# Proof that $\sqrt{2}$ is irrational is not convincing. Please help.

I understand the irrationality of $$\sqrt{2}$$ in the following way:

To prove: $$\sqrt{2}$$ is irrational

Proof: Assume $$\sqrt{2}$$ is rational.

i.e. $$\sqrt{2}=\dfrac{a}{b}$$

Assume $$a$$ and $$b$$ are co-prime

...... (the usual steps)

Hence $$a$$ and $$b$$ cannot be co-prime.

So first assumption is wrong.

So $$\sqrt{2}$$ is irrational.

MY CONFUSION:

We are making two different assumptions. This is not the way proof by contradiction works. If the second assumption gets contradicted, for what reason will the first assumption be false?

• Any rational number can be written in the form $\frac ab$ with $\gcd(a,b)=1$. The second "assumption" really just directs us to select that form. – lulu Jul 20 '19 at 11:54
• There is no second assumption if you want. Just point out the fractions you'll use are always given in reduced form... – DonAntonio Jul 20 '19 at 11:56

The second assumption can be avoided if it bothers you. You can say:

Suppose $$\sqrt 2=\frac{a}{b}$$ ($$a,b$$ not necessarily co-prime).

Let $$c=\frac{a}{(a,b)}$$ and $$d=\frac{b}{(a,b)}$$, where $$(a,b)$$ denotes the highest common factor of $$a$$ and $$b$$.

Then $$\sqrt 2=\frac{c}{d}$$, and $$c$$ and $$d$$ are co-prime.

Now proceed with the proof as above, with $$c,d$$ in place of $$a,b$$.

The first assumption implies the second. If a number is rational, then we can write it as $$\frac ab$$ with $$a,b$$ coprime. Thus if we cannot write a number as $$\frac ab$$ with $$a,b$$ coprime, then it is not rational.

This works since if $$q$$ is assumed to be rational, then it's always true that it can be expressed as a fraction of coprime integers. Thus if you disprove this, it means that $$q$$ is not rational.

Well, you can always assume that in the quotient of integers, $$q=a/b$$, numerator and denominator are coprime.

The proof makes use of the main theorem of arithmetic that each positive integer can be uniquely written as a product of prime powers. This will lead of a contradiction:

If $$\sqrt 2 = \frac{a}{b}$$ then $$2 =\frac{a^2}{b^2}$$, i.e., $$2b^2 = a^2$$. Now think about $$2$$ as a prime factor on the left-hand and the right-hand side and you will have the contradiction.