Prove $1+\sqrt2$ is irrational I am trying to prove $1+\sqrt 2$ is an irrational number.
I start with contradiction 
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!
 A: At the third step, there's already a contradiction (that is, if you've proven $\sqrt2$ is irrational already).
A: 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 $\,\sqrt{2}+1\not\in\Bbb  Q$
A: 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 $2 \pmod 4$
A: 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.
A: 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 $n+(m-2n)=m-n<m+n$. 
A: 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. 
