 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Commutative Exponentiation (Posted on 2017-07-07) In general exponentiation is not commutative. However most people are aware of the pair 2 and 4 as a pair of unequal numbers for which exponentiation is commutative.

Show that there are an infinite number of pairs of unequal numbers for which exponentiation is commutative.

Go further and show that there are an infinite number of pairs of unequal rational numbers for which exponentiation is commutative.

 Submitted by Brian Smith Rating: 4.0000 (1 votes) Solution: (Hide) Assume x is a multiple of y in the equation x^y=y^x. Then let y=v*x and substitute. After a bit of algebra a parametric solution emerges: x=v^(1/(v-1)) and y=v^(v/(v-1)) v=2 or v=1/2 generates the classic pair of 2 and 4. Assume v is rational. Then 1/(v-1) = p/q for some coprime integers p and q. Then x=((p+q)/p)^(p/q) and y=((p+q)/p)^((p+q)/q). If q is not equal to 1 then irrational roots pop back in, so make q=1. Then x=((p+1)/p)^p and y=((p+1)/p)^(p+1) form a nontrivial rational solution to x^y=y^x for any positive integer. This page at cut-the-knot.org discusses this problem. Subject Author Date re(2): One unusual solution hellenjos 2018-10-31 03:07:22 re: One unusual solution warren zephaniah 2017-08-18 05:07:48 One unusual solution Steve Herman 2017-07-07 19:15:29 rest of the solution Paul 2017-07-07 15:09:38 rational solutions Paul 2017-07-07 14:53:40 exploration Charlie 2017-07-07 14:22:36 Real solutions Jer 2017-07-07 12:20:19 Please log in:

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