# Prove $1+\sqrt2$ is irrational

I am trying to prove $1+\sqrt 2$ is an irrational number.

Proof: assume that $1+\sqrt 2$ is a rational number such that $1+\sqrt 2=\frac{m}{n}$ where m and n are some integers. Then, $$1+\sqrt 2=\frac{m}{n}$$ $$\implies \sqrt 2 =\frac{m}{n}-1$$ $$\implies \sqrt 2=\frac{m-n}{n}$$ $$\implies \sqrt{2} n=m-n$$ $$\implies 2n^2=(m-n)^2$$

I get stuck at this step, can anyone give a hint or a suggestion? Thanks!

• At the third step of your work, you have $\sqrt{2}=\frac{m-n}n$. But $\frac{m-n}n \in \mathbb{Q}$ whereas $\sqrt{2} \not\in \mathbb{Q}$. This is your contradiction. Your proof should stop at that third step. Feb 28 '15 at 5:22

At the third step, there's already a contradiction (that is, if you've proven $\sqrt2$ is irrational already).

A striking way to prove such irrationality is via Euclid's gcd algorithm, which works for rationals. Namely, if $\,w=\sqrt{2}+1\in\Bbb Q\,$ then $\,w-2 =\sqrt{2}-1\in \Bbb Q\,$ therefore

\begin{align} d = \gcd(1,\sqrt{2}+1)\ &=\, (\sqrt{2}+1)\,\gcd(\sqrt{2}-1,\,1)\quad {\rm by}\ \ \gcd(ab,ac)\, =\, a\gcd(b,c)\\ &=\, (\sqrt{2}+1)\, \gcd(\sqrt{2}+1,\,1)\quad {\rm by}\ \ \gcd(a,b)\, =\, \gcd(a\!+\!2b,b)\\\end{align}\quad

hence the gcd $\,d\,$ satisfies $\, d = (\sqrt{2}+1)d,\,$ contradiction! Therefore $\,\sqrt{2}+1\not\in\Bbb Q$

• This is the first time I see this type proof for irrationality. Very neat. Aug 14 '15 at 5:23

If you can't use that $\sqrt 2$ is irrational, assume $m$ and $n$ are coprime and consider the last $\pmod 4$ They can't both be even, if one is odd the right side is odd, if both are odd the right is a multiple of $4$, but the right is $2 \pmod 4$

If $x = 1 + \sqrt{2} \to (x-1)^2 - 2 = 0 \to x^2 - 2x - 1 = 0$. An application of the well-known Eisenstein's rational root test shows this equation can't have a rational root, hence the conclusion.

• Eisenstein's test? He's done great stuff, but I would be surprised that this was not known by Gauss for instance. The WP page gives a "see also" link to the Schönemann–Eisenstein theorem, but for no apparent reason. Feb 28 '15 at 16:57

I suppose that you want to use this before having proved the well known fact that $\sqrt2$ is irrational (because it is obviously equivalent to what you ask: adding or subtracting the rational number $1$ from some rational number would give another rational number), so that you don't want to use that fact. And in order to be able to obtain an interesting alternative proof of the fact that $\sqrt2$ is irrational, you want to also avoid using unique factorisation of integers, or even just Euclid's lemma for the prime$~2$ (if a product of numbers is even, at least one of them is even). Then, even though your hands thus tied behind your back, you could still proceed as follows.

Assume as you did that $\sqrt2+1=\frac mn$ for positive integers $m,n$; if such integers exist one can take a pair where $m+n$ is minimal. Now since $\sqrt2>1$ one has $0<\sqrt2-1=\frac mn-\frac nn=\frac{m-2n}n$. Multiplying gives $$\frac mn\times\frac{m-2n}n = (\sqrt2+1)(\sqrt2-1)=(\sqrt2)^1-1^2=2-1=1.$$ Therefore $$\frac n{m-2n}=1\left/\frac {m-2n}n\right. =\frac mn,$$ But this contradicts minimality of $m+n$, since $n+(m-2n)=m-n<m+n$.

• Readers may find it instructive to compare this way of proof to the proof using gcds of rationals in my answer. They are essentially the same at the heart, but the use of fractions simplifies matters - as is often true. Feb 28 '15 at 17:08

Consider this, Prove that $\sqrt{2}$ is irrational.

Assume $\sqrt{2} = m/n$ then, suppose $m$ is odd, $n$ is even (without loss of generality), and $\gcd(m, n) = 1$ and $m, n$ are integers.

$$2n^2 = m^2$$

Since $m$ was odd, $m^2$ is odd, but since $n$ is even, $2n^2$ is also even. So $m$ is both odd an even, a contradiction.

Then, since $1$ is rational. Give a general proof.

If $x$ is irrational and $y$ is rational then $x + y$ is irrational. Proof:

$x = x + y - y = (x + y) - y$, which means $x$ is rational, contradictory.

Let $x = \sqrt{2}$ and let $y = 1$.

$$x + y = \sqrt{2} + 1$$ Is irrational.