# Is square root of an integer, either an integer or an irrational number, but never (non-integer) rational? [duplicate]

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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?

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

• 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
• This is a duplicate of many prior questions, e.g. math.stackexchange.com/q/4467/242 and math.stackexchange.com/q/5/242 and math.stackexchange.com/q/26499/242 and math.stackexchange.com/q/22423/242 and math.stackexchange.com/q/11872/242 – Bill Dubuque Sep 5 '12 at 20:39
• @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 $x = \frac{\sqrt{m}}{\sqrt{n}}$