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$\sqrt a$ is either an integer or an irrational number.

$\sqrt{2}$ is irrational number, but $\sqrt{9} = 3$ is an integer. Are there such integers whose square root is a (non-integer)rational number?

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marked as duplicate by Ross Millikan, Thomas Andrews, Henning Makholm, Pedro Tamaroff, Bill Dubuque Sep 5 '12 at 20:46

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Well, integers are rational numbers. You presumably mean non-integer rational numbers. – Thomas Andrews Sep 5 '12 at 20:37
Every integer is a rational number. I suppose you mean "Are there integers whose square root is rational, but not an integer?" In fact, the answer is no. – Geoff Robinson Sep 5 '12 at 20:38
@GeoffRobinson (you should) write it as an answer. – user2468 Sep 5 '12 at 20:45
@tzador, please read the duplicate question. It says that the square root of every (positive) integer is either an integer or irrational. There is no integer whose square root is a rational fraction. – user2468 Sep 5 '12 at 20:47

Assume $x^2 =\frac{m}{n}$ where $m \in \mathbb Z$ and $ n \in \mathbb N$

this implies $x = \frac{\sqrt{m}}{\sqrt{n}}$

which means that for an integer to have a rational root it's root must be the ratio of two integer roots. For this ratio to not be an integer m and n must be distinct which implies that x is not an integer quantity.

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To be clear this means that there isn't an integer rational with a rational root that isn't an integer. – Vilid Sep 5 '12 at 20:44
"For this ratio to not be an integer m and n must be distinct which implies that x is not an integer quantity." You need to prove this, as nothing you have said makes it obvious this is true. In fact I can't think of any proof of this sentence which doesn't prove the whole theorem along the way. Thus your answer is incomplete/wrong. – Alex Becker Sep 5 '12 at 20:51

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