{"id":6886,"date":"2015-06-18T20:25:50","date_gmt":"2015-06-19T00:25:50","guid":{"rendered":"https:\/\/spinor.info\/weblog\/?p=6886"},"modified":"2015-06-18T20:26:11","modified_gmt":"2015-06-19T00:26:11","slug":"recreational-mathematics","status":"publish","type":"post","link":"https:\/\/spinor.info\/weblog\/?p=6886","title":{"rendered":"Recreational mathematics"},"content":{"rendered":"

Having just finished work on a major project milestone, I took it easy for a few days, allowing myself to spend time thinking about other things. That’s when I encountered an absolutely neat problem on Quora<\/a>.<\/p>\n

\"pi\"<\/p>\n

Someone asked a\u00a0seemingly innocuous number theory question: are there two positive integers such that one is exactly the\u00a0\u03c0-th power of the other?<\/p>\n

Now wait a minute, you ask… We know that\u00a0\u03c0 is a transcendental number. How can an integer raised to a transcendental power be another integer?<\/p>\n

But then you think about \\(\\alpha=\\log_2 3\\) and realize that although \\(\\alpha\\) is a transcendental number, \\(2^\\alpha=3\\). So why can’t we have \\(n^\\pi=m\\), then?<\/p>\n

As it turns out, we (probably) cannot, but the reason is subtle and it relies on a very important, but unproven conjecture from transcendental number theory.<\/p>\n

But first, let us rewrite the equation by taking its logarithm:<\/p>\n

$$\\pi\\log n = \\log m.$$<\/p>\n

We can also divide both sides by \\(\\log n\\), which leads to<\/p>\n

$$\\pi = \\frac{\\log m}{\\log n}=\\log_n m,$$<\/p>\n

but it turns out to be not very helpful. However, squaring the equation will help, as we shall shortly see:<\/p>\n

$$\\pi^2\\log^2 n=\\log^2 m.$$<\/p>\n

Can this equation ever by true for positive integers \\(n\\) and \\(m\\), other than the trivial solution \\(n=m=1\\), that is?<\/p>\n

To see why it cannot be the case, let us consider the following triplet of numbers:<\/p>\n

$$(i\\pi,\\log n,\\log m),$$<\/p>\n

and their exponents,<\/p>\n

$$(e^{i\\pi}=-1, e^{\\log n}=n, e^{\\log m}=m).$$<\/p>\n

The three numbers \\((i\\pi,\\log n,\\log m)\\) are linearly independent over \\({\\mathbb Q}\\) (that is, the rational numbers). What this means is that there are no rational numbers \\(A, B, C, D\\) such that \\(Ai\\pi+B\\log n+C\\log m + D=0\\). This is easy to see as the ratio of \\(\\log n\\) and \\(\\log m\\) is supposed to be transcendental but both numbers are real, whereas \\(i\\pi\\) is imaginary.<\/p>\n

On the other hand, their exponents are all rational numbers (\\(-1, n, m\\)). And this is where the unproven conjecture, Schanuel’s conjecture<\/a>, comes into the picture. Schanuel’s conjecture says that given \\(n\\) complex numbers \\((\\alpha_1,\\alpha_2,…,\\alpha_n)\\) that are linearly independent over the rationals, out of the \\(2n\\) numbers \\((\\alpha_1,…,\\alpha_n,e^{\\alpha_1},…,e^{\\alpha_n})\\), at least \\(n\\) will be\u00a0transcendental numbers that are algebraically independent over\u00a0\\({\\mathbb Q}\\). That is, there is no algebraic expression involving roots and powers of the \\(\\alpha_i\\), \\(e^{\\alpha_i}\\), and rational numbers that will yield 0.<\/p>\n

The equation \\(\\pi^2\\log^2 n=\\log^2 m\\), which we can rewrite as<\/p>\n

$$(i\\pi)^2\\log^2 n + \\log^2 m=0,$$<\/p>\n

is just such an equation, and it can never be true.<\/p>\n

I wish I could say that I came up with this solution but I didn’t. I was this<\/strong> close: I was trying to apply Schanuel’s conjecture, and I was of course using the fact that \\(\\pi=-i\\log -1\\). But I did not fully appreciate the implications and meaning of Schanuel’s conjecture, so I was applying it improperly. Fortunately, another Quora user saved the day<\/a>.<\/p>\n

Still I haven’t had this much fun with pure math (and I haven’t learned this much pure math all at once) in years.<\/p>\n<\/fb:like>","protected":false},"excerpt":{"rendered":"

Having just finished work on a major project milestone, I took it easy for a few days, allowing myself to spend time thinking about other things. That’s when I encountered an absolutely neat problem on Quora. Someone asked a\u00a0seemingly innocuous number theory question: are there two positive integers such that one is exactly the\u00a0\u03c0-th power […]<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[30],"tags":[],"class_list":["post-6886","post","type-post","status-publish","format-standard","hentry","category-mathematics","category-30-id","post-seq-1","post-parity-odd","meta-position-corners","fix"],"_links":{"self":[{"href":"https:\/\/spinor.info\/weblog\/index.php?rest_route=\/wp\/v2\/posts\/6886","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/spinor.info\/weblog\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/spinor.info\/weblog\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/spinor.info\/weblog\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/spinor.info\/weblog\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=6886"}],"version-history":[{"count":15,"href":"https:\/\/spinor.info\/weblog\/index.php?rest_route=\/wp\/v2\/posts\/6886\/revisions"}],"predecessor-version":[{"id":6902,"href":"https:\/\/spinor.info\/weblog\/index.php?rest_route=\/wp\/v2\/posts\/6886\/revisions\/6902"}],"wp:attachment":[{"href":"https:\/\/spinor.info\/weblog\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=6886"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/spinor.info\/weblog\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=6886"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/spinor.info\/weblog\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=6886"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}