((x/y)^(n+1)) > ((x/y)^n) for all n ≥ 1 and x > y > 0.
To prove this statement by induction, we will use the principle of mathematical induction.
Base case: When n = 1, we have:
((x/y)^(1+1)) = ((x/y)^2) = (x^2)/(y^2)
((x/y)^1) = (x/y)
Since x > y > 0, we have x/y > 1. Therefore, (x^2)/(y^2) > (x/y), which means that the base case is true.
Inductive step: Assume that ((x/y)^(k+1)) > ((x/y)^k) for some arbitrary positive integer k. We want to prove that this implies that ((x/y)^(k+2)) > ((x/y)^(k+1)).
Starting with ((x/y)^(k+2)), we can write:
((x/y)^(k+2)) = ((x/y)^(k+1)) * ((x/y)^1)
Using the induction hypothesis, we know that ((x/y)^(k+1)) > ((x/y)^k), and we also know that x/y > 1. Therefore, we have:
((x/y)^(k+2)) > ((x/y)^k) * (x/y)
Simplifying this expression, we get:
((x/y)^(k+2)) > ((x/y)^k) * (x/y)
((x/y)^(k+2)) > ((x^k)/(y^k)) * (x/y)
((x/y)^(k+2)) > ((x^(k+1))/(y^(k+1)))
Therefore, we have shown that ((x/y)^(k+2)) > ((x/y)^(k+1)) for all positive integers k, which completes the inductive step.
By the principle of mathematical induction, we have proven that ((x/y)^(n+1)) > ((x/y)^n) for all n ≥ 1 and x > y > 0.