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Possible Duplicate:
$\sqrt a$ 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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    @Jennifer: It is a duplicate, so no real need.2012-09-05

1 Answers 1

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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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    "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.2012-09-05