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Thread: Complicated induction proof, where you are given a function of two variables

  1. #1
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    Unhappy Complicated induction proof, where you are given a function of two variables

    Let N denote the natural numbers. Let f: NxN -> N be defined by
    f(a,0) = a for all natural numbers a
    f(0,b) = 2b for all natural number b
    f(a,b) = [f(a-1, b) + f(a, b-1)+3]/2 for all positive integers a,b

    Compute f(a,b) for all 0 <= a,b <= 3
    Prove that f(a,b)<= 2(a+b) for all natural numbers a,b by induction

    Any guidance and help is much appreciated I am not sure where to begin with this problem. Thank you.
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  2. #2
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    Re: Complicated induction proof, where you are given a function of two variables

    Quote Originally Posted by ak12 View Post
    Let N denote the natural numbers. Let f: NxN -> N be defined by
    f(a,0) = a for all natural numbers a
    f(0,b) = 2b for all natural number b
    f(a,b) = [f(a-1, b) + f(a, b-1)+3]/2 for all positive integers a,b

    Compute f(a,b) for all 0 <= a,b <= 3
    Prove that f(a,b)<= 2(a+b) for all natural numbers a,b by induction

    Any guidance and help is much appreciated I am not sure where to begin with this problem. Thank you.
    If $f$ is defined on $N \times N$ then $f(a,0)$ and $f(0,b)$ aren't defined so the problem makes no sense.
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  3. #3
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    Re: Complicated induction proof, where you are given a function of two variables

    I highly doubt my professor would give us a problem that isnt possible (i hope) I am attaching a screen shot of the question please let me know if it still doesnt make sense.Complicated induction proof, where you are given a function of two variables-screen-shot-2018-10-03-7.31.31-pm.png
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  4. #4
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    Re: Complicated induction proof, where you are given a function of two variables

    Normally "N" is the set of positive integers so doesn't include 0. Just use the non-negative integers so that 0 is included. You need to do a "double induction". First prove that, for n a non-negative integer, $\displaystyle f(m, 0)\ge m$, then that $\displaystyle f(0, n)= n$. Then continue the indutions on a and b.
    Thanks from ak12
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