# [SOLVED] Proof: uniqueness of a natural number

• Feb 16th 2009, 08:50 PM
jzellt
[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.
• Feb 16th 2009, 08:56 PM
Chris L T521
Quote:

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.

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

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?
• Feb 16th 2009, 08:57 PM
Jhevon
Quote:

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!