# Proof using method of contradiction. Use the method of contradiction to prove that √2 is irrational. [duplicate]

Use the method of contradiction to prove that √2 is irrational. I don't understand how to prove that √2 is irrational using this method. And I feel difficult to form the contradiction.

• Write the steps of the proof and highlight what you're having difficulty with. – Deepak Aug 26 '19 at 4:42
• The contradiction would be: $\sqrt 2$ is rational. Or in other words that there is a rational number $q$ so that $q^2 = 2$. Or in other words there are two integers $m$ and $n$ in lowest terms so that $2 =\frac {m^2}{n^2}$. I'll give you a boost to get started: That would mean $2n^2 = m^2$. Can you get a contradiction from that? Second hint: Can you determine whether $m$ and/or $n$ are even or odd? Or not? – fleablood Aug 26 '19 at 6:01

If you want to proof $$A\Rightarrow B$$, a proof by contradiction reads as follows:

Suppose $$B$$ is false, then $$A$$ has to be false.

So suppose $$\sqrt{2}$$ is rational, then $$\sqrt{2}=\frac{p}{q}$$ for $$p\in\mathbb{Z}$$, $$q\in\mathbb{N}$$ where $$q\neq 0$$.

Without loss of generality we can assume that the fraction $$\frac{p}q$$ is completly reduced.That means $$\operatorname{gcd}(p,q)=1$$ (greatest common divisor) This is our assumption we take to the contradiction! We can assume that, because if the fraction is not completly reduced, we can reduce it. :)

Now it is a little bit of equivalent manipulations:

$$\sqrt{2}=\frac{p}{q}\Leftrightarrow 2=\frac{p^2}{q^2}\Leftrightarrow 2q^2=p^2$$.

From this equality we get that $$2$$ divides $$p^2$$. Since $$2$$ is prime we have $$2$$ divides $$p$$.

[Indeed: $$2\mid n^2$$ iff $$2\mid n$$.

$$\Leftarrow$$:

When $$2\mid n$$ then $$n=2m$$ for some $$m\in\mathbb{Z}$$. Then $$n^2=(2m)^2=4m^2=2\cdot 2m^2$$ so $$2\mid n^2$$.

$$\Rightarrow$$:

Let $$2\mid n^2$$. Since $$2$$ is prime 2 has to divide a factor of $$n^2=n\cdot n$$. So $$2\mid n$$.

]

From this little lemma we can conclude that $$2\mid p$$. So we can write $$p=2l$$ for some $$l\in\mathbb{N}$$.

Then $$2q^2=p^2\Leftrightarrow 2q^2=4l^2\Leftrightarrow q^2=2l^2$$. But now we can conclude from this equation that $$2\mid q^2$$ so $$2\mid q$$. Which is a contradiction, because $$\operatorname{gcd}(p,q)=1$$, but we just showed that $$p$$ and $$q$$ both contain the factor $$2$$!