I'm trying to do this proof by contradiction. I know I have to use a lemma to establish that if $x$ is divisible by $3$, then $x^2$ is divisible by $3$. The lemma is the easy part. Any thoughts? How should I extend the proof for this to the square root of $6$?

  • $\begingroup$ Suppose $\sqrt{3}$ is rational, then $p/q = \sqrt{3}$ for some coprime integers $p$ and $q$. Square both sides... $\endgroup$
    – angryavian
    Sep 14, 2014 at 4:06
  • $\begingroup$ Wait, what does this have to do with the real analysis? I suppose there is a presence of a measure? $\endgroup$
    – awllower
    Sep 14, 2014 at 4:12
  • $\begingroup$ This question is in my real analysis book. It's in the first section titled preliminaries. I guess this have to do more with mathematical logic than real analysis. Sorry for the mistake. My bad. $\endgroup$ Sep 14, 2014 at 4:18
  • $\begingroup$ A questino in a book on real analysis might pertain in nature to a totally different subject, and, in that case, I find no reason to classify such a question under the tag of real-analysis. Per chance an edit is under consideration? $\endgroup$
    – awllower
    Sep 14, 2014 at 4:46

6 Answers 6


Say $ \sqrt{3} $ is rational. Then $\sqrt{3}$ can be represented as $\frac{a}{b}$, where a and b have no common factors.

So $3 = \frac{a^2}{b^2}$ and $3b^2 = a^2$. Now $a^2$ must be divisible by $3$, but then so must $a $ (fundamental theorem of arithmetic). So we have $3b^2 = (3k)^2$ and $3b^2 = 9k^2$ or even $b^2 = 3k^2 $ and now we have a contradiction.

What is the contradiction?

  • $\begingroup$ @The-Duderino By the way, the proof for $ \sqrt{6} $ follows in the same steps almost exactly. $\endgroup$
    – user109879
    Sep 14, 2014 at 16:28
  • 3
    $\begingroup$ Could you please point out, what is the contradiction here? Is it this - > $b^2$ must be divisible by 3, therefore $b$ must be divisible by 3. Which then means that, $a$ and $b$ are not coprime. Is this analysis correct or were you alluding to something else? $\endgroup$
    – Oiler
    Mar 28, 2017 at 16:18
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    $\begingroup$ @Rio1210 Your analysis is correct. $\endgroup$
    – user109879
    Mar 28, 2017 at 17:03
  • $\begingroup$ This seems correct $\endgroup$
    – Cyrusmith
    Feb 8, 2018 at 15:04
  • 1
    $\begingroup$ It is also by Euclid's Lemma, by the way $\endgroup$
    – Mr Pie
    May 11, 2018 at 10:49

suppose $\sqrt{3}$ is rational, then $\sqrt{3}=\frac{a}{b} $ for some $(a,b) $ suppose we have $a/b$ in simplest form.
\begin{align} \sqrt{3}&=\frac{a}{b}\\ a^2&=3b^2 \end{align} if $b$ is even, then a is also even in which case $a/b$ is not in simplest form.
if $b$ is odd then $a$ is also odd. Therefore:
\begin{align} a&=2n+1\\ b&=2m+1\\ (2n+1)^2&=3(2m+1)^2\\ 4n^2+4n+1&=12m^2+12m+3\\ 4n^2+4n&=12m^2+12m+2\\ 2n^2+2n&=6m^2+6m+1\\ 2(n^2+n)&=2(3m^2+3m)+1 \end{align} Since $(n^2+n)$ is an integer, the left hand side is even. Since $(3m^2+3m)$ is an integer, the right hand side is odd and we have found a contradiction, therefore our hypothesis is false.

  • $\begingroup$ How does $4n^{2}+4n+1=2n^{2}+2n?$ $\endgroup$ Feb 3, 2019 at 22:29
  • $\begingroup$ @JamesWarthington both sides are subtracted by 1, and both sides are then divided by 2. Here is that step expanded even more. \begin{align} 4 n^2 + 4 n + 1 &= 12 m^2 + 12 m + 3 \\ 4 n^2 + 4 n &= 12 m^2 + 12 m + 2 \\ 2 n^2 + 2 n &= 6 m^2 + 6 m + 1 \end{align} $\endgroup$ Feb 14, 2019 at 4:53
  • $\begingroup$ Just a quick question, couldn't you do the following, which I think is valid, but provides an opposite result: \begin{align} 4n^2 + 4n &= 12m^2 + 12m + 2 \\ 2(2n^2 + 2n) &= 2(6m^2 + 6m) + 2 \end{align} I think that's totally valid, but it left both sides as even, and therefore breaks the contradiction. Is something wrong here? $\endgroup$ Apr 26, 2020 at 0:02
  • $\begingroup$ I guess we have to take it to its simplest form for the proof to be considered complete? $\endgroup$ Apr 26, 2020 at 0:15
  • $\begingroup$ A recent question concerning the same approach: Novel (?) proof of the irrationality of sqrt(3) $\endgroup$ Jul 3, 2021 at 6:06

A supposed equation $m^2=3n^2$ is a direct contradiction to the Fundamental Theorem of Arithmetic, because when the left-hand side is expressed as the product of primes, there are evenly many $3$’s there, while there are oddly many on the right.

  • $\begingroup$ Can this exact argument be used for any irrational number aside from square root of 3 as well? It seems like it should apply to everything. $\endgroup$ Sep 16, 2022 at 13:40
  • $\begingroup$ Yea, absolutely, @MathandML . This shows that the strength of the FTA is not the existence of the prime decomposition of any number, but the uniqueness thereof. $\endgroup$
    – Lubin
    Sep 17, 2022 at 0:29

The number $\sqrt{3}$ is irrational ,it cannot be expressed as a ratio of integers a and b. To prove that this statement is true, let us Assume that it is rational and then prove it isn't (Contradiction).

So the Assumptions states that :

(1) $\sqrt{3}=\frac{a}{b}$

Where a and b are 2 integers

Now since we want to disapprove our assumption in order to get our desired result, we must show that there are no such two integers.

Squaring both sides give :



(Note : If $b$ is odd then $b^2$ is Odd, then $a^2$ is odd because $a^2=3b^2$ (3 times an odd number squared is odd) and Ofcourse a is odd too, because $\sqrt{odd number}$ is also odd.

With a and b odd, we can say that :



Where x and y must be integer values, otherwise obviously a and b wont be integer.

Substituting these equations to $3b^2=a^2$ gives :


$3(4y^2 + 4y + 1) = 4x^2 + 4x + 1$

Then simplying and using algebra we get:

$6y^2 + 6y + 1 = 2x^2 + 2x$

You should understand that the LHS is an odd number. Why?

$6y^2+6y$ is even Always, so +1 to an even number gives an ODD number.

The RHS side is an even number. Why? (Similar Reason)

$2x^2+2x$ is even Always, and there is NO +1 like there was in the LHS to make it ODD.

There are no solutions to the equation because of this.

Therefore, integer values of a and b which satisfy the relationship = $\frac{a}{b}$ cannot be found.

Therefore $\sqrt{3}$ is irrational

  • $\begingroup$ $3(4y^{2}+4y+1)=12y^{2}+12y+3$. Factor out 2 gives you $2(6y^2+6y+\frac{3}{2})$ How does this equal to $6y^2+6y+1$ $\endgroup$ Feb 3, 2019 at 23:42
  • 1
    $\begingroup$ @JamesWarthington I was confused initially too but it actually makes sense. You're not doing "$3(4y^2 + 4y + 1)$" but "$3(4y^2 + 4y + 1) = 4x^2 + 4x + 1$". So... \begin{align}3(4y^2 + 4y + 1) = 4x^2 + 4x + 1\\12y^2 + 12y + 3 = 4x^2 + 4x + 1\\12y^2 + 12y = 4x^2 + 4x - 2\\12y^2 + 12y = 4x^2 + 4x - 2\\6y^2 + 6y = 2x^2 + 2x - 1\\6y^2 + 6y + 1 = 2x^2 + 2x\\2(3y^2 + 3y) + 1 = 2(x^2 + x)\\2n + 1 = 2n\end{align} $\endgroup$
    – TigerYT
    Jun 16, 2022 at 5:46
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    $\begingroup$ I must say out of all the contradiction proofs... this one actually makes sense to me. Entirely. $\endgroup$
    – TigerYT
    Jun 16, 2022 at 5:51

The following is a somewhat non-classical and less elementary proof.
Suppose that $\sqrt3\in\mathbb Q,$ and write $\alpha:=\sqrt3=p/q$ where $p$ and $q$ are relatively prime. Then we have $q\alpha=p,$ so in the ring $\mathbb Z[\alpha], \alpha$ divides $p,$ hence $\alpha^2\mid p^2,$ i.e. $3\mid p^2.$ So $p^2/3\in R:=\mathbb Z[\alpha]\cap\mathbb Q.$
In fact, this ring $R$ is nothing else but $\mathbb Z,$ by the lemma below, therefore $3$ divides $p^2,$ hence $p,$ in $\mathbb Z.$ So we conclude that $3$ divides $q\alpha$ as an integer; then $9\mid 3q^2.$ Since $p$ and $q$ are relatively prime, $9$ is prime to $q^2,$ thus $9\mid 3,$ a contradiction.

Lemma: $R:=\mathbb Z[\alpha]\cap\mathbb Q=\mathbb Z.$

For those who know some algebraic number theory already, this lemma needs no proof, but, for the sake of reference, I shall provide a brief proof of the above-used fact that $R=\mathbb Z,$ assuming known that $\mathbb Z[\alpha]$ is the integral closure of $\mathbb Z$ in $\mathbb Z(\alpha).$

Since $\mathbb Z[\alpha]$ is the integral closure of $\mathbb Z$ in $\mathbb Q(\alpha),$ $R$ consists of the elements in $\mathbb Q$ which are integral over $\mathbb Z,$ namely, of the elements $m/n\in\mathbb Q$ such that there exists a polynomial $f(x)=x^k+a_1x^{k-1}+\cdots+a_k$ with $a_i\in\mathbb Z, \forall i$ such that $f(m/n)=0.$ Then we might assume that $m$ and $n$ are relatively prime, and obtain the relation $m^k+a_1m^{k-1}n\cdots+a_kn^k=0,$ so every prime divisor of $n$ also divides $m,$ contradicting our hypothesis, unless $n=1,$ i.e. $m/n\in\mathbb Z.$ This proves that $R=\mathbb Z.$

P.S. This works for every square-free integer, after a slight modification.
Hope this may be of some interest, though not necessarily of any use. ;P


Here is a proof of mine by contradiction that if $n$ is a positive integer that is not a perfect square then $\sqrt{n}$ is irrational: $\sqrt{17}$ is irrational: the Well-ordering Principle


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