# 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. Commented Nov 13, 2014 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
Commented Nov 13, 2014 at 16:50
• Every rational number can be written uniquely as a fraction in "lowest terms". Commented Nov 13, 2014 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. Commented Nov 13, 2014 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. Commented Nov 13, 2014 at 18:06

What is a rational number? The definition I use is that it's a number that can be written as a ratio of two integers. If your book doesn't say explicitly that the integers can be coprime, it's because the notion of a reduced fraction is fundamental to working with fractions. You're not allowed to work with fractions until you believe that they can be written in a unique reduced form, and that your numerator and denominator will be coprime.

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.

In the proof that you're trying, I don't see the need for contradiction. Statement X and Y imply each other. If you have a ratio of non-coprime numbers, reduce it. If you have a ratio of coprime numbers, multiply them both by 2 and now you have a ratio of non-coprime numbers. (The statements you have written are the contrapositive of this.)

• How X directly implies Y? Do you mean that X can be reduced to Y without any proof? Commented Nov 13, 2014 at 17:43
• Whether or not something requires proof depends on the reader. If you can agree with me that the contrapositive of a statement is logically equivalent to the statement, then in my last paragraph, I've proved that X implies Y. Commented Nov 14, 2014 at 17:51
• Actually I heard the term "contrapositive" for the first time. I will reply you soon once I read about it. Commented Nov 15, 2014 at 8:43
• Thanks a lot for introducing me the term contrapositive. Had I read about it before I would not have ask any question. The wikipedea page specifically answer what my questions is, en.wikipedia.org/wiki/Contraposition#Application Commented Nov 15, 2014 at 14:46
• Yes, that section is very relevant to your original question! Commented Nov 15, 2014 at 20:04

Let $$\sqrt2 = a/b$$ where $$a$$ and $$b$$ are not co-prime, that is they have a common divisor greater than or equal to $$2$$, so let $$d = gcd(a, b)$$, $$a’=a/d$$, $$b’=b/d$$. Now $$a’/b’ = a/b = \sqrt2$$, and $$a’$$, $$b’$$ are co-prime.