# How can you prove that the square root of two is irrational?

I have read a few proofs that $\sqrt{2}$ is irrational.

I have never, however, been able to really grasp what they were talking about.

Is there a simplified proof that $\sqrt{2}$ is irrational?

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You use a proof by contradiction. Basically, you suppose that $\sqrt{2}$ can be written as $p/q$. Then you know that $2q^2 = p^2$. However, both $q^2$ and $p^2$ have an even number of factors of two, so $2q^2$ has an odd number of factors of 2, which means it can’t be equal to $p^2$.

Another method is to use continued fractions (which was used in one of the first proofs irrationality of $\displaystyle \pi$).

Instead of $\displaystyle \sqrt{2}$, we will consider $\displaystyle 1 + \sqrt{2}$.

Now $\displaystyle v = 1 + \sqrt{2}$ satisfies

$$v^2 – 2v – 1 = 0$$

i.e

$$v = 2 + \frac{1}{v}$$

This leads us to the following continued fraction representation

$$1 + \sqrt{2} = 2 + \cfrac{1}{2 + \cfrac{1}{2 + \dots}}$$

Any number with an infinite simple continued fraction is irrational and any number with a finite simple continued fraction is rational and has at most two such simple continued fraction representations.

Thus it follows that $\displaystyle 1 + \sqrt{2}$ is irrational, and so $\displaystyle \sqrt{2}$ is irrational.

Exercise: Show that the Golden Ratio is irrational.

Assume that $\sqrt{2}$ is rational. Then there exists some rational $R=\sqrt{2}=\frac{Q}{D}$, where $Q$ and $D$ are positive integers and relatively prime (since $R$ can be expressed in simplified form).

Now consider $R^2 = 2 = \frac{Q^2}{D^2}$. Since $Q$ and $D$ are relatively prime, this means that only $Q^2$ can have $2$ in its prime decomposition, and the exponent must be one. Thus, $Q^2 = 2^1 x$, for some odd integer $x$. But $Q^2$ is a square, and thus the exponents for all of its prime factors must be even. Here we have a contradiction.

Thus, $\sqrt{2}$ must be irrational.

If $\sqrt 2$ were rational, we could write it as a fraction $a/b$ in
lowest terms. Then
$$a^2 = 2 b^2.$$
Look at the last digit of $a^2$. It has to be $0$, $1$, $4$, $5$, $6$ or $9$.
Now look at the last digit of $2b^2$. It has to be $0$, $2$ or $8$.
As $a^2$ and $2b^2$ are the same number, its last digit must be $0$.
But that’s only possible if $a$ ends in $0$ and $b$ ends in $0$ or $5$.
Either way both $a$ and $b$ are multiples of $5$ contradicting $a/b$
being in lowest terms.

The continued fraction proof in Aryabhata’s answer can be recast into an elementary form that requires no knowledge of continued fractions. Below is a variant of such that John Conway (JHC) often mentions, followed by my (WGD) reinterpretation of it to highlight the key role played by the principality of (denominator) ideals in $\:\mathbb Z\:$ (which I call unique fractionization).

THEOREM (JHC) $\quad \rm r = \sqrt{n}\ \:$ is integral if rational,$\:$ for $\:\rm n\in\mathbb{N}$

Proof $\ \ \$ Put $\ \ \displaystyle\rm r = \frac{A}B ,\;$ least $\rm\; B>0\:.\;$ $\ \displaystyle\rm\sqrt{n}\; = \frac{n}{\sqrt{n}} \ \Rightarrow\ \frac{A}B = \frac{nB}A.\ \:$ Taking fractional parts yields $\rm\displaystyle\ \frac{b}B = \frac{a}A\$ for $\rm\ 0 \le b < B\:.\$ But $\rm\displaystyle\ B\nmid A\ \Rightarrow\:\ b\ne 0\ \:\Rightarrow\ \frac{A}B = \frac{a}b\$ contra $\rm B$ least. $\:$ QED

Abstracting out the Euclidean descent at the heart of the above proof yields the following

THEOREM (WGD) $\quad \rm r = \sqrt{n}\ \:$ is integral if rational,$\:$ for $\:\rm n\in\mathbb{N}$

Proof $\ \$ Put $\ \ \displaystyle\rm r = \frac{A}B ,\;$ least $\rm\; B>0\:.\;$ $\ \displaystyle\rm\sqrt{n}\; = \frac{n}{\sqrt{n}} \ \Rightarrow\ \frac{A}B = \frac{nB}A\ \Rightarrow\ B\:|\:A\$ by this key result:

Unique Fractionization $\$ The least denominator $\rm\:B\:$ of a fraction divides every denominator.

Proof $\rm\displaystyle\ \ \frac{A}B = \frac{C}D\ \Rightarrow\ \frac{D}B = \frac{C}A \:.\$ Taking fractional parts $\rm\displaystyle\ \frac{b}B = \frac{a}A\$ where $\rm\ 0 \le b < B\:.\$ But

$\rm\displaystyle\ \:B\nmid D\ \Rightarrow\ b\ne 0\ \Rightarrow\ \frac{A}B = \frac{a}b\ \$ contra leastness of $\rm\:B.\,$ Thus $\rm\,B\mid D\,$ as claimed $\quad$ QED

Thus JHC’s proof essentially “inlines” the above proof – which can be more conceptually viewed as the principality of (denominator) ideals in $\mathbb Z,\,$ cf. my post here. See also this sci.math discussion between John Conway and I (click “plain text” to get correct format).

You can also use the rational root test on the polynomial equation $x^2-2=0$ (whose solutions are $\pm \sqrt{2}$). If this equation were to have a rational solution $\frac{a}{b}$, then $a \vert 2$ and $b \vert 1$, hence $\frac{a}{b}\in \{\pm 1, \pm 2\}$. However, it’s straightforward to check that none of $1,-1,2,-2$ satisfy the equation $x^2-2=0$. Therefore the equation has no rational roots and $\sqrt{2}$ is irrational.

There is also a proof of this theorem that uses the well ordering property of the set of positive integers, that is in a non empty set of positive integers there is always a least element. The proof follows the approach of proof by contradiction but uses the well ordering principle to find the contradiction 🙂 –

Let us assume $\sqrt{2}$ is rational, hence it can be written down in
the form $\sqrt{2}=a/b$ assuming that both $a$ and $b$ are positive
integers in that case if we look at the set $S = \{k\sqrt{2} \mid k, k\sqrt{2}\text{ are integers}\}$ we find that it’s a non empty set of
positive integers, it’s non empty because $a = b\sqrt{2}$ is in the
above set. Now using the Well ordering principle we know that every
set of positive integers which is non-empty has a minimum element, we
assume that smallest element to be $s$ and let it equal to $s =t\sqrt{2}$. Now an interesting thing happens if we take the difference
between the following quantities $s\sqrt{2} – s = (s-t)\sqrt{2} = s(\sqrt{2} – 1)$ which is a smaller element than $s$ itself, hence
contradicting the very existence of $s$ being the smallest element.
Hence we find that $\sqrt{2}$ is irrational.

I know the proof but I am still amazed at how the author came up with the set assumption. Sometimes such assumptions make you feel kinda dumb :). If anyone has some insight regarding how to come up with such assumptions kindly post your answer in the comment, otherwise I would just assume that it was a workaround.

Another one that is understandable by high schoolers and below.

We will use the following lemma:

If $n$ is an integer, $n^2$ is even (resp. odd) iff $n$ is even (resp. odd).

For the high-schoolers, the proof is about writing $(2k)^2 = 2(2k^2)$ and $(2k+1)^2=2(2k^2+2k)+1$ …

Now, assume $\sqrt 2 = \frac{a}{b}$ with $a$ and $b$ strictly positive integers.

Then $a^2=2b^2$, $\implies a^2$ is even ($=2b^2$), $\implies a$ is even (from the lemma), $\implies a=2a_1$ with $a_1 \in \mathbb N^*$, $\implies b^2=2a_1^2$.

Repeat with $b$ to find that $b=2b_1$ with $b_1 \in \mathbb N^*$ and $(a_1,b_1)$ verifies $a_1^2=2b_1^2$.

By repeating these two steps, we build two sequences $(a_n)_{n\in \mathbb N}$ and $(b_n)_{n\in \mathbb N}$ with values in $\mathbb N^*$ and strictly decreasing, which is impossible, ergo $\sqrt{2}$ is irrational.

(Here of course we use the well-ordering principle which most high schoolders would not know about, but the intuition that the sequence would hit $0$ after at most $a_0=a$ steps is easy to get).

Let $x^2-2=0$ be the polynomial equations this have a possibles rational solutions $\pm1,\pm2$ and no one of this is a solution then the solution is irrational and we now that this are $\pm \sqrt2$

Here are some of my favorite (sketches) of proofs for the irrationality of $\sqrt{2}$.

• Using Newton’s method to approximate roots of the polynomial $f(x) = x^2 – 2$, then showing that the sequence does not converge to a rational number.
• Proof by contradiction, assume that $\sqrt{2} = \frac{n}{m}$ for some $n,m \in \mathbb{Z}$ with $m \neq 0$, then $2m^{2} = n^2$, hence $n$ must be even and we can let $n = 2k$ for some $k \in \mathbb{Z}$, but then $m^2 = 2k^2$ will also be even, which is impossible if $\frac{n}{m}$ is reduced. Therefore, $\sqrt{2}$ cannot be expressed as a ratio of integers.
• Since $f(x) = x^2 -2$ is irreducible over $\mathbb{Q}[x]$, its roots must lie in some finite extension field $\mathbb{Q}(\sqrt{2})$ over the rationals.

[Reposted from closed topicProve the square root of 2 is irrational

The irrationality of the square root of 2 follows from our knowledge of how Pythagorean triples behave, specifically, that for positive integers x, y, and z, if x^2 + y^2 = z^2, then x is not equal to y. But if the square root of 2 were rational, then there would exist positive integers a and b such that a/b = the square root of 2. Then a^2/b^2 = 2. Then a^2 = 2b^2. Then b^2 + b^2 = a^2, and so we would have a Pythagorean triple with x = y, a contradiction.