There is a very simple proof by means of divisibility that $\sqrt 2$ is irrational. I have to prove that $\sqrt 3$ is irrational too, as a homework. I have done it as follows, ad absurdum:

Suppose $$\sqrt 3= \frac p q$$

with $p/q$ irreducible, then

$$\begin{align} & 3q^2=p^2 \\ & 2q^2=p^2-q^2 \\ &2q^2=(p+q)(p-q) \\ \end{align}$$

Now I exploit the fact that $p$ and $q$ can't be both even, so it is the case that they are either both odd, or have different parity. Suppose then that $p=2n+1$ and $q=2m+1$

Then it is the case that

$$\begin{align} &p-q=2(n-m) \\ &p+q=2(m+n+1) \\ \end{align}$$

Which means that

$$\begin{align} &2q^2=4(m-n)(m+n+1) \\ &q^2=2(m-n)(m+n+1) \\ \end{align}$$

Then $q^2$ is even, and so is then $q$, which is absurd. Similarly, suppose $q=2n$ and $p=2m+1$.

Then $p+q=2(m+n)+1$ and $p-q=2(m-n)+1$. So it is the case that

$$\begin{align} &2q^2=(2(m-n)+1)(2(m+n)+1)\\ &2q^2=4(m^2+m-n^2)+1 \\ \end{align}$$

So $2q^2$ is odd, which is then absurd.

Is this valid?

  • 3
    $\begingroup$ There is a very simple proof by means of divisibility that $\sqrt3$ is irrational, too! Why not use that? This proof looks fine. $\endgroup$ – The Chaz 2.0 Apr 13 '12 at 17:56
  • 3
    $\begingroup$ The far simpler proof is use the fact the $p$ and $q$ cannot both divide by 3 leading to a contradiction in two lines. $\endgroup$ – nbubis Apr 13 '12 at 17:56
  • $\begingroup$ @nbubis Hint me, then. $\endgroup$ – Pedro Tamaroff Apr 13 '12 at 17:57
  • 1
    $\begingroup$ @David That'd be a WLOG quintessential. $\endgroup$ – Pedro Tamaroff Apr 13 '12 at 17:59
  • 2
    $\begingroup$ @PeterT.off: You didn't deal explicitly with the case of opposite parity, though you mentioned it. It is clearly impossible. Very nice proof, creative. I had not seen it before. $\endgroup$ – André Nicolas Apr 13 '12 at 18:09

10 Answers 10


It works, but can be simplified: $\rm\:mod\ 2\!: p\equiv p^2 = 3q^2 \equiv q,\:$ so $\rm\:p,q,\:$ being coprime, are odd. $\rm\:mod\ 4\!:\ odd\equiv \pm 1,\:$ so $\rm\:odd^2\equiv 1,\:$ so $\rm\: 1\equiv p^2 = 3q^2\equiv 3\ \Rightarrow\ 4\:|\:3-1\:\Rightarrow\Leftarrow$

  • 2
    $\begingroup$ Peter, the symbolic approach is Bill's style... $\endgroup$ – The Chaz 2.0 Apr 13 '12 at 18:59
  • 2
    $\begingroup$ @Bill It isn't that I can't follow, but rather that I need a middle input{symbol}=output{word} process to read it. $\endgroup$ – Pedro Tamaroff Apr 13 '12 at 20:01
  • 3
    $\begingroup$ this is one way to show that using more symbols isn't helping.. $\endgroup$ – stefan Apr 13 '12 at 23:21
  • 5
    $\begingroup$ @Peter Alas, I have no clue what your prior comment means. What I wrote above is very simple modular arithmetic. It is well-worth your effort to become proficient at such. It's impossible for me to say more unless you can be more specific about what is problematic. $\endgroup$ – Bill Dubuque Apr 13 '12 at 23:28
  • 4
    $\begingroup$ @Peter That is part of the natural learning process. When you become more proficient you'll no longer need to do this translation process (similar to when you master a foreign language). $\endgroup$ – Bill Dubuque Apr 13 '12 at 23:59

$3q^2 = p^2$, so $3|p$ (as in the case of $\sqrt2$).

Hence $q^2 = 3k$ for some $k$, and then $3|q$


  • $\begingroup$ This is the correct way to do it (in my opinion), because as it is just the same as the usual method for $\sqrt{2}$ and carries across almost unchanged to $\sqrt{p}$ for any prime $p$. It will also carry across without too much work to $\sqrt{n}$ for any non-square $n$. $\endgroup$ – George Lowther May 9 '12 at 19:57
  • $\begingroup$ @George: I would hope that the plenitude of answers to this question would suggest that there is not just one right way to go about this :) $\endgroup$ – The Chaz 2.0 May 9 '12 at 20:05

Here are some proofs I've found (link at bottom):

If $\sqrt 3 = m/n$: $$ \frac{m}{n} = \sqrt 3 \frac{\sqrt 3 - 1}{\sqrt 3 - 1} = \frac{3-\sqrt 3}{\sqrt 3 - 1} = \frac{3-m/n}{m/n-1} = \frac{3 n - m}{m-n}$$ and the right side has a smaller denominator, since $m < 2n$ (i.e., $\sqrt 3 < 2$).

$x = \sqrt{3} - 1$ is a root of the equation $x^2 + 2x - 2 = 0$, thus: $$x(3+x) = 2+x$$

$$x = \frac{2+x}{3+x} = \cfrac{1}{1 + \cfrac{1}{2 + x}}$$

And thus

$$x = [1,2,1,2,\dots]$$

So $\sqrt{3}$ os irrational.

These proofs and others: How to prove that $\sqrt 3$ is an irrational number?


There is also a really nice proof using what is called reductio ad absurdum (or infinite regress), which can also be framed as a simple contradiction using the minimality property of the natural numbers.

Suppose WLOG (without loss of generality) that $$\sqrt3=\frac{u}{v}$$ for $u,v\in\mathbb{N}$ relatively prime (any positive rational number in $\mathbb{Q}$ can be expressed as a fraction in lowest terms). The reason we can assume $u,v\in\mathbb{N}$ rather than $\mathbb{Z}$ without losing the generality of our argument is because any case in the latter category furnishes one in the former by observing that $3>0$ so that $u$ and $v$ must have the same sign, and if they are negative, then $-u,-v\in\mathbb{N}$ also has the same ratio. So then $$u^2=3\,v^2.$$ But $3$ is prime and divides the RHS, hence it divides the LHS, and that means it must divide $u$ (it is a fact, known as Euclid's Lemma, that if $p$ is prime, then $p|ab \implies p|a$ or $p|b$). But then $3|u=u_0$ means that $u=3u_1$ for some $u_1\in\mathbb{N}$, and consequently, $$9u_1^2=(3u_1)^2=3v^2 \quad\implies\quad 3u_1^2=v^2.$$ At this point, if we have not assumed that $u,v$ are relatively prime, we continue by noting that $v=v_0=3v_1$ for some $v_1\in\mathbb{N}$, whence $u_1^2=3v_1^2$ $\implies\cdots\implies$ $$\forall n\in\mathbb{N}:u_n=u\cdot3^{-n},~v_n=v\cdot3^{-n}\in\mathbb{N}$$ which is an impossible infinite regress, also called a reductio ad absurdum (reduction to absurdity). The absurdity, impossibility, or contradiction it leads to is that, from the hypothesis, it shows that the natural numbers $u$ and $v$ are infinitely divisible, while we know that for some $N\in\mathbb{N}$ (eventually, sufficiently large), $\frac{u}{3^n}$ and $\frac{v}{3^n}$ are obviously less than $1$ and thus not whole numbers for all $n\ge N$.

A more elegant variation (avoiding these "infinite gymnastics") is to use the stipulation, which we can make without loss of generality, that $u$ and $v$ are relatively prime. Then, we can stop as soon as we deduce that $3|v$, since at this point we already knew that $3|u$.

An even more elegant variation uses the well-orderedness of $\mathbb{N}$, where we also suppose to begin with that $u$ and $v$ are minimal (or if in doubt about whether we can simultaneously require this of both variables, suppose that either $u$ or $v$ is minimal). Then, as soon as we discover our first extra factor of $3$, we already reach a contradiction.

  • $\begingroup$ Great answer! But I do use reductio ad absurdum, don't I? I like the fact the proof you give arrives to $\dfrac u {3^n}$, since then we can use Arichimede's priniciple $x <n$ for every $x \in \mathbb R$ and some appropriately large enough $n \in \mathbb N$. $\endgroup$ – Pedro Tamaroff Apr 13 '12 at 19:59
  • 3
    $\begingroup$ Who downvoted?! $\endgroup$ – Pedro Tamaroff Apr 13 '12 at 20:02
  • 1
    $\begingroup$ Yes, of course you knew and used reductio ad absurdum, I just thought this was a proof worth knowing for its simplicity and didactic versatility. One downvote doesn't matter; we can rely on the law of large numbers to determine the post's merit over time. $\endgroup$ – bgins Apr 13 '12 at 20:04

I'll just add to this mayhem with this bit of logic.

Assume $\sqrt{3}=\frac{a}{b}\implies3=\frac{a^2}{b^2}\implies3b^2=a^2$. Now, this cannot be true! When you square something, you double all of its prime factors. Then we know $a^2$ and $b^2$ have an even number of factors, therefor $3b^2$ has an odd number of factors because we have added a factor of $3$. So $3b^2 \neq a^2$.

  • 1
    $\begingroup$ And, for similar reasons, the $\sqrt[r]{n}$ is irrational for any $n$ which isn't an $r$'th power. $\endgroup$ – George Lowther May 9 '12 at 19:59

Here is a proof using some results from field theory: Suppose that $\sqrt{3}$ is rational. Then the polynomial $x^2 - 3$ has a root in $\Bbb{Q}$. However we know for degree 2 polynomials that having a root in $\Bbb{Q}$ is equivalent to the polynomial being reducible over $\Bbb{Q}$. But then by Eisenstein's criterion with $p=3$ we have that $x^2 -3$ is irreducible over $\Bbb{Q}$ which is a contradiction.

$\textbf{Edit:}$ Here are some applications of this method. Suppose you want to prove that $\sqrt{2} + \sqrt{3}$ is irrational. For a contradiction suppose it is not. Then you can write $\sqrt{2} +\sqrt{3} = p$ for some rational number $p$, so that squaring both sides we have that $\frac{p^2 -5}{2} = \sqrt{6}$. This is saying that $\sqrt{6}$ is a rational number. However if it is, then the polynomial $x^2 - 6$ is reducible over $\Bbb{Q}$. But then this is a contradiction because Eisenstein's Criterion with $p=3$ or $p=2$ shows that $x^2 - 6$ is irreducible over $\Bbb{Q}$. Done.

Generalisation of this by Eisenstein's Criterion: The square root of any square free positive integer is irrational.

  • $\begingroup$ What is Eisenstein's criterion? What does it mean that a polynomial is irreducible over a set of numbers? Sorry, but I don't know about that. However, I'd be glad if you gave me the basics. $\endgroup$ – Pedro Tamaroff Apr 13 '12 at 22:54
  • 1
    $\begingroup$ @PeterT.off First let $f(x)$ be a polynomial with rational coefficients. We say that $f$ is irreducible over $\Bbb{Q}$ if it is not possible to write $f = gh$ for $g$ and $h$ polynomials of degree at least $1$ with coefficients in $\Bbb{Q}$. Eisenstein's criterion (which you can check in wikipedia) is a test to tell if a polynomial is irreducible over $\Bbb{Q}$ or not. In fact I think it even works for the fraction field of just any unique factorisation domain. $\endgroup$ – user38268 Apr 13 '12 at 23:19
  • 2
    $\begingroup$ @Peter: Eisenstein's is, very crudely, a more elaborate version of the rational roots theorem... :) $\endgroup$ – J. M. is a poor mathematician Apr 14 '12 at 4:45
  • $\begingroup$ @PeterT.off Please see edit above. $\endgroup$ – user38268 Apr 19 '12 at 1:25
  • $\begingroup$ @BenjaminLim Thanks for that add! $\endgroup$ – Pedro Tamaroff Apr 19 '12 at 1:49

A proof of R. Dedekind, from 1901.

(He proves that if $n\in\mathbb{N}$ is not a square number, then $\sqrt{n}$ is irrational.)

$\sqrt{3}$ is irrational.

Proof: Suppose that $\sqrt{3}=\frac{x}{y}$, $x$ and $y$ are integers, when $x$ is the smallest integer which satisfies that equation. (If there exist $x$ and $y$ such that $\sqrt{3}=\frac{x}{y}$ then exist that minimal $x$, $\mathbb{N}$ is well-ordered set and $x$ is there). $\frac{x}{y}$ is not an integer ($3$ is not a perfect square), hence exist $k\in\mathbb{N}$, such that:


Now, let us look at

$$x'=(k-\frac{x}{y})x=kx-3y \ \ \ \ ; \ \ \ \ y'=(k-\frac{x}{y})y=ky-x $$

Furthermore, $x$ and $y$ are integers, and:


But, $k-\frac{x}{y}<1$, therefor $x'<x$, and that is a contradiction to the fact that out $x$ chosen to be the minimal which satisfies $\sqrt{3}=\frac{x}{y}$.



Here's another proof:

Binomial expansion shows that $(\sqrt{3}- 1)^n = p_n + q_n \sqrt{3}$ where $p_n, q_n$ are integers. The sequence $(\sqrt{3}-1)^n = p_n + q_n \sqrt{3}$ is always nonzero, but it tends to $0$ as $n \to \infty$. Thus for any $\delta > 0$, for sufficiently large $n$ (depending on $\delta$), the number $\sqrt{3}$ admits a "good" rational approximation $-p_n/q_n$ (note the minus sign), in the sense that $$0 < |q_n \sqrt{3} + p_n| < \delta .$$ This shows that $\sqrt{3}$ must be irrational.

The proof is, of course, not original; it is perhaps folklore.

  • $\begingroup$ I have made the post CW since it does not really answer the question. $\endgroup$ – Srivatsan Apr 19 '12 at 2:27
  • $\begingroup$ This should have more upvotes... could you add some more? $\endgroup$ – Pedro Tamaroff May 10 '12 at 19:48

One can show a more general result, that is for any prime number $p$ we must have $\sqrt{p}$ is irrational. see my answer here.


Your proof is simple and easy to be understood, but I think the last equation $$ 2q^2 = 2(m^2 + 2m - n^2) + 1 $$ is wrong, which should be $$ 2q^2 = 4(m^2 + m - n^2) + 1 $$


protected by cactus314 Aug 31 '18 at 18:35

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.