/sci/ - Science & Math

>>7497051

So we did the "forward" half of the above biconditional, and to get at the other half, we consider first the INVERSE situation

<span class="math"> \displaystyle (a,b,c) \notin EB \rightarrow (ka,kb,kc) \notin EB [/spoiler]

and, in this rendering, what does it mean to say that some triple of naturals (a,b,c) is NOT "an Euler brick"? Quite simply, that ANY ONE (or two, or all three) of the involved equations >>7496579 don't give up some integer hypotenuse d,e or f.

So WLOG we can consider the first case, in the situation where d is NOT a natural number. This is enough to say that d^2 is not a perfect square, but we know something even better about perfect squares, and naturals that aren't: perfect squares by definition have natural square roots, and any natural number that ISN'T a perfect square, has for its positive square root an IRRATIONAL number (due to FTA, roots of primes etc), so the "not-natural-but-maybe-just-rational" business that a close reader could have picked up on as a possibility a clause or two ago, vanishes. Since non-zero d is not-natural by assumption, it is forced to be irrational due to the above. And, bringing natural k into the situation and considering the term kd (the crux here, again), we know from pic related that the product of a non-zero irrational and a non-zero rational number is ALWAYS irrational (to suppose otherwise is immediately absurd).

Thus etc, and we have an even stronger BICONDITIONAL about scalings of Euler bricks. As a reminder, the reason for being so careful about this result is that (questions about) scalings of Euler bricks are directly related to the conjecture early in the thread, conjecturing that edge lengths can't be evenly spaced!

>>7495902

The general question (except for OP's case!) is of course simple, and I had occasion to describe the general case recently, for the reals.

In the picture, we describe multiplication of 0, some random real nonzero rational r, and some random real, irrational i (not to be confused with the imaginary unit of complex numbers). Because multiplication of real numbers is commutative (order of elements doesn't matter), our chart is symmetric about the main diagonal. We include the trivial 0 cases in the interest of simple completeness, since 0 is an obvious "spoiler" for various versions of OP's question, which bears mentioning.

0 times any real number is 0, and five of our nine cases are done. 5/9

Any rational number (zero or otherwise, but we already covered that case) times any other rational number (zero etc.), always gives a rational number. A little check is all that's needed to know this. 6/9.

Meanwhile, any NON-ZERO rational number, times any NON-ZERO irrational number, always returns an irrational number. A proof involves supposing the opposite (that their product is rational) and showing that this is absurd. 8/9.

It's only the latter case, which OP is wondering about, which is interesting, since the situation can go either way. But OP shouldn't feel too bad (nor should he be scolded too harshly), for there are elementary products of irrational numbers which are not known to be rational or irrational e.g. π x e ! Similar open questions entail other elementary arithmetic and functional operations involving pi and e, viz.

https://en.wikipedia.org/wiki/Irrational_number#Open_questions