[SOLVED] Proof: uniqueness of a natural number

**jzellt** · February 16th 2009, 08:50 PM

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.

**Chris L T521** · February 16th 2009, 08:56 PM

Originally Posted by jzellt

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.

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

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

Does this make sense?

**Jhevon** · February 16th 2009, 08:57 PM

Originally Posted by jzellt

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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