# Uniquness of integer values between a rational

• Sep 29th 2010, 11:59 PM
lllll
Uniquness of integer values between a rational
If $x \in \mathbb{Q}$ show that there is a uniquely determined $n \in \mathbb{Z}$, such that $n \leq x \leq n+1$

I was thinking of simply showing this by contradiction, that if $m \neq n$ then the integer values would be different, but I got nowhere. I have a feeling that I should assume that $m \neq n$, and by manipulating it getting $n-m=0$, but I'm not too sure on how to go about it.
• Sep 30th 2010, 01:58 AM
HappyJoe
Have you proved yet that there does exist an integer n, such that $n\leq x\leq n+1$? For this kind of problem, it is generally hard to know exactly what you may and may not use to solve it.

By the Archimedean property, there exists an integer n, such that x<n. Hence the set $\{n | n integer, n\leq x\}$ is finite. Let n be the largest integer in this set. Then $n\leq x$, but x<n+1, because otherwise n+1 would be in the set, contradicting maximality of n. This show existence.

Suppose both m and n have the desired property, and suppose contrariwise that m<n. Then $m+1\leq n$, and so

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