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

Question

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.

This contradicts our second assumption.

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?

Answer 1

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 $a,b$.

Answer 2

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.

Answer 3

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.

Answer 4

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.