I am studying the proof of $\sqrt 2$ is irrational number. I understand most of the proof, but I lack an understanding of the main idea which is:

we assume $\frac{m^2}{n^2} = 2$ then not both $m$ and $n$ can be even. I do not understand, why not both $m$ and $n$ can be even?

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    $\begingroup$ You start by assuming that $\sqrt 2=\frac mn$ where $m$ and $n$ are co-prime. $\endgroup$ – Sahiba Arora Aug 4 '17 at 12:14
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    $\begingroup$ Asserting that your are studying the proof that $\sqrt2$ is irrational shows a serious misunderstanding. What you are studying is just one of the many known proofs of that statement. $\endgroup$ – José Carlos Santos Aug 4 '17 at 12:16
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    $\begingroup$ If $\sqrt2$ were rational, then $\sqrt2=\frac mn$ for a fraction in simplest form i.e. where $m,n$ are not both even. If they were both even, then you should see there is no simplest form of the fraction $\frac mn$ equal to $\sqrt2$, and since all fractions can be written in simplest form, $\sqrt2$ is not rational. $\endgroup$ – Simply Beautiful Art Aug 4 '17 at 12:19
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    $\begingroup$ as Sahiba said, you choose $n,m$ to be co-prime, equivalently you can choose $n,m$ be minimal. $\endgroup$ – Yanko Aug 4 '17 at 12:20
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    $\begingroup$ If $m$ and $n$ are both even then the fraction is not in lowest terms. There is no pair of numbers (m,n) that can not ever be put into lowest terms. Dividing both m and n by two will still result in $\frac m2$ and $\frac n2$ being even so those can be divided in two... forever. So each n and m must be divisible by an infinite power of 2. Which is not possible. $\endgroup$ – fleablood Aug 4 '17 at 16:44

10 Answers 10


we can assume that $$\gcd(m,n)=1$$ and $$2=\frac{m^2}{n^2}$$ then $$2n^2=m^2$$ thus the left-hand side is even and so $$m^2$$ this is a contradiction, both numbers $m,n$ can not be even


Before you continue the argument you write the fraction $m/n$ in lowest terms. Then the numerator and denominator can't both be even.


Because if $n=2n_1$ and $m=2m_1$ then $\frac{m_1^2}{n_1^2}=2$... and we get an infinite series $n>n_1>n_2>...$ of natural numbers, which is impossible.


An another way

Be $\sqrt{2}\in \mathbb{Q}$, and $\mathcal{Q}=\bigg\{q\in\mathbb{N}^*,\quad q\sqrt{2}\in\mathbb{N}\bigg\}\implies \mathcal{Q}\neq\varnothing$

Let $q_s:=\min \mathcal{Q} $

Let $p:=q_s\sqrt{2}-q_s$

$p=q_s\sqrt{2}-q_s<2q_s-q_s=q_s\iff \boxed{p<q_s}\quad (1)$

$p=q_s\sqrt{2}-q_s\iff\sqrt{2}p=2q_s-\sqrt{2}q_s \iff \boxed{\sqrt{2}p\in\mathbb{N}}\quad (2)$

$(1)\land(2)\implies$ we have found $p<q_s$ such that $\sqrt{2}p\in\mathbb{N}$ so we have a contradiction thus $\sqrt{2}\notin \mathbb{Q}$

  • $\begingroup$ However for this proof, you have first to prove that $\sqrt{2}<2$. $\endgroup$ – celtschk Feb 20 '18 at 7:57

Okay, this may be a bit pendantic.

There are a few fundamentals assumed but not stated.

1) Given that $a = k*n; a,k,n \in \mathbb Z$ and $b = k*m; b, m \in Z$ then then rational value of the ratio $\frac ab $ is the same as the ratio $\frac mn$ and we say $\frac ab = \frac nm$.

Why? Well... because. (If someone has a better answer feel free to leave a comment). I feel this is the basic concept of ratio or proportion and ratio of "$n$ units to $m$ units" is axiomatically the same value regardless of the unit. And if the "units" is a measure of $k$ value then "$n$ $k$-s are in proportion to $m$ $k$-s" as "$n$ is in proportion to $m$" is simply an axiom.

If anyone has a better or more correct way of putting this, let me know.

2) Given any two integers $a,b$ there is some common factor $k$ so that if $a\div k =n$ and $b \div k = m$ and $m,n$ have only $1$ as a common (positive integer) factor.

Why? If $a,b $ don't have any common factor other than then $k=1$ is precisely that common factor. If $a, b$ have a common $j$ other than $1$ then $j > 1$ and $a\div j = n$ then $a > n$ and if $b\div j = m$ then $b > n$. If $a$ and $b$ were such that there is no such common factor, then $j$ can not be that factor so $m$ and $n$ have a common factor $j_2 > 1$ and if $m =j_2*m_2$ and $n = j_2*n_2$, we have $a > n > n_2$ and $b > m > m_2$. If there is no such terminating common factor we can do this infinite so we can get an infinite chain of $a > n > n_2> n_3>.....$ and $b > m > m_2 > m_3> .....$.

This is clearly impossible. Why? Because these are integers the difference between $n_i$ and $n_{i+1}$ is at least $1$ and so $a$ and $b$ must be larger than an infinite $1$s. (i.e. $a$ and $b$ are infinite.)

From those two assumptions, we can conclude:

A) If $q = \frac ab$ is any rational number with $a,b$ are integers. Then $a,b$ have a common factor $k$ so that $n = \frac ak; m = \frac bk$ and $n$ and $m$ have no common factor other than $1$. Thus we can state $q =\frac nm$ where $n$ and $m$ have no factors other than $1$ in common.

We can express any rational $q$ in such a way.


So now you can start your proof:

Assume $\sqrt{2}$ is rational. Then $\sqrt{2} = \frac ab$ for integers $a,b$ so that $a$ and $b$ have no common factor other than $1$.

Then...... < < details omitted > >... $2$ is a factor of $a$ and ..... < < details omitted > > .... $2$ is a factor of $b$.

And therefore $a$ and $b$ have a factor of $2$ in common, which contradicts that $a$ and $b$ have no factors in common other than $1$.

So $\sqrt{2}$ is not rational.

  • $\begingroup$ “Why? Well... because. (If someone has a better answer feel free to leave a comment).” — Because $q=\frac mn$ is constructed to be the solution of the equation $nq=m$. And clearly for $k\ne 0$, the equations $nq=m$ and $knq=km$ are equivalent. Thus any construction where $\frac{km}{kn}$ were a different number than $\frac mn$ would not be suitable for that purpose. $\endgroup$ – celtschk Feb 20 '18 at 8:07
  • $\begingroup$ That assumes $nq=m $ has a unique solution. Which... is either so practical it has to be true; or axiomatic. Thing I wanted to avoid axioms (which are "because we said so") and to have a bit more vigor than "well, just look at it; of course it's true". I think if one way or another if we accept that a ration $n:m $ is meaningful it's basic that $nk:mk $ is the same. But as to why? Eventually you have say by definition and/or axiom. I guess I wanted to be honest about that. $\endgroup$ – fleablood Feb 20 '18 at 9:10
  • $\begingroup$ I did write this a long time ago. The proof hinges on the assumption a rational number has a fraction representation in lowest terms, and that we can divide common terms out. That's perfectly reasonable but if the op doesn't see it... well, I feel I have to spell it out that we are assuming those. $\endgroup$ – fleablood Feb 20 '18 at 9:15
  • $\begingroup$ My point is that rational numbers were invented for a purpose. And that purpose implies that relation. That's axiomatic, but it's not really “because I say so”, it's “because it fulfils the purpose”. $\endgroup$ – celtschk Feb 20 '18 at 9:19
  • $\begingroup$ Perhaps. I think basically we are both saying the same thing. But... here we are. We have an OP asking why we can't have a rational number where the numerator are both even. "Well, we could divide it out". What if it's still even. "Then we do it again; it can't go forever and it has to have a representation in lowest terms". Why? Well, because that's how numbers work. $\endgroup$ – fleablood Feb 20 '18 at 9:26

Okay thanks I understand it now after seeing that the fraction m/n in the lowest terms, in this situation both m and n cannot both be even.

Here is my proof
Prove that $√2$ is irrational
Let's assume $√2$ is rational then $2=m^2/n^2$
by writing it in the lowest terms not both $m$ and $n$ can be even
We obtain $m^2=2n^2$
Since any integer $n^2$ even or odd multiplied by $2$ is even, then $m^2$ must be even as well, hence m is also even because any square root of an even number is even.
Hence $m=2k$ for some positive integer $k$
Hence $m^2=(2k)^2$
so $4k^2=2n^2$
therefore $n^2=2k^2$ for some positive integer $k$, which indicates that $n^2$ is even.
However from the assumption $2=m^2/n^2$ we know that not both m and n can be even, therefore $m^2/n^2=2$ cannot be true

  • $\begingroup$ "any square root of an even number is even." that is wrong for example $\sqrt{6}$ but a square is even only if its square root is even $\endgroup$ – Stu Aug 4 '17 at 17:40
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    $\begingroup$ @Stu, I think he's implicitly talking about integers that are perfect squares. $\endgroup$ – user1717828 Aug 4 '17 at 18:09
  • $\begingroup$ Mmmm... He is implicitly talking about perfect squares but he is also being imprecise. I'd forgive such imprecision except this proof is specifically about a square root of an even number that is itself not an even integer. I think stu is right to correct the error but I think rather that stating "that is wrong", it's more accurate to say "that is incorrectly stated". A correct statement would be "the square root of any even square is even". $\endgroup$ – fleablood Aug 4 '17 at 21:08
  • $\begingroup$ @fleablood sorry but I don't understand what you mean.We are doing math so an assertion is right or wrong, false or true.whatever and I provided the right assertion "a square is even if and only if its square root is even". May you be more explicit about your intention? $\endgroup$ – Stu Aug 4 '17 at 21:18
  • $\begingroup$ We aren't actually doing math. We are offering constructive criticism to someone else who is doing math. It is clear he is implicitly talking about integer square roots (roots of perfect squares) is which case his statement "integer square roots of even numbers are even" is perfectly correct. The error was not that the OP thinks $\sqrt {12}$ is an even number-- the error is in thinking that the phrase "square root of an even number" refers to $\sqrt{16}$ or $\sqrt{36}$ but not to $\sqrt {12}$ or ... ahem... $\sqrt{2}$. $\endgroup$ – fleablood Aug 4 '17 at 21:28

In more concise terms : if $\sqrt{2}$ is rational number It can be written like $\sqrt{2}=\dfrac{m}{n}, \quad m\in\mathbb{Z},n\in \mathbb{Z}^*\quad\gcd(m,n)=1$

$\sqrt{2}=\dfrac{m}{n}\iff 2n^2=m^2\implies 2|m^2\iff\boxed{2|m}\;(1)\iff4|m^2\implies 4|2n^2\iff2|n^2\iff \boxed{2|n}\;(2)\qquad (1)\land(2)\implies\gcd(m,n)\geq2$

So $\sqrt{2}$ is not a rational number

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    $\begingroup$ "Concise terms" seem deeply antipedagogical to me... $\endgroup$ – Miguel Aug 4 '17 at 14:09
  • $\begingroup$ @ Miguel, you are right, but more beautiful!! but don't worry,he had understood, when I posted my answer. $\endgroup$ – Stu Aug 4 '17 at 14:42
  • $\begingroup$ What is the meaning of gcd(m,n)=1 $\endgroup$ – user420309 Aug 4 '17 at 14:45
  • $\begingroup$ greatest common divisor =1 <=>$\frac{m}{n}$is an Irreducible fraction $\endgroup$ – Stu Aug 4 '17 at 14:46

In addition to the other answers, I like the following way of proving it: Assume there are $a\in \mathbb Z$ and $b\in \mathbb N$ such as

$(\ast)\qquad \sqrt 2 = \frac a b $

with $\gcd(a,b)=1$.

We know that $\sqrt2>1$ as $1^2=1$ and $\sqrt2<2$ as $2^2=4$.

Squaring $(\ast)$, we obtain

$(+)\qquad 2=\frac{a^2}{b^2}$

As 2 ist a natural number, the numerator in equation $(+)$ must be a multiple of the denominator. Because of $\gcd(a,b)=1$, we know that $a$ and $b$ do not have any common prime factors. Thus, $a^2$ and $b^2$ cannot have any common prime factors, either. We can deduce $b^2=1$ and thus $b=1$. As a consequence, we get $a^2=2$ or $a=\sqrt2$, but $a$ is meant to be an integer.

So $\sqrt2$ being rational would imply $\sqrt2\in\mathbb Z$ and even $\sqrt2\in\mathbb N$, as roots are not negative. So $\sqrt2$ would be natural, greater than 1 and smaller than 2. However, there is no natural number between 1 and 2. Contradiction.

  • $\begingroup$ "As 2 is a natural number, the numerator in equationmust be a multiple of the denominator. Because of gcd(a,b)=1,that means we have $b^2=1$"... I think $b^2|a^2$ and $\gcd(b,a)$ implies $b = 1$ needs to be justified. At heart this is the same proof but distanced a bit. Also... how do we know that $0 < b < c$ implies $b^2 < c^2$? So why can't $\sqrt{2}$ be an integer? .... Everything you say is true but I think assuming those probositions is begging the question. $\endgroup$ – fleablood Aug 4 '17 at 20:58
  • $\begingroup$ If a and b do not have common prime factors, this must be true for their squares as well. So - given the initial conditions - the fraction $a^2/b^2$ can only be an integer if $b^2$ is 1. As b is natural, this means b is 1. And $sqrt2$ cannot be an integer, because the square function is monotonous: 1^2 is too small, 2^2 is too big, any other square will be even bigger and there are no other (positive) integers left. $\endgroup$ – Philipp Imhof Aug 4 '17 at 21:16
  • $\begingroup$ Also consider positive integers b and c with b<c. Multiplying both sides by b gives $b^2$<bc$ and multiplying by c yields $bc<c^2$. I would say these things are really too trivial to be explicitly stated in the proof, unless asked for -- then, of course, one should be able to justify that stuff. $\endgroup$ – Philipp Imhof Aug 4 '17 at 21:26
  • $\begingroup$ My point wasn't that I don't see them, my point is I think for the purpose of a proof at this level they will need to be justified. (Okay, $1< \sqrt{2}<2$ was a bit of a quibble.) Knowing that if $p$ divides $b$ and $b^2|a^2$ implies $p|a^2$ really is the point of this proof. prime divisors and relative prime really haven't been developed for this level of proof. They are simple and easy. But I think the need justification. $\endgroup$ – fleablood Aug 4 '17 at 21:36
  • $\begingroup$ it might be better to write it with more details: $\gcd(a,b)=1\iff\gcd(a^2,b^2)=1\implies b^2\not\mid\;a^2$ but $\frac{a^2}{b^2}=2\implies b^2=1\implies b=1\implies 2=a^2\implies\sqrt{2}=a\implies \sqrt{2}\in\mathbb{Z}$ but $1<2<4\iff \sqrt{1}<\sqrt{2}<\sqrt{4}\iff 1<\sqrt{2}<2\implies$ there is an integer between 1 and 2, that is the contradiction $\endgroup$ – Stu Aug 5 '17 at 0:44

$\sqrt{2}$ is not a rational number using $\mod 3$:

Suppose $\sqrt{2}$ is rational. note that $\sqrt{2}>1$, so w.l.o.g. I can take $a,b\in \mathbb{N}$ such that $\sqrt{2}=\frac{a}{b}$

Then $a^2=2b^2$. Note that $a^2\equiv 0,1(\mod 3)$

Case 1:

If $a^2\equiv 1(\mod 3)$, then $b^2\equiv 1(\mod 3)\Rightarrow 2b^2\equiv 2(\mod 3)$, but $a^2=2b^2$, then $a^2\equiv 2(\mod 3)\Rightarrow 1\equiv 2(\mod 3)$, which is impossible.

Case 2:

If $a^2\equiv 0(\mod 3)$, then $b^2\equiv 0(\mod 3)$, then $a=3c$, $b=3d$ where $c,d \in \mathbb{N}$.

Now we again get $c^2=2d^2$, since $9c^2=2\times9d^2$. This lead to a loop. The loop is we can not go to case 1(since that case is impossible). Then we have stay in case 2, which will lead to $3\mid c$ and $3\mid d$ and this process will continue until we end up to $1=2$ which is again impossible.

So there does not exist $a,b\in \mathbb{N}$ such that $a^2=2b^2$. Hence $\sqrt{2}$ is not a rational number$.\space\space\space\blacksquare$


The usual proof assumes from the beginning that $m$ and $n$ don't have a common divisor. Then finding one is a contradiction to that assumption.

However you can write the proof also without that assumption; it just gets a bit more complicated.

Assume there exist integers $m,n$ such that $\left(\frac mn\right)^2 = 2$, that is, $m^2=2n^2$. Now be $k$ the largest natural number such that $2^k|m$, and $l$ the largest natural number such that $2^l|n$. Then we can write $m=2^kM$ and $n=2^lN$ with $M$ and $N$ odd.

Therefore $m^2 = 2^{2k} M^2$ and $n^2 = 2^{2l} N^2$, where of course $M^2$ and $N^2$, as products of odd numbers, are also odd.

Thus the claim is that $2^{2k}M^2 = 2\cdot 2^{2l} N^2 = 2^{2l+1} N^2$. Since $M^2$ and $N^2$ are both odd, this implies $2^{2k} = 2^{2l+1}$, which is equivalent to $2k = 2l+1$. But $2k$ is even and $2l+1$ is odd, therefore they cannot be equal.

  • $\begingroup$ At last phase how you have cancelled $M^2$ and $N^2$ in both sides. Please tell me. $\endgroup$ – NewBornMATH Dec 11 '18 at 14:10
  • $\begingroup$ I have taken that in unique factorisation method. And equating the exponents of 2 of both sides. Did you mean this or anything else ? Please let me know. $\endgroup$ – NewBornMATH Dec 11 '18 at 15:31
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    $\begingroup$ @NewBornMATH: Since $M$ and $N$ are both odd, by definition they don't contain any factor $2$. So all the factors $2$ are those explicitly written. And yes, you need the unique factorization property of the integers to argue that the exponents have to be equal. $\endgroup$ – celtschk Dec 13 '18 at 20:50

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