I am not a mathematician either, I always maintained the university made a mistake granting make a math teacher degree :) It was also long ago enough a lot of y

I am not a mathematician either, I always maintained the university made a mistake granting make a math teacher degree :) It was also long ago enough a lot of you weren't even alive and I haven't done any such work in a quarter century. Nonetheless... > but x^y will always be larger than y^x, for y>x and x^y > y^x (prove this by taking the first derivative to get rate of change. Do you use Laplace z domain for discrete, instead of s for continuous?) I am struggling to follow what are you saying here > Why resort to the continuous domain to solve a problem n the discrete, is this even a valid approach? I can't make heads or tails of this question. The proof says, correctly, for all (real) x != y, x < y solutions it is true that 0<x<e and e<y. He found this by investigating the derivative and establishing the monotonically increasing / decreasing nature of the function. Only after finding this out using does he go back to the original question: what integer x could be? Since he restricted x to be 0 < x < e , we only need to investigate the case of 1 and 2.