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    Induction on a Sum of Natural Numbers

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    Let f:N x N -> N be the function defined recursively as follows:

    f(0, 0) = 6

    f(i, j) = f(i - 1, j) + 2 if i > 0 and j = 0

    f(i, j) = f(i, j - 1) + 1 if j > 0

    Use induction on the sum i + j to prove that f(i, j) = 2i + j + 6 for all (i, j) in N x N.

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    https://brainmass.com/math/number-theory/induction-sum-natural-numbers-512746

    Solution Preview

    Please see the attached .pdf file for a complete solution.

    Since i >= 0 and j >= 0, we have that i + j >= 0.

    Base case: i + j = 0
    In this case, (i, j) = (0, 0). By definition, f(0, 0) = 6, so
    f(i, j) = f(0, 0) = 6 = 0 + 0 + 6 = 2(0) + 0 + 6 = 2i + j + 6

    Induction step: Let k >= 1, let i, ...

    Solution Summary

    A complete, detailed proof is provided.

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