# Understanding the proof of “$\sqrt{2}$ is irrational” by contradiction.

I have some difficulties in understanding the proof of "$\sqrt{2}$is irrational" by contradiction. I am reading it in 10th class(in India) Mathematics book( available online, here )

This is the snapshot of it:  The proof starts with assuming that $\sqrt{2}$ can be written as a ratio of two integers and then that this fraction can be reduced to its lowest terms i.e. $\sqrt{2}=\dfrac ab$, where gcd(a,b)=1 . Then at last we reach at the contradiction that gcd$(a,b)\neq1$. Then they say that because of this contradiction $\sqrt{2}$ cannot be a rational number.

What I do not understand is that how the contradiction proves that $\sqrt{2}$ cannot be a rational number. The contradiction only proves that $\sqrt{2}$ cannot be written as the ratio of two coprime numbers. But can't we write $\sqrt{2}$ as the ratio of two non-coprime numbers?

Let us consider two statements, X and Y as:

X: $\sqrt{2}$ cannot be written as the ratio of two coprime numbers.

Y: $\sqrt{2}$ cannot be written as the ratio of two non-coprime numbers.

The contradiction proves only the statement X not the statement Y.

I guess that we can prove statement Y from X as: Let us suppose that $\sqrt{2}$ can be written as the ratio of two non-coprime numbers, i.e. $\sqrt{2}=\dfrac RS$, where $R$ and $S$ are mutually non-coprime. But every rational number can be written as a fraction in lowest terms. So let's say $\dfrac RS$ in its lowest terms is $\dfrac rs$, but this means that $\sqrt{2}$ is also equal to $\dfrac rs$, where $r$ and $s$ are coprime. This eventually contradicts the statement X, hence by contradiction $\sqrt{2}$ cannot be written as the ratio of two non-coprime numbers, or the statement Y is true.

Question:

1. Did I prove the statement Y from X correctly ?

2. Why does the book directly mention "$\sqrt{2}$ is irrational" without justifying statement Y? Is the justification too trivial to be mentioned?

3. Is there any other way than mine(proof by contradiction) to deduce Y from X?

I only want to clarify these three doubts, nothing else.

• Y can (trivially) be reduced to statement X. – Frank Vel Nov 13 '14 at 16:44
• possible duplicate of Query about Reductio Ad Absurdum. (The title does't suggest it, but that person's question is just like yours, and my answer to that question is just what I would say to you. Key point: “The claim that $a$ and $b$ have no common factor is not an assumption.”) – MJD Nov 13 '14 at 16:50
• Every rational number can be written uniquely as a fraction in "lowest terms". – Chazz Nov 13 '14 at 16:55
• @user31782 If it could be written as two non-coprime numbers, that would be equivalent to being possible to write it as two coprime numbers. Thus, it is enough to consider the case where $a$ and $b$ are coprime. – Frank Vel Nov 13 '14 at 18:00
• @user31782 Examples: $\frac{12}{16}$ can be reduced to $\frac{3}{4}$, and this is true for all non-coprime numbers. Thus if it can't be written as fraction of coprime numbers, it cannot be written as a fraction with non-coprime numbers. E.g. if it cannot be written as $\frac{1}{2}$ it cannot be written as $\frac{12}{16}$ nor any reducible fraction of the form $\frac{1\times k}{2\times k}$ for $k \in \mathbb{Z}$. And yes it is trivial, although nothing can ever be too trivial not to question. – Frank Vel Nov 13 '14 at 18:06

If you wanted to, you can prove this with the fundamental theorem of arithmetic. Suppose your rational number $q$ can be written as the ratio of integers $\frac{a}{b}$. Then by the fundamental theorem of arithmetic, $a$ and $b$ have unique factorizations into primes. So factorize them, and the factors that appear in both will cancel. Then you have a coprime numerator and denominator.