better than the
first strategy. It also fails badly if the class contains an odd
number of people.</p>
<p>Thanks to Brian Lee for figuring out <a href="https://www.moderndescartes.com/essays/2n_choose_n/">the asymptotic value of
<img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%244%5e%7b%2dn%7d%5cbinom%7b2n%7d%7bn%7d%24"></a>
so I didn't have to.</p></li>
<li><p>Just because this was the best strategy I could think of in no way
means that it is the best there is. There might have been
something much smarter that I did not think of, and if there is
then my strategy will sabotage everyone else.</p>
<p>Game theorists do think of all sorts of weird strategies that you
wouldn't expect could exist.
<a href="https://blog.plover.com/math/envelope.html">I wrote an article about one a few years back</a>.</p></li>
<li><p>Going in the other direction, even if <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24n%2d1%24"> of the smartest
people all agree on the smartest possible strategy, if the <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24n%24">th
person is Leeroy Jenkins, he is going to ruin it for everyone.</p></li>
<li><p>If I were grading this exam, I might give full marks to anyone who
wrote down either <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%2410%24"> or <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%2d10%28n%2d1%29%24">, even if the average came
out to something else. <!-- (Slightly reminds me of the brilliant
answer to the exam question "Give a definition of risk": "No.") --></p></li>
<li><p>For a similar and also interesting but less slippery question, see
Wikipedia's article on
<a href="https://en.wikipedia.org/wiki/Guess_2%2f3_of_the_average">Guess ⅔ of the average</a>. Much of
the discussion there is directly relevant. For example, “For Nash
equilibrium to be played, players would need to assume both that
everyone else is rational and that there is common knowledge of
rationality. However, this is a strong assumption.” LEEROY
JENKINS<img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%0a%0a6%2e%20People%20sometimes%20suggest%20that%20the%20real%20Schelling%20point%20is%20for%0a%20%20%20everyone%20to%20write%20%24">\infty<img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%2e%20%20%28Or%20perhaps%20%24">-\infty<img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%2e%29%0a%20%20%20%0a%20%20%20Feh%2e%0a%0a7%2e%20If%20the%20class%20knows%20ahead%20of%20time%20what%20the%20question%20will%20be%2c%20the%0a%20%20%20strategy%20becomes%20a%20great%20deal%20more%20complicated%21%20Say%20there%20are%20six%0a%20%20%20students%2e%20%20At%20most%20five%20of%20them%20can%20win%2e%20So%20they%20get%20together%20and%0a%20%20%20draw%20straws%20to%20see%20who%20will%20make%20a%20sacrifice%20for%20the%20common%20good%2e%0a%20%20%20Vidkun%20gets%20the%20%28unique%29%20short%20straw%2c%20and%20agrees%20to%20write%20%24">-50<img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%2e%20%20The%0a%20%20%20others%20accordingly%20write%20%24">10<img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%2c%20but%20they%20discover%20that%20instead%20of%0a%20%20%20%24">-50<img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%2c%20Vidkun%20has%20written%20%24">22<img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%20and%20is%20the%20only%20person%20to%20have%20guessed%0a%20%20%20correctly%2e%0a%20%20%20%0a%20%20%20I%20would%20be%20interested%20to%20learn%20if%20there%20is%20a%20playable%20Nash%0a%20%20%20equilibrium%20under%20these%20circumstances%2e%20%20It%20might%20be%20that%20the%0a%20%20%20optimal%20strategy%20is%20for%20everyone%20to%20play%20as%20if%20they%20_didn%27t_%20know%0a%20%20%20what%20the%20question%20was%20beforehand%21%0a%20%20%20%0a%20%20%20Suppose%20the%20players%20agree%20to%20follow%20the%20strategy%20I%20outlined%2c%20each%0a%20%20%20rolling%20a%20die%20and%20writing%20%24">-50<img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%20with%20probability%20%24">\frac16<img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%2c%20and%0a%20%20%20%24">10<img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%20otherwise%2e%20And%20suppose%20that%20although%20the%20others%20do%20this%2c%0a%20%20%20Vidkun%20skips%20the%20die%20roll%20and%20unconditionally%20writes%20%24">10<img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%2e%20%20As%0a%20%20%20before%2c%20%24">n-1!! players (including Vidkun) win if exactly one of
them rolls zero. Vidkun's chance of winning increases. Intuitively,
the other players' chances of winning ought to decrease. But by
how much? I think I keep messing up the calculation because I keep
getting zero. If this were actually correct, it would be a
fascinating paradox!</p></li>
</ol>
-
Everyday examples of morphisms with one-sided inverses
https://blog.plover.com/2024/08/05#inverses
<p>Like almost everyone except Alexander Grothendieck, I understand
things better with examples. For instance, how do you explain that</p>
<p>$$(f\circ g)^{-1} = g^{-1} \circ f^{-1}?$$</p>
<p>Oh, that's easy. Let <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24f%24"> be putting on your shoes
and <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24f%5e%7b%2d1%7d%24"> be taking off your shoes.
And let <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24g%24"> be
putting on your socks and <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24g%5e%7b%2d1%7d%24"> be taking off your socks.</p>
<p>Now <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24f%5ccirc%20g%24"> is putting on your socks and then your shoes. And
<img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24g%5e%7b%2d1%7d%20%5ccirc%20f%5e%7b%2d1%7d%24"> is taking off your shoes <em>and then</em> your
socks. You can't <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24f%5e%7b%2d1%7d%20%5ccirc%20g%5e%7b%2d1%7d%24">, that says to take your
socks off before your shoes.</p>
<p>(I see a topologist jumping up and down in the back row, desperate to
point out that the socks were never inside the shoes to begin with.
Sit down please!)</p>
<p>Sometimes operations commute, but not in general. If you're teaching
group theory to high school students and they find nonabelian
operations strange, the shoes-and-socks example is an unrebuttable
demonstration that not everything is abelian.</p>
<p>(Subtraction is <em>not</em> a good example here, because subtracting <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24a%24">
and then <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24b%24"> is the same as subtracting <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24b%24"> and then <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24a%24">. When
we say that subtraction isn't commutative, we're talking about
something else.)</p>
<p>Anyway this weekend I was thinking about very very elementary
category theory (the only kind I know) and about left and right
inverses. An arrow <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24f%20%3a%20A%5cto%20B%24"> has a left inverse <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24g%24"> if</p>
<p>$$g\circ f = 1_A.$$</p>
<p>Example of this are easy. If <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24f%24"> is putting on your shoes, then
<img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24g%24"> is taking them off again. <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24A%24"> is the state of shoelessness
and <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24B%24"> is the state of being shod. This <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24f%24"> has a left inverse
and no right inverse. You can't take the shoes off before you put
them on.</p>
<p>But I wanted an example of an <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24f%24"> with right inverse and no left inverse:</p>
<p>$$f\circ h = 1_B$$</p>
<p>and I was pretty pleased when I came up with one involving pouring the
cream pitcher into your coffee, which has no left inverse that gets
you back to black coffee. But you can ⸢unpour⸣ the cream if you do it
<em>before</em> mixing it with the coffee: if you first put the cream back
into the carton in the refrigerator, then the pouring does get you to
black coffee.</p>
<p>But now I feel silly. There is a trivial theorem that if <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24g%24"> is a
left inverse of <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24f%24">, then <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24f%24"> is a right inverse of <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24g%24">. So
the shoe example will do for both. If <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24f%24"> is putting on your shoes,
then <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24g%24"> is taking them off again. And just as <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24f%24"> has a left
inverse and no right inverse, because you can't take your shoes off
before putting them on, <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24g%24"> has a right inverse (<img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24f%24">) and no left inverse,
because you can't take your shoes off before putting them on.</p>
<p>This reminds me a little of the time I tried to construct an example
to show that “is a blood relation of" is not a transitive relation. I
had this very strange and elaborate example involving two sets of
sisters-in-law. But the right example is that almost everyone is the
blood relative of both of their parents, who nevertheless are not
(usually) blood relations.</p>
-
A triviality about numbers that look like abbc
https://blog.plover.com/2024/07/05#multiple-of-37
<p>Looking at license plates the other day I noticed that if you have a
four-digit number <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24N%24"> with digits <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24abbc%24">, and <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24a%2bc%3db%24">, then
<img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24N%24"> will always be a multiple of <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%2437%24">. For example, <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%244773%20%3d%0a37%5ccdot%20129%24"> and <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%241776%20%3d%2037%5ccdot%2048%24">.</p>
<p>Mathematically this is uninteresting. The proof is completely
trivial. (Such a number is simply <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%241110a%20%2b111c%24">, and
<img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24111%3d3%5ccdot%2037%24">.)</p>
<p>But I thought that if someone had pointed this out to me when I was
eight or nine, I would have been very pleased. Perhaps if you have a
mathematical eight- or nine-year-old in your life, they will be
pleased if you share this with them.</p>
-
Well, I guess I believe everything now!
https://blog.plover.com/2024/04/23#k2
<p>The <em>principle of explosion</em> is that in an inconsistent system
<em>everything</em> is provable: if you prove both <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24P%24"> and not-<img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24P%24"> for
<em>any</em> <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24P%24">,
you can then conclude <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24Q%24"> for <em>any</em> <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24Q%24">: </p>
<p>$$(P \land \lnot P) \to Q.$$</p>
<p>This is, to put it briefly, not intuitive. But it is <em>awfully</em> hard
to get rid of because it appears to follow immediately from two
principles that <em>are</em> intuitive:</p>
<ol>
<li><p>If we can prove that <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24A%24"> is true, then we can prove that at least
one of <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24A%24"> or <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24B%24"> is true. (In symbols, <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24A%5cto%28A%5clor%20B%29%24">.)</p></li>
<li><p>If we can prove that at least one of <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24A%24"> or <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24B%24"> is true, and we
can prove that <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24A%24"> is false, then we may conclude that that <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24B%24"> is
true. (Symbolically, <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%28A%5clor%20B%29%20%5cto%20%28%5clnot%20A%5cto%20B%29%24">.).</p></li>
</ol>
<p>Then suppose that we have proved that <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24P%24"> is both true and false.
Since we have proved <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24P%24"> true, we have proved that at least one of
<img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24P%24"> or <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24Q%24"> is true. But because we have also proved that <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24P%24"> is
false, we may conclude that <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24Q%24"> is true. Q.E.D.</p>
<p>This proof is as simple as can be. If you want to get rid of this, you
have a hard road ahead of you. You have to follow Graham Priest into
the wilderness of <a href="https://en.wikipedia.org/wiki/paraconsistent_logic">paraconsistent logic</a>.</p>
<p>Raymond Smullyan observes that although logic is supposed to model
ordinary reasoning, it really falls down here. Nobody, on discovering
the fact that they hold contradictory beliefs, or even a false one,
concludes that therefore they must believe <em>everything</em>. In fact,
says Smullyan, almost everyone <em>does</em> hold contradictory beliefs. His
argument goes like this:</p>
<ol>
<li><p>Consider all the things I believe individually, <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24B_1%2c%20B_2%2c%0a%20%20%20%5cldots%24">. I believe each of these, considered separately, is true.</p></li>
<li><p>However, I <em>also</em> believe that I'm not infallible, and that at
least one of <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24B_1%2c%20B_2%2c%20%5cldots%24"> is false, although I don't know
which ones.</p></li>
<li><p>Therefore I believe both <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5cbigwedge%20B_i%24"> (because I believe each
of the <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24B_i%24"> separately) and <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5clnot%5cbigwedge%20B_i%24"> (because I
believe that not all the <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24B_i%24"> are true).</p></li>
</ol>
<p>And therefore, by the principle of explosion, I ought to believe that
I believe absolutely everything.</p>
<p>Well anyway, none of that was exactly what I planned to write about.
I was pleased because I noticed a very simple, specific example of
something I believed that was clearly inconsistent. Today I learned
that <a href="https://en.wikipedia.org/wiki/K2">K2</a>, the second-highest mountain in the world, is in
Asia, near the border of Pakistan and westernmost China. I was
surprised by this, because I had thought that K2 was in Kenya
somewhere.</p>
<p>But I <em>also</em> knew that the highest mountain in Africa was
Kilimanjaro. So my simultaneous beliefs were flatly contradictory:</p>
<ol>
<li>K2 is the second-highest mountain in the world.</li>
<li>Kilimanjaro is not the highest mountain in the world, but it is the
highest mountain in Africa</li>
<li>K2 is in Africa</li>
</ol>
<p>Well, I guess until this morning I must have believed everything!</p>
-
3-coloring the vertices of an icosahedron
https://blog.plover.com/2024/04/13#colored-icosahedron
<p>I don't know that I have a point about this, other than that it makes me sad.</p>
<p><a href="https://math.stackexchange.com/q/4897997/25554">A recent Math SE post (since deleted) asked</a>:</p>
<blockquote>
<p>How many different ways are there to color the vertices of the
icosahedron with 3 colors such that no two adjacent vertices have
the same color?</p>
</blockquote>
<p>I would love to know what was going on here. Is this homework? Just
someone idly wondering?</p>
<p>Because the interesting thing about this question is (assuming that
the person knows what an icosahedron is, etc.) it should be solvable
in sixty seconds by anyone who makes the least effort. If you don't
already see it, you should try. Try what? Just take an icosahedron,
color the vertices a little, see what happens. Here, I'll help you
out, here's a view of part of the end of an icosahedron, although I
left out most of it. Try to color it with 3 colors so that no two
adjacent vertices have the same color, surely that will be no harder
than coloring the whole icosahedron.</p>
<p><img src="https://pic.blog.plover.com/math/colored-icosahedron/pentagon.svg" class="center" /></p>
<p>The explanation below is a little belabored, it's what OP
would have discovered in seconds if they had actually tried the
exercise.</p>
<p>Let's color the middle vertex, say blue.</p>
<p><img src="https://pic.blog.plover.com/math/colored-icosahedron/pentagon-2.svg" class="center" /></p>
<p>The five vertices around the edge can't be blue, they must be the
other two colors, say red and green, and the two colors must alternate:</p>
<p><img src="https://pic.blog.plover.com/math/colored-icosahedron/pentagon-3.svg" class="center" /></p>
<p>Ooops, there's no color left for the fifth vertex.</p>
<p><img src="https://pic.blog.plover.com/math/colored-icosahedron/pentagon-4.svg" class="center" /></p>
<p>The phrasing of the question, “how many” makes the problem sound
harder than it is: the answer is zero because we can't even color
<em>half</em> the icosahedron.</p>
<p>If OP had even tried, even a little bit, they could have discovered
this. They didn't need to have had the bright idea of looking at a a
partial icosahedron. They could have grabbed one of the pictures from
Wikipedia and started coloring the vertices. They would have gotten
stuck the same way. They didn't have to try starting in the middle of
my diagram, starting at the edge works too: if the top vertex is blue,
the three below it must be green-red-green, and then the bottom two
are forced to be blue, which isn't allowed. If you <em>just try it</em>, you
win immediately. The only way to lose is not to play.</p>
<p>Before the post was deleted I suggested in a comment “Give it a try,
see what happens”. I genuinely hoped this might be helpful. I'll
probably never know if it was.</p>
<p>Like I said, I would love to know what was going on here. I think
maybe this person could have used a dose of Lower Mathematics.</p>
<p>Just now I wondered for the first time: what would it look like if I
were to try to list the principles of Lower Mathematics? “Try it and
see” is definitely in the list.</p>
<div class="bookbox"><table align=right width="20%" bgcolor="#ffffdd"
border=1><tr><td align=center><A HREF="https://bookshop.org/a/93187/9780691164076"><font
size="-2">Buy</font><br><cite><font>How to Solve It</font></cite><br><IMG SRC="https://images-us.bookshop.org/ingram/9780691164076.jpg?height=250&v=v2" BORDER="0"
ALIGN="center" ALT="(How to Solve It cover missing)" ><br>
<font size="-2">from Bookshop.org<br>(with kickback)</a><br><a href="https://bookshop.org/a/00000/9780691164076">(without kickback)</a></font></a>
</td></tr></table></div>
<p>Then I thought: <em>How To Solve It</em> has that sort of list and something like “try it
and see” is
probably on it. So I took it off the shelf and found: “Draw a
figure”, “If you cannot solve the proposed problem”, “Is it possible
to satisfy the condition?”. I didn't find anything called “fuck
around with it and see what you learn” but it is probably in there
under a different name, I haven't read the book in a long time. To
this important principle I would like to add “fuck around with it and
maybe you will stumble across the answer by accident” as happened
here.</p>
<p>Mathematics education is too much method, not enough heuristic.</p>
-
Optimal boxes with and without lids
https://blog.plover.com/2024/03/06#optimal-boxes
<p>Sometime around 1986 or so I considered the question of the dimensions
that a closed cuboidal box must have to enclose a given volume
but use as little material as possible. (That is, if its surface area
should be minimized.) It is an elementary calculus exercise and it is
unsurprising that the optimal shape is a cube.</p>
<p>Then I wondered: what if the box is open at the top, so that
it has only five faces instead of six? What are the optimal
dimensions then?</p>
<p>I did the calculus, and it turned out that the optimal lidless box has
a square base like the cube, but it should be exactly half as tall.</p>
<p>For example the optimal box-with-lid enclosing a cubic meter is a
1×1×1 cube with a surface area of <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%246%24">.</p>
<p>Obviously if you just cut off the lid of the cubical box and throw it
away you have a one-cubic-meter lidless box with a surface area of
<img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%245%24">. But the <em>optimal</em> box-without-lid enclosing a cubic meter is
shorter, with a larger base. It has dimensions $$2^{1/3} \cdot
2^{1/3} \cdot \frac{2^{1/3}}2$$</p>
<p>and a total surface area of only <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%243%5ccdot2%5e%7b2%2f3%7d%20%5capprox%204%2e76%24">. It
is what you would get if
you took an optimal complete box, a cube, that enclosed <em>two</em> cubic
meters, cut it in half, and threw the top half away.</p>
<p>I found it striking that the optimal lidless box was the same
proportions as the optimal complete box, except half as tall. I asked
Joe Keane if he could think of any reason why that should be obviously
true, without requiring any calculus or computation. “Yes,” he said.
I left it at that, imagining that at some point I would consider it at
greater length and find the quick argument myself.</p>
<p>Then I forgot about it for a while.</p>
<p>Last week I remembered again and decided it was time to consider it at
greater length and find the quick argument myself. Here's the explanation.</p>
<p>Take the cube and saw it into two equal halves. Each of these is a
lidless five-sided box like the one we are trying to construct. The
original cube enclosed a certain volume with the minimum possible material.
The two half-cubes each enclose half the volume with half the
material.</p>
<p>If there were a way to do better than that, you would be able to make a
lidless box enclose half the volume with <em>less</em> than half the
material. Then you could take two of those and glue them back
together to get a complete box that enclosed the original volume with
less than the original amount of material. But we already knew that
the cube was optimal, so that is impossible.</p>
-
Math SE report 2023-10: Peano's definition of addition is not a tautology, and what was great about Ramanujan?
https://blog.plover.com/2023/12/02#2023-10
<p>Content warning: grumpy complaining. This was a frustrating month.</p>
<h3><a href="https://math.stackexchange.com/q/4781798/25554">Need an intuitive example for how "P is necessary for Q" means "Q⇒P"?</a></h3>
<p>This kind of thing comes up pretty often. Why are there so many ways that the logical
expression <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24Q%5cimplies%20P%24"> can appear in natural language?</p>
<ul>
<li>If <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24Q%24">, then <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24P%24"></li>
<li><img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24Q%24"> implies <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24P%24"></li>
<li><img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24P%24"> if <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24Q%24"></li>
<li><img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24Q%24"> is sufficient for <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24P%24"></li>
<li><img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24P%24"> is necessary for <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24Q%24"></li>
</ul>
<p>Strange, isn't it? <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24Q%5cland%20P%24"> is much simpler: “Both <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24Q%24"> and <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24P%24"> are true” is pretty much it.</p>
<p>Anyway this person wanted an intuitive example of “<img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24P%24"> is necessary for <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24Q%24">”</p>
<p>I suggested:</p>
<blockquote>
<p>Suppose that it is <strong>necessary</strong> to have a ticket (<img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24P%24">) in order to
board a certain train (<img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24Q%24">). That is, <strong>if</strong> you board the train
(<img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24Q%24">), <strong>then</strong> you have a ticket (<img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24P%24">).</p>
</blockquote>
<p>Again this follows the principle that <em>rule enforcement</em> is a good
thing when you are looking for intuitive examples. Keeping ticketless
people off the train is something that the primate brain is wired up
to do well.</p>
<p>My first draft had “board a train” in place of “board a certain
train”. One commenter complained:</p>
<blockquote>
<p>many people travel on trains without a ticket, worldwide</p>
</blockquote>
<p>I was (and am) quite disgusted by this pettifogging.</p>
<blockquote>
<p>I said “<strong>Suppose that</strong>…”. I was not claiming that the condition applies to every train in all of history.</p>
</blockquote>
<p>OP had only asked for an example, not some universal principle.</p>
<h3><a href="https://math.stackexchange.com/q/4792771/25554">Does ...999.999... = 0?</a></h3>
<p>This person is asking one of those questions that often puts Math
StackExchange into the mode of insisting that the idea is completely
nonsensical, when it is actually very close to perfectly mundane
mathematics. (Previously:
<a href="https://blog.plover.com/misc/half-baked.html">[1]</a>
<a href="https://blog.plover.com/math/se/2023-04.html#half-baked">[2]</a>
<a href="https://blog.plover.com/math/se/2023-02.html#half-baked">[3]</a> ) That <em>didn't</em> happen
this time, which I found very gratifying.</p>
<p>Normally, decimal numerals have a finite integer part on the left of
the decimal point, and an infinite fractional part on the right of the
decimal point, as with (for example) <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5cfrac%7b13%7d%7b3%7d%20%3d%204%2e333%5cldots%24">.
It turns out to work surprisingly well to reverse this, allowing an
infinite integer part on the left and a finite fractional part on the
right, for example <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5cfrac25%20%3d%20%5cldots%20333%2e4%24">. For technical
reasons <a href="https://en.wikipedia.org/wiki/P%2dadic_number">we usually do this in base <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24p%24"> where <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24p%24"> is
prime</a>; it doesn't work as well in base <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%2410%24">.
But it works well enough to use: If we have the base-10 numeral
<img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5cldots%209999%2e0%24"> and we add <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%241%24">, using the ordinary
elementary-school right-to-left addition algorithm, the carry in the
units place goes to the tens place as usual, then the next carry goes
to the hundreds place and so on to infinity, leaving us with <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5cldots%0a0000%2e0%24">, so that <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5cldots%209999%2e0%24"> can be considered a representation
of the number <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%2d1%24">, and that means we don't need negation signs.</p>
<p>In fact this system is fundamental to the way numbers are represented
in computer arithmetic. Inside the computer the integer <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%2d1%24"> is
literally represented as the base-2 numeral
<img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%2411111111%5c%3b11111111%5c%3b11111111%5c%3b11111111%24">, and when we add <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%241%24"> to
it the carry bit wanders off toward infinity on the left. (In the
computer the numeral is finite, so we simulate infinity by just
discarding the carry bit when it gets too far away.)</p>
<p>Once you've seen this a very reasonable next question is whether you
can have numbers that have an infinite sequence of digits on both
sides. I think something goes wrong here — for one thing it is no
longer clear how to actually do arithmetic. For the
infinite-to-the-left numerals arithmetic is straightforward
(elementary-school algorithms go right-to-left anyway) and for the
standard infinite-to-the-right numerals we can sort of fudge it. (Try
multiplying the infinite decimal for <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5csqrt%202%24"> by itself and see
what trouble you get into. Or simpler: What's <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%244%2e666%5cldots%20%5ctimes%0a3%24">?)</p>
<p>OP's actual question was: If <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5cldots%209999%2e0%20%24"> can be considered to
represent <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%2d1%24">, and if <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%240%2e9999%5cldots%24"> can be considered to
represent <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%241%24">, can we add them and conclude that <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5cldots%0a9999%2e9999%5cldots%20%3d%200%24">?</p>
<p>This very deserving question got <a href="https://math.stackexchange.com/a/4792785/25554">a good answer from someone who was
not me</a>. This was a
relief, because <em>my</em> shameful answer was pure shitpostery. It should have been
heavily downvoted, but wasn't. The gods of Math SE karma are capricious.</p>
<p><a name="addition-with-successor">‌</a></p>
<h3><a href="https://math.stackexchange.com/q/4795154/25554">Why define addition with successor?</a></h3>
<p>Ugh, so annoying. OP had read (Bertrand Russell's explanation of) the
Peano definition of addition, and did not understand it. Several
people tried hard to explain, but communication was not happening.
Or, perhaps, OP was more interested in having an argument than in
arriving at an understanding. I lost a bit of my temper when
they claimed:</p>
<blockquote>
<p>Russell's so-called definition of addition (as quoted in my
question) is nothing but a tautology: ….</p>
</blockquote>
<p>I didn't say:</p>
<blockquote>
<p>If you think Bertrand Russell is stupid, it's because you're stupid.</p>
</blockquote>
<p>although I wanted to at first. The reply I did make is still not as
measured as I would like, and although it leaves this point implicit,
the point is still there. I did at least shut up after that. I had
answered OP's question as well as I was able, and carrying on a
complex discussion in the comments is almost never of value.</p>
<h3><a href="https://math.stackexchange.com/q/4799789/25554">Why is Ramanujan considered a great mathematician?</a></h3>
<p>This was easily my best answer of the month, but the question was
deleted, so you will only be able to see it if you have enough Math SE
reputation.</p>
<p>OP asked a perfectly reasonable question: Ramanujan gets a lot of
media hype for stuff like this:</p>
<p>$${\sqrt {\phi +2}}-\phi ={\cfrac {e^{{-2\pi /5}}}{1+{\cfrac
{e^{{-2\pi }}}{1+{\cfrac {e^{{-4\pi }}}{1+{\cfrac {e^{{-6\pi
}}}{1+\,\cdots }}}}}}}}$$</p>
<p>which is not of any obvious use, so “why is it given such high regard?”</p>
<p>OP appeared to be impugning a famous mathematician, and Math SE always
responds badly to that; their heroes must not be questioned. And even
worse, OP mentioned the notorious non-fact that $$1+2+3+\ldots
=-\frac1{12}$$ which drives Math SE people into a frothing rage.</p>
<p>One commenter argued:</p>
<blockquote>
<p>Mathematics is not inherently about its "usefulness". Even if you
can't find practical use for those formulas, you still have to admit
that they are by no means trivial</p>
</blockquote>
<p>I think this is fatuous. OP is right here, and the commenter is
wrong. Mathematicians are not considered great because they produce
wacky and impractical equations. They are considered great because
they solve problems, invent techniques that answer previously
impossible questions, and because they contribute insights into deep
and complex issues.</p>
<p>Some blockhead even said:</p>
<blockquote>
<p>Most of the mathematical results are useless. Mathematics is more like an art.</p>
</blockquote>
<p>Bullshit. Mathematics is about trying to understand stuff, not about
taping a banana to the wall. I replied:</p>
<blockquote>
<p>I don't think “mathematics is not inherently about its usefulness"
is an apt answer here. Sometimes mathematical results have
application to physics or engineering. But for many mathematical
results the application is to other parts of mathematics, and
mathematicians do judge the ‘usefulness’ of results on this
basis. Consider for example Mochizuki's field of “inter-universal
Teichmüller theory”. This was considered interesting only as long as
it appeared that it might provide a way to prove the <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24abc%24">
conjecture. When that hope collapsed, everyone lost interest in it.</p>
</blockquote>
<p>My answer to OP elaborated on this point:</p>
<blockquote>
<p>The point of these formulas wasn't that they were useful in
themselves. It's that in order to find them he had to have a deep
understanding of matters that were previously unknown. His
contribution was the deep understanding.</p>
</blockquote>
<p>I then discussed Hardy's book on the work he did with Ramanujan and
<a href="https://archive.org/details/dli.csl.3438">Hardy's own estimation of Ramanujan's work</a>:</p>
<blockquote>
<p>The first chapter is somewhat negative, as it summarizes the parts
of Ramanujan's work that he felt didn't have lasting value — because
Hardy's next eleven chapters are about the work that he felt <em>did</em>
have value.</p>
</blockquote>
<p>So if OP wanted a substantive and detailed answer to their question,
that would be the first place to look.</p>
<p>I also did an arXiv search for “Ramanujan” and found many recent
references, including one with “applications to the Ramanujan
<img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%cf%84%24">-function”, and concluded:</p>
<blockquote>
<p><a href="https://en.wikipedia.org/wiki/Ramanujan_tau_function">The
<img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5ctau%24">-function</a>
is the subject of the entire chapter 10 of Hardy's book and appears
to still be of interest as recently as last Monday.</p>
</blockquote>
<p>The question was closed as “opinion-based” (a criticism that I
think my answer completely demolishes) and then it was deleted. Now
if someone else trying to find out why Ramanujan is held in high regard
they will not be able to find my factual, substantive answer.</p>
<p>Screw you, Math SE. This month we both sucked.</p>
-
Uncountable sets for seven-year-olds
https://blog.plover.com/2023/11/27#uncountable-hotel
<p>I was recently talking to a friend whose seven-year old had been
reading about
<a href="https://en.wikipedia.org/wiki/Hilbert%27s_paradox_of_the_Grand_Hotel">the Hilbert Hotel paradoxes</a>.
One example: The hotel is completely full when a bus arrives with 53
passengers seeking rooms. Fortunately the hotel has a countably
infinite number of rooms, and can easily accomodate 53 more guests
even when already full.</p>
<p>My friend mentioned that his kid had been unhappy with the associated
discussion of <em>uncountable</em> sets, since the explanation he got
involved something about people
whose names are infinite strings, and it got confusing. I said yes,
that is a bad way to construct the example, because names are <em>not</em>
infinite strings, and even one infinite string is hard to get your
head around. If you're going to get value out of the hotel metaphor,
you waste an opportunity if you abandon it for some weird mathematical
abstraction. (“Okay, Tyler, now let <img src="https://chart.apis.google.com/chart?chf=bg,s,00000000&cht=tx&chl=%24%5cmathscr%20B%24"> be a projection from a
vector bundle onto a compact Hausdorff space…”)</p>
<p>My first attempt on the spur of the moment involved the guests
belonging to clubs, which meet in an attached convention center with a
countably infinite sequence of meeting rooms. The club idea is good
but my original presentation was overcomplicated and after thinking
about the issue a little more I sent this email with my ideas for
how to explain it to a bright seven-year-old.</p>
<hr />
<p>Here's how I think it should go. Instead of a separate hotel and
convention center, let's just say that during the day the guests
vacate their rooms so that clubs can meet in the same rooms. Each
club is assigned one guest room that they can use for their meeting
between the hours of 10 AM to 4 PM. The guest has to get out of the
room while that is happening, unless they happen to be a member of the
club that is meeting there, in which case they may stay.</p>
<p>If you're a guest in the hotel, you might be a member of the club that
meets in your room, or you might not be a member of the club that
meets in your room, in which case you have to leave and go to a
meeting of one of your clubs in some other room.</p>
<p>We can paint the guest room doors blue and green: blue, if the guest
there is a member of the club that meets in that room during the day,
and green if they aren't. Every door is now painted blue or green,
but not both.</p>
<p>Now I claim that when we were assigning clubs to rooms, there was a
club we missed that has nowhere to meet. It's the Green Doors Club of
all the guests who are staying in rooms with green doors.</p>
<p>If we did assign the Green Doors Club a guest room in which to meet,
that door would be painted green or blue.</p>
<p>The Green Doors Club isn't meeting in a room with a blue door. The
Green Doors Club only admits members who are staying in rooms with
<em>green</em> doors. That guest belongs to the club that meets in
their room, and it isn't the Green Doors Club because the guest's door is
blue.</p>
<p>But the Green Doors Club isn't meeting in a room with a green door.
We paint a door green when the guest is <em>not</em> a member of the club
that meets in their room, and this guest <em>is</em> a member of the Green
Doors Club.</p>
<p>So however we assigned the clubs to the rooms, we must have missed out
on assigning a room to the Green Doors Club.</p>
<p>One nice thing about this is that it works for finite hotels too. Say
you have a hotel with 4 guests and 4 rooms. Well, obviously you can't
assign a room to each club because there are 16 possible clubs and
only 4 rooms. But the blue-green argument still works: you can assign
any four clubs you want to the four rooms, then paint the doors, then
figure out who is in the Green Doors Club, and then observe that, in
fact, the Green Doors Club is not one of the four clubs that got a
room.</p>
<p>Then you can reassign the clubs to rooms, this time making sure that
the Green Doors Club gets a room. But now you have to repaint the
doors, and when you do you find out that membership in the Green Doors
Club has changed: some new members were admitted, or some former
members were expelled, so the club that meets there is no longer the
Green Doors Club, it is some other club. (Or if the Green Doors Club
is meeting somewhere, you will find that you have painted the doors
wrong.)</p>
<hr />
<p>I think this would probably work. The only thing that's weird about
it is that some clubs have an infinite number of members so that it's
hard to see how they could all squeeze into the same room. That's
okay, not every member attends every meeting of every club they're in,
that would be impossible anyway because everyone belongs to multiple
clubs.</p>
<p>But one place you could go from there is: what if we only guarantee
rooms to clubs with a <em>finite</em> number of members? There are only a
countably infinite number of clubs then, so they do all fit into the
hotel! Okay, Tyler, but what happens to the Green Door Club then?
I said all the finite clubs got rooms, and we know the Green Door Club
<em>never</em> gets a room, so what can we conclude?</p>
<p>It's tempting to try to slip in a reference to Groucho Marx, but I
think it's unlikely that that will do anything but confuse matters.</p>
<p>[ <a href="https://blog.plover.com/math/russell.html">Previously</a> ]</p>
<p>[ Update: My friend said he tried it and it didn't go over as well as I thought it might. ]</p>