6a. Prove that if 0 is less than or equal to x which is less than y, then x^n < y^n, n = 1, 2, 3...
b. Prove that if x < y and n is odd, then x^n < y ^n.
c. Prove that if x^n = y^n and n is odd, then x = y.
d. Prove that if x^n = y^n and n is even, then x = y or x = -y.
The Attempt at a Solution
My problem is I'm so confused no how to start this proving something for all n. My attempt at 6a is 6a. I'm not sure how I would do this but my start: x^n < y^n. N = 1, so x < y. If n is two, you can start with x < y. Then you square both sides to get x^n < y^n. Etc. But...how do I prove this for all n?
Logically, I understand why b-d is true, but I have no idea how to prove this. I can probably prove this for specific numbers, but how do I do this for all n? Can someone please provide me with some assistance? I'm supposed to hand in a problem explaining this tomorrow.