My only exposure to proofs was in a math logic class I took in University. I was wondering if my attempt at proving that $\sqrt{12}$ is irrational is OK.

$$\Big(\frac{m}{n}\Big)^2 = 12$$ $$\Big(\frac{m}{2n}\Big)^2 = 3$$ $$m^2=3*(2n)^2$$

This implies $m$ is even and so $n$ must be odd.

The problem can be reduced to:

$$\Big(\frac{p}{n}\Big)^2 = 3$$

Because $n$ is odd, $p^2$ is odd, so $p$ is odd.

This implies: $$4a+1 = 3(4b+1)$$ $$4a - 12b = 2$$ $$2a - 6b = 1$$

I'm kind of stuck at this point. I know that this can't be true but I don't know how to state it. Any critiques or suggestions? Thanks!

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    $\begingroup$ "... and so $n$ must be odd." doesn't follow. However, you presumably meant to start the proof with "assume $\sqrt{12}$ is rational, and write it as $m/n$ is lowest terms". If so, then it does follow that $n$ must be odd (because otherwise $m$ and $n$ would have a common factor). What you say after this point seems to come out of thin air. Where did $p$ come from? How can the problem be reduced? Then where do $a$ and $b$ and those equations come from? $\endgroup$ – Hurkyl Feb 16 '13 at 15:28
  • $\begingroup$ I see your point. What I had in mind was m = 2p. For the last equation, if p and n are both odd then there is an integer k such that 4*k+1 = p^2 same can be said for n. $\endgroup$ – user21154 Feb 16 '13 at 16:24
  • $\begingroup$ What I would have done is to write $\sqrt{12}$ as $2\sqrt{3}$ and then show that this implies that $\sqrt{12}$ is irrational since rational times irrational is irrational! $\endgroup$ – LoveFood Mar 11 '14 at 18:50

10 Answers 10


You made it too complicated in my opinion.

First we show that a rational number (different from $0$) times an irrational is irrational.

Proof by contradiction: Let $x\in \mathbb{R}\setminus\mathbb{Q}$, $a,c\in\mathbb{Z}\setminus\{0\}$, $b,d\in \mathbb{N}, d \neq 0$. $$x\cdot \frac{a}{b}=\frac{c}{d} \iff x=\frac{bc}{ad},$$ so $x$ would be rational.

Use $$\sqrt{12}=\sqrt{4\cdot 3} = \sqrt{4} \cdot \sqrt{3} = 2 \cdot \sqrt{3}$$ So $\sqrt{12}$ irrational $\iff \sqrt{3}$ is irrational.

Now, we show $3|p^2 \implies 3|p$: We know 3 is prime so with Euclid's lemma we have $$3|p^2 \implies 3|p \wedge 3|p \implies 3|p$$ To prove that $\sqrt{3}$ is irrational, you derive a contradiction: $$\sqrt{3}=\frac{p}{q}\iff 3q^2=p^2 \implies \exists k \in \mathbb{N}: 3k =p$$ $$q^2=3k^2$$ So $q$ and $p$ both have the divisor $3$. Now there are two different ways to use this information. The first proceeds by contradiction, assuming that $p$ and $q$ don't have a common divisor, but as you can show they always have the divisor $3$, you can't write $\sqrt{3}$ as a fraction of numbers without common divisors. This is the more elegant way in my opinion. The other way is to show that both $p$ and $q$ can't be finite, because by repeating this argument we see $3^n|p$ for all $n \in \mathbb{N}$ and similarly for $q$.
But because of $p,q\in \mathbb{N}$, $3^n> 1+2n$ and the Archimedean principle we get a contradiction.

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    $\begingroup$ OK great. Thanks. $\endgroup$ – user21154 Feb 16 '13 at 16:04
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    $\begingroup$ You might also want to prove that a rational number times an irrational one is irrational. $\endgroup$ – PyRulez Feb 16 '13 at 16:36
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    $\begingroup$ Two points: one needs to justify the claim $\, 3\mid p^2,q^2 \Rightarrow 3 \mid p,q\,$ and one needs to make some initial assumption in order for $\rm\,3\mid p,q\,$ to terminate the proof (e.g. assume that $\,p/q\,$ is in lowest terms, hence $\,p,q\,$ are coprime) $\endgroup$ – Math Gems Feb 16 '13 at 19:23
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    $\begingroup$ @Dominic There are many ways to deduce a contradiction from $\,3\mid p,q\,$ but it is crucial to say which way your proof does this if you wish to convince the reader that what you have in mind is correct. Ditto for the justification of the claim that $\,3\mid p^2\Rightarrow 3\mid p.\,$ A rigorous proof leaves the reader no doubts about such. Rigor is especially important here, since many students make errors on these matters, and for millenia there was much confusion about such. $\endgroup$ – Math Gems Feb 16 '13 at 19:55
  • $\begingroup$ Oh I see i am sry i misunterstood you. Thanks, i added the stuff $\endgroup$ – Dominic Michaelis Feb 16 '13 at 20:01

If you’re willing to use the Fundamental Theorem of Arithmetic, which says that the decomposition of any nonzero integer as a product of primes is unique, then this proof, and all others for irrationality of $r$-th roots, drops right out.

Write $m^2=12n^2$. This contradicts FTA because there are evenly many $3$’s on the left but oddly many on the right.

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    $\begingroup$ This one is cute. $\endgroup$ – Ali Caglayan Feb 16 '14 at 14:30

The following is different in style. It avoids the use of the unique factorization property that arguments about the divisibility by the prime $3$ use.

Assume $\sqrt{12}$ is rational. Choose among all equivalent fractions the one with the least positive denominator; let it be $m/n$. Thus $m^2 = 12 n^2$, and we have $$9 n^2 < m^2 < 16 n^2$$ $$3n < m < 4 n$$ $$0 < m -3n < n$$ Now $$\begin{align} \left({12n - 3m \over m - 3n}\right)^2 &= { 9(16n^2 -8mn+m^2)\over m^2-6mn+9n^2}\\ &= { 9(16n^2 - 8mn + 12n^2)\over 12n^2 - 6mn + 9n^2} \\ &= { 36(7n-2m)n\over 3(7n-2m)n} = 12\,, \end{align}$$ which is to say that ${12n - 3m \over m -3n}$ equals $\sqrt{12}$ and has a lesser denominator. This is impossible since we chose $m/n$ to be in least terms.

In case you're wondering how to find the fraction, it's from the continued fraction expansion of $\sqrt{12}$. One has $$\begin{align} x=\sqrt{a^2+b}&=a+{b \over 2a + \displaystyle{b \over 2a + \displaystyle {b \over 2a + \cdots}}} \\ &= a + {b \over a + x} \end{align}$$ In this case $a=b=3$ and if $x = M/N = m/n$ are two square-roots of $12$, then $$ {m\over n} = 3 + {3 \over 3 + {M \over N}} = {12 N + 3M \over 3N + M} $$ Now solve the system $$ m = 12N + 3M, \quad n = 3N + M$$ for $M$, $N$, and get the new fraction in terms of $m$, $n$: $$ {M \over N} = {12n - 3m \over m -3n} $$ You then have to check that it's still a square-root of $12$ and the denominator is positive and has decreased.

The procedure works in general as long as $a^2+b$ is not a square (and $a$ and $b$ are positive).

  • $\begingroup$ Wait: I found something strange: in your last 2 line, before QED, the fraction ought to be $\frac{9(28n^2-8mn)}{3(7n^2-3mn)}$, not as you claimed. Per chance I missed something? Regards. $\endgroup$ – awllower Apr 7 '13 at 9:45
  • $\begingroup$ Thanks! There was a typo earlier, too. Fixed now. $\endgroup$ – Michael E2 Apr 7 '13 at 11:49
  • $\begingroup$ Thanks for this great answer thus. $\endgroup$ – awllower Apr 9 '13 at 0:33

If you know that $\sqrt{3}$ is irrational then we have easier method as follows:

If $\sqrt{12}$ want to be rational so it should be at form $\frac{m}{n}$ but we know $\sqrt{12}=\sqrt{2^{2}.3}=2\sqrt{3}$ so $\sqrt{3}=\frac{m}{2n}$ and should be rational too which is contradiction. So $\sqrt{12}$ can not be rational.

And if you don't know $\sqrt{3}$ is irrational you can prove it as usual way that is described by others and then use this method to conclude $\sqrt{12}$ is irrational according to $\sqrt{3}$ is irrational.


Yet another simple way to show this!
The question is equivalent with the irrationality of $\sqrt3$, which is the same as showing that there is no rational solution to $x^2-3=0$. By Eisenstein's criterion, the polynomial is indeed irreducible over $\mathbb Q$. So this finishes the proof.


To do it directly ignore the prime 2 altogether, and go for the prime 3 which appears to an odd power in the equation $$m^2=12n^2$$ (assume lowest terms)

The right hand side is divisible by 3, so the left hand side must be divisible by 3, so we must have $m=3r$. Our equation becomes $$9r^2=12n^2 \text{ or }3r^2=4n^2$$

Now we see similarly that $n$ must be divisible by 3, contradicting our lowest terms assumption.

To show that $3\mid m^2\implies 3\mid m$ let $m=3r+d$ with $d\in \{-1,0,1\}$ then $m^2=9r^2+6dr+d^2=3(3r^2+2dr)+d^2$ and this is not divisible by $3$ if $d^2=1$ so we must have $d=0$.

This establishes the result for the prime $3$ using the division algorithm and cases (which is pretty much equivalent to using the concepts odd and even when dealing with the prime $2$).

This can be done by cases for small primes, but to prove the general theorem for all primes does require more complex machinery as Pete Clark notes in the comments.

  • $\begingroup$ You are using "$3 \mid m^2 \implies 3 \mid m$". At this level, many instructors would want to see justification for this. $\endgroup$ – Pete L. Clark Feb 16 '13 at 19:47
  • $\begingroup$ @PeteL.Clark Isn't it trivial though? $\endgroup$ – Thomas Feb 16 '13 at 19:49
  • $\begingroup$ @Thomas: No, it isn't. If you know Euclid's Lemma or the Fundamental Theorem of Arithmetic, it is easy. But neither of these are trivial results and in fact most first courses in proof techniques do not establish them. A lower level proof uses the contrapositive and the division algorithm. $\endgroup$ – Pete L. Clark Feb 16 '13 at 19:53
  • $\begingroup$ @PeteL.Clark I see, thanks. $\endgroup$ – Thomas Feb 16 '13 at 19:57
  • $\begingroup$ @PeteL.Clark I have added some extra - several of the other answers assume the same. I could try to be naughty and define a prime by the property that $p\mid mn$ then $p\min m$ or $p\mid n$, but then I would have to show $3$ is prime. $\endgroup$ – Mark Bennet May 1 '15 at 15:54

Another angle on this is to prove that no integer is the square of a ratio. All squares are the squares of integers. Then, since $12$ does not have an integer square root, its square root cannot be rational, either.

To show that no integer is the square of a ratio, suppose $(\frac{n}{m})^2 = k$ where $m, n$ and $k$ are integers, $n/m$ is in lowest terms, $m\neq 1$, and all are integers. But that situation is impossible.

$$(\frac{n}{m})^2 = k$$

$$\frac{n\cdot n}{m\cdot m} = k$$

If $n/m$ is in lowest terms, as we assumed, that means that $m$ does not divide $n$. This implies that $m$ and $n$ have completely distinct prime factors. Which implies that no multiple of $m$ divides any multiple of $n$, because multiples of a number are just combinations of its prime factors. Thus $\frac{n\cdot n}{m\cdot m}$ cannot be an integer, unless $m = 1$ which we ruled out.

Therefore, $\sqrt 12$ cannot be a ratio. It must either be an integer ($12$ is a square), or else irrational. Our goal is therefore to show that $12$ isn't a square.

Observe that $12$ factors into $2\cdot 2\cdot 3$. A square has prime factors that are all of even duplicity, so that these factors can be divided into two identical groups. There is no way to separate the factors $2\cdot 2\cdot 3$ into two identical groups because $3$ occurs only once. (By contrast, consider $36 = 2\cdot 2\cdot 3\cdot 3$ whose factors are each of duplicity 2, and so can be split into two groups $(2\cdot 3)(2\cdot 3) = 6\cdot 6$.)

Since $\sqrt 12$ isn't an integer, and no integer has a square root which is a ratio, $\sqrt 12$ must be irrational.


You could have done this way too. \ Assume $\displaystyle\frac{m}{n}$ is written in its simplest form. Then $m^2=2(6n^2)\Rightarrow m$ is even. Substituting $m=2k\Rightarrow n$ is even. Thus both $m$ and $n$ have a common factor of 2 contradicting you statement that $\displaystyle\frac{m}{n}$ is not in its simplest form.

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    $\begingroup$ How did you deduce "$m=2k \Rightarrow n$ is even"? $\endgroup$ – P.. Feb 16 '13 at 15:38

I'll assume you know how to show that $\sqrt 2$ is irrational, in the same way you can show that for any prime $p$ then $\sqrt p$ is irrational, and we know that if $a$ is irrational and $b\ne 0$ is rational then $ab$ is irrational then we have$$\sqrt12=2\sqrt3$$now $2\ne 0$ is rational and 3 is prime which implies that $\sqrt3$ is irrational thus we have $\sqrt12=2\sqrt3$ is irrational.


Just call $\sqrt {12}$ a rational number. What can you say about it? $12$ must be a square of rational number.

$12= k^2$ where ($k =\dfrac{p}{q}$)


$12= 2^2 \cdot 3=\dfrac{p^2}{q^2} \implies 3= \dfrac{p^2}{q^2 \cdot 2^2} \implies 3$ is a square. Which implies it has odd number of divisors, but it doesn't.


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