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.
Actually, I was just being lazy. If you consider the pair (a,b) = ( (n/n1)^n1 and (n/n1)^n ), a little bit of algebra reveals that ln(a)/a = ln(b)/b = [ (n1)^n / n^(n1) ] * (ln(n)  ln(n1)), and if ln(a)/a = ln(b)/b then ln(a^b) = ln(b^a) and a^b = b^a.
So there *are* an infinite number of distinct rational pairs, but what I don't know is whether there are any such (rational) pairs that aren't of this form.

Posted by Paul
on 20170707 15:09:38 