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Getting started on proving this

  1. Sep 13, 2009 #1
    1. The problem statement, all variables and given/known data

    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.

    2. Relevant equations



    3. 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.
     
  2. jcsd
  3. Sep 13, 2009 #2

    LCKurtz

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    To prove it by induction the first step is to check whether the statement:

    If 0 <= x < y then x^n < y^n

    is true for n = 1. That statement is:

    If 0 <= x < y then x^1 < y^1, which is obviously true.

    I think setting up the induction step is what is bothering you. For the induction step you must show that if the statement is true for n = k then it is true for n = k + 1. So to do the induction step you assume the statement is true for n = k, so you are given:

    (*) If 0 <= x < y then x^k < y^k

    and what you want to prove is that it is true for n = k + 1:

    (**) If 0 <= x < y then x^(k+1) < y^(k+1)

    So see if you can use that fact that x < y and that (*) is assumed to be true to get the step that (**) must then be true.
     
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