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Math Help - proof by induction

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    proof by induction

    show that (a^m)*(a^n)=a^(n+m) knowing that a^0=1 and a^n=a^(n-1)a. I know it's not a hard proof but i just find myself using the property i'm trying to prove to solve the proof which is deffinatly wrong, just need someone elses input if you know what i mean
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    Quote Originally Posted by action259 View Post
    show that (a^m)*(a^n)=a^(n+m) knowing that a^0=1 and a^n=a^(n-1)a. I know it's not a hard proof but i just find myself using the property i'm trying to prove to solve the proof which is deffinatly wrong, just need someone elses input if you know what i mean
    We know that
    1. a^n = a^(n-1) * a ==> a^(n + 1) = a^n * a
    2. a^0 = 1
    as givens.

    So n = 0 case:

    a^m * a^0 = a^m * 1, by 2
    = a^m = a^(m + 0)

    Now assume that a^m * a^n = a^(m + n) for some n. We need to show that a^m * a^(n + 1) = a^(m + n + 1).

    So
    a^m * a^(n + 1) = a^m * a^n * a, by 1
    = a^(m + n) * a, by hypothesis
    Now, m + n is just some integer x.
    = a^x * a = a^(x + 1), by 1
    = a^(m + n + 1)

    Thus a^m * a^n = a^(m + n) for n = 0 and thus for all integer n >= 0.

    -Dan

    Frankly proof of this by induction is a bit of overkill as far as I'm concerned. We can prove it to be true just by applying the definition of a^n.
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    Quote Originally Posted by action259 View Post
    show that (a^m)*(a^n)=a^(n+m) knowing that a^0=1 and a^n=a^(n-1)a. I know it's not a hard proof but i just find myself using the property i'm trying to prove to solve the proof which is deffinatly wrong, just need someone elses input if you know what i mean
    For any m we note that for any m,
    (a^m)(a^0)=a^m=a^(m+0)

    Thus, there is a k such as,
    (a^m)(a^k)=a^(m+k)

    Multiply both sides by "a":
    (a^m)(a^k)(a)=a^(m+k)(a)
    Thus,
    a^m(a^{k+1})=a^{m+k+1}
    Q.E.D.
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