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Math Help - Math Induction

  1. #1
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    Math Induction

    Let a be a non zero number and m and n be integers. Prove by induction that:

    (1) a^{m+n} = a^{m} a^{n}

    (2). {(ab)}^{n} = {a^n}{b^n}

    can anyone explain me the basis and inductive steps here?

    Do I have to start by supposing n=0 in (1) and

    n=1 in (2)?

    what about the inductive step?
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  2. #2
    Senior Member Tinyboss's Avatar
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    Exactly what properties of exponentiation are you given to work with?
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  3. #3
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    Quote Originally Posted by Tinyboss View Post
    Exactly what properties of exponentiation are you given to work with?
    the question doesnt state anything about properties of exponentiation. However, induction has to be used here
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  4. #4
    Senior Member Tinyboss's Avatar
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    If you don't know what exponentiation is, you have no hope of proving anything about it. (Not saying you don't know how to exponentiate--just that your book/prof is not expecting you to prove this without giving you something to start from. Look in your chapter or ask your prof what you're allowed to assume about exponentiation to begin with.)
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  5. #5
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    Quote Originally Posted by serious331 View Post
    Let a be a non zero number and m and n be integers. Prove by induction that:

    (1) a^{m+n} = a^{m} a^{n}

    (2). {(ab)}^{n} = {a^n}{b^n}

    can anyone explain me the basis and inductive steps here?

    Do I have to start by supposing n=0 in (1) and

    n=1 in (2)?

    what about the inductive step?
    When n=0 \text{ and } m=0, a^{m+n} = a^{m} a^{n} holds since a^{0+0}=1=a^{0 \cdot 0}

    For induction hypothesis, assume a^{p+q} = a^{p} a^{q}, where p and q are integers.

    We begin by a^{p+q} = a^{p} a^{q}.

    Multiplying through by a, we obtain a \cdot (a^{p+q}) = a \cdot a^{p} a^{q}.

    Plug in the assumption for the left handside in the parethesis, we obtain

    a \cdot (a^{p} a^{q})= a \cdot a^{p} a^{q}. The result is as desired.

    Consequently, by induction hypothesis a^{m+n} = a^{m} a^{n} for all integers m and n.
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