So here's the question:

Prove or disprove: For every $x \in \mathbb{Q}$, there is a unique $n \in \mathbb{N}$ which is the closest natural number to $x$.

I know we can define a rational number as $\frac{a}{b}$, with $a,b$ as natural numbers, but I'm not sure where to go from there. Any hints/help would be greatly appreciated!

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    $\begingroup$ The uniqueness part has a problem with e.g. $\frac{5}{2}$. $\endgroup$ – Daniel Fischer Dec 11 '14 at 16:30
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    $\begingroup$ The claim is wrong. For $x=\frac32$, there is no unique closest natural number $\endgroup$ – Hagen von Eitzen Dec 11 '14 at 16:30
  • $\begingroup$ so how would you prove its wrong? just with counter example? $\endgroup$ – Zoë Soriano Dec 11 '14 at 16:39
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    $\begingroup$ @quid, my professor wrote the book. it says exactly: Prove or disprove that: For every x∈Q, there is a unique n∈N which is the closest natural number to x. It doesn't say anything about what it means to be a closest integer. $\endgroup$ – Zoë Soriano Dec 11 '14 at 21:49
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    $\begingroup$ @quid, thank you very much! I appreciate your help. $\endgroup$ – Zoë Soriano Dec 11 '14 at 21:56

No, there is not always a unique closest natural number, as mentioned in comments. In more detail:

If $q$ is a rational number of the form $q= a/2$ with $a$ an odd positive integer, then the two natural number $(a-1)/2$ and $(a+1)/2$ are both at distance $1/2$ from $q$. Every natural number other than these two is either less than $(a-1)/2$ or greater than $(a+1)/2$. Thus its distance to $q$ is strictly greater than $1/2$.

Therefore, the smallest distance from a natural number to such a $q$ is $1/2$. This distance is however attained by two natural numbers, $(a-1)/2$ and $(a+1)/2$, and therefore there is not always a unique closest natural number.


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