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Thread: [SOLVED] Proof: uniqueness of a natural number

  1. #1
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    [SOLVED] Proof: uniqueness of a natural number

    Let m, n e N. If n >= m, then there is a unique b e N such that n = m + b.

    How do I prove that b is unique.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by jzellt View Post
    Let m, n e N. If n >= m, then there is a unique b e N such that n = m + b.

    How do I prove that b is unique.

    $\displaystyle m,n\in\mathbb{N},n\geqslant m\implies\exists!\,b\in\mathbb{N}:n=m+b$

    Suppose not. Let $\displaystyle m,n\in\mathbb{N}$ and let $\displaystyle b,c\in\mathbb{N}$ such that $\displaystyle n=m+b$ and $\displaystyle n=m+c$. By transitivity, we see that $\displaystyle m+b=m+c$. Thus, by cancellation, we have $\displaystyle b=c$. This shows that $\displaystyle b$ is unique.

    Does this make sense?
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by jzellt View Post
    Let m, n e N. If n >= m, then there is a unique b e N such that n = m + b.

    How do I prove that b is unique.
    assume there is some other natural number, b' such that n = m + b'. then we have m + b = m + b'. subtracting m from both sides, we get b = b'. so our other element was in fact, b. thus we have uniqueness.


    EDIT: too late! i shall defeat you yet, Chris!
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