Wikipedia:Reference desk/Archives/Mathematics/2024 January 6

= January 6 =

Cutting up Regular Polytopes.
Is the following true from the definition of a Regular Polytope? For any n dimensional Polytope make up of k n-1 dimensional cells (so a Octahedron is make up of 8 2 dimensional triangles and a 24-cell is made of 24 3 dimensional octohedrons) and s*t = k that *any* selection of s cells out of the k in the polytope can be replicated t times and "assembled" into the polytope. This would mean, for example, that the faces of an icoasahedron could be colored with 5 cells each of one of four colors where each color is the same three dimensional shape.Naraht (talk) 10:48, 6 January 2024 (UTC)
 * With an icosahedron you can put a band of ten cells around in a ring so the other tenform two groups of five. NadVolum (talk) 12:58, 6 January 2024 (UTC)
 * Pentagonal pyramid.png me try to give a full proof that the regular icosahedron is a counterexample (if I understand the conjecture correctly). Let a pyramidal cap be the figure formed by the lateral faces of a pyramid, obtained by removing its base. Consider the pyramidal cap of the pyramid whose base is a regular pentagon and whose five lateral faces are equilateral triangles (see the figure). Let us call it the pentacap. Any vertex of the icosahedron is the apex of a pentacap. Before getting to the actual proof, here is a lemma:
 * Lemma. In a convex polyhedron with $$V$$ vertices and $$E$$ edges, the inequality $$2E\geq3V$$ holds.
 * Proof. The degree $$\deg(v)$$ of each vertex $$v$$ is at least $$3.$$ So, by the degree sum formula, $$\textstyle{2E=\sum_{v}\deg(v)\geq\sum_{v}3=3V.}$$
 * Now the actual proof, which is by contradiction. Suppose we can glue four copies of a pentacap together to form an icosahedron. Now consider the polyhedron obtained by replacing each pentacap by the pentagon that is the base of the pyramid of which it is a cap. This polyhedron has 4 faces and 10 edges. (The latter can be seen by considering that each of the 5 edges of the 4 pentagonal faces is shared by 2 faces. Alternatively, we can argue that in replacing the 4 caps by the bases, we removed 4 × 5 edges from the 30 edges of the polyhedron.) Using Euler's formula $$V-E+F=2$$ we obtain $$V=E-F+2=8.$$ The combination $$E=10,V=8$$ contradicts the inequality of the lemma. --Lambiam 20:18, 6 January 2024 (UTC)

Division by zero on a mechanical calculator
There are a few YouTube videos demonstrating the practical effect of attempting division by zero on a mechanical calculator (it tries to do repeated subtractions of zero and gets into an infinite loop), but I'm wondering if anyone can help me dig up a "reliable source" explicitly describing this (could be e.g. a calculator manual, a book on arithmetic, a book of mathematical curiosities, a peer-reviewed paper about ??? mentioning this in passing, ...). I did some keyword searches around in corpora of book scans and academic literature, but didn't turn up anything. –jacobolus (t) 20:19, 6 January 2024 (UTC)


 * I don't have a reliable source, but that is what they did. My father's office had desktop calculators (four arithmetic operations) from the early 1960s to the 1980s.  They were mechanical but powered by an electric motor. If you did a division by zero, it would go on cycling indefinitely.  Bubba73 You talkin' to me? 20:32, 6 January 2024 (UTC)


 * I would assume that only some models of calculator would have this issue. In other words a video showing this occurred with one mechanical calculator doesn't mean that it occurred with all of them. One the video I saw it didn't seem like much of a problem since the human operating the thing could just turn it off if it started to cycle. Before electronic calculators came out many people used slide rules, and division by zero isn't an issue with them because 0 isn't on them. --RDBury (talk) 04:05, 7 January 2024 (UTC)


 * On the ones I saw, you could do something short of unplugging it to stop the cycle. Bubba73 You talkin' to me? 04:18, 7 January 2024 (UTC)
 * Yes, it's not important precisely which models of calculator this applies to. I'm just hoping to find a reliable source which I can cite in addition to an amateur YouTube video. –jacobolus (t) 05:40, 7 January 2024 (UTC)


 * Here are some manuals to look at. None of the parts I read mentioned this situation, though.  I think some intelligence on the user's part was assumed. --142.112.220.136 (talk) 03:50, 7 January 2024 (UTC)
 * Perhaps this will do, in a book from 1959, when calculators typically still were mechanical calculators:
 * Otherwise, division by zero would result (as it does in mechanical calculators) in a divide cycle which would never stop. The divide stop control on mechanical calculators serves the same purpose.
 * Other sources call this the "divide stop key":
 * When the divide key is depressed, division is performed. Touching the divide stop key stops the division at any desired position of the carriage and clears the keyboard.
 * --Lambiam 08:36, 7 January 2024 (UTC)
 * That's great, thanks! –jacobolus (t) 09:47, 7 January 2024 (UTC)
 * Interesting. I wonder if the devide stop key was also needed when a zero was devided by zero, in which case - any default (e.g. "1") chosen by the calculator as an initial result and being multiplied by the second zero - would immediately give the first zero (just as when A is devided by a non-zero B then multiplying the result by B would give A), so the default becomes the final result - hence no infinite loop is expected. My aunt (who passed away a year and a half ago) had an old mechanical calculator, so I'm going to ask her children what about it and what happens when a zero is devided by zero. I promise to let you know about the result as soon as possible. HOTmag (talk) 10:36, 7 January 2024 (UTC)
 * One I used first left-shifted the carriage as far as possible . Then it repeatedly subtracted until the borrow ran out, upon which it would do one addition. After a right shift of the carriage it started subtracting again, and so on, until the carriage could not right-shift any further. (For hand operation with non-automatic calculators you'd follow the same procedure, where borrow running out was signalled by a bell.) For example, dividing 68 by 03, representing the borrow running out with an *, you'd get: 68 = 68 + 0 × 30 = 38 + 1 × 30 = 08 + 2 × 30 = *78 + 3 × 30 = 08 + 2 × 30 = 08 + 20 × 3 = 05 + 21 × 3 = 02 + 22 × 3 = *99 + 23 × 3 = 02 + 22 × 3. So the result is 22, with a remainder of 2. With a zero divisor there are no borrows, so they cannot run out and subtraction will go on until interrupted.  --Lambiam 20:12, 7 January 2024 (UTC)
 * One I used first left-shifted the carriage as far as possible . Then it repeatedly subtracted until the borrow ran out, upon which it would do one addition. After a right shift of the carriage it started subtracting again, and so on, until the carriage could not right-shift any further. (For hand operation with non-automatic calculators you'd follow the same procedure, where borrow running out was signalled by a bell.) For example, dividing 68 by 03, representing the borrow running out with an <tt>*</tt>, you'd get: <tt>68 = 68 + 0 × 30 = 38 + 1 × 30 = 08 + 2 × 30 = *78 + 3 × 30 = 08 + 2 × 30 = 08 + 20 × 3 = 05 + 21 × 3 = 02 + 22 × 3 = *99 + 23 × 3 = 02 + 22 × 3</tt>. So the result is <tt>22</tt>, with a remainder of <tt>2</tt>. With a zero divisor there are no borrows, so they cannot run out and subtraction will go on until interrupted.  --Lambiam 20:12, 7 January 2024 (UTC)