Any odd integer squared is always odd, and likewise, any even integer squared is always even. Therefore, the square root of an odd number must be odd and the square root of an even number must be even. For it can never be the case the square root of an odd number is even, or the square root of an even number is odd. Furthermore, only integers can be said of as being either even or odd.


$${a \over b} = \sqrt 2 $$


$${{{a^2}} \over {{b^2}}} = 2$$

$${a^2} = 2{b^2}$$ ${a^2}$ must be even owing to the two present in the equation. Therefore, $a$ must be even. Suppose $a = 2$ then, $${2^2} = 2({b^2})$$ $${{{2^2}} \over 2} = {b^2}$$ $$2 = {b^2}$$ Therefore, $b$ must be even. But, $$b = \sqrt 2 $$ The square root of two is not an even integer, because it's not an integer at all.

In more general terms where $n$ is any integer, $$\sqrt {{{{n^2}} \over 2}} = n\sqrt {{1 \over 2}} $$ So am I right in saying the square root of a half is also irrational?

  • $\begingroup$ An odd integer squared is odd. $\endgroup$
    – David P
    Jan 18, 2015 at 0:27
  • $\begingroup$ Consider the prime factorization of a number. (I.E. The prime factorization of $60$ is $2\cdot2\cdot3\cdot5$.) Now, square it—how many $2$'s are in the prime factorization now? (Answer: It turns out there are always an even number of 2's in the prime factorization of a square. Remember that $0$ is an even number. Do you know why? How does this information help solve your problem?) $\endgroup$ Jan 18, 2015 at 0:38
  • $\begingroup$ You're on the path to a proof by contradiction. Your assumption is that is it possible to write $\sqrt{2}=\frac{a}{b}$ with $a$ and $b$ integers. However such an expression is not possible. $\endgroup$
    – Joffan
    Jan 18, 2015 at 1:23
  • $\begingroup$ It's not clear to me, are you trying to prove $\sqrt 2$ is irrational? Because you are on the right track if so. Let me just give you some hints: 1) You can assume that $a$ and $b$ are relatively prime. 2) If you have $a^2 = 2b^2$ can you conclude that $a$ is even? Can you conclude the same for $b$? $\endgroup$
    – Ennar
    Jan 18, 2015 at 1:29

4 Answers 4


A square must be divisible by an even number of powers of $2$.


You can't say “suppose $a=2$”: you don't know what $a$ is, you just know it must be even (if it exists). What you're proving is that $a=2$ leads a contradiction, but it's not sufficient.

You have to use some more properties of the integers, in order to finish your proof.

A possibility is to take $a/b$ as the fraction with minimum possible positive numerator.

You proved that $a$ is even, so $a=2A$ for a positive integer $A$. But then $$ 4A^2=2b^2 $$ and so $b^2=2A^2$ and $b$ is even as well. Thus $b=2B$ and so $$ \frac{a}{b}=\frac{2A}{2B}=\frac{A}{B} $$ contradicting the minimality of $a$, as $A<a$.


You can do so by proof by contradiction.

Suppose that $\sqrt{2}$ were a rational number, so by definition $\sqrt{2} = {a\over b}$ where $a$ and $b$ are non-zero integers with no common factor. Thus, $b\sqrt{2} = a$. Squaring both sides yields $2b^2 = a^2$. Since $2$ divides the left hand side, $2$ must also divide the right hand side (as they are equal and both integers). So $a^2$ is even, which implies that a must also be even. So we can write $a = 2c$, where $c$ is also an integer. Substitution into the original equation yields $2b^2 = (2c)^2 = 4c^2$. Dividing both sides by $2$ yields $b2 = 2c^2$. But then, by the same argument as before, $2$ divides $b2$, so $b$ must be even. However, if $a$ and $b$ are both even, they share a factor, namely $2$. This contradicts our assumption, so we are forced to conclude that $\sqrt{2}$ is an irrational number.



Every positive real number is the square of some real number. So, they are kind of all perfect squares.

In your post the only case when $b$ is an integer is when $a=b\sqrt{2}$. Otherwise both $a$ and $b$ are not integers.


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