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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?

Community
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tzador
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    Well, integers are rational numbers. You presumably mean non-integer rational numbers. – Thomas Andrews Sep 05 '12 at 20:37
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    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 05 '12 at 20:38
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    This is a duplicate of many prior questions, e.g. http://math.stackexchange.com/q/4467/242 and http://math.stackexchange.com/q/5/242 and http://math.stackexchange.com/q/26499/242 and http://math.stackexchange.com/q/22423/242 and http://math.stackexchange.com/q/11872/242 – Bill Dubuque Sep 05 '12 at 20:39
  • @GeoffRobinson (you should) write it as an answer. –  Sep 05 '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. –  Sep 05 '12 at 20:47
  • One way of looking at this is to try taking the square of a rational number $q$ which is not an integer. If this is in its lowest terms, then the denominator of $q$ will have a non-trivial prime factor $p$ and $q^2$ will have denominator divisible by $p^2$. I mention this because the ideas here, though seemingly trivial, have fruitful and interesting generalisations. – Mark Bennet Sep 05 '12 at 21:01
  • @Jennifer: It is a duplicate, so no real need. – Geoff Robinson Sep 05 '12 at 21:20

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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.

Vilid
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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 05 '12 at 20:44
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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. – Alex Becker Sep 05 '12 at 20:51