I watched a video https://www.youtube.com/watch?v=isbt-7DQBy0 in which the instructor gives an example of a non-rational number.

He says, if the square root of 2 would be a rational number, it could be expressed as the fraction of two integers:

${\sqrt 2} = \frac a b$

Which can be rewritten to

$2 = \frac{a^2}{b^2}$

$2{b^2} = {a^2}$

From which he concludes that ${a^2}$ is even, because it would be two times some integer.

He then goes on to describe why a is even, using this:

${a^2} = 2(2{k^2})$

$a \cdot a = (2k) \cdot (2k) $

$a = 2k$

I don't understand the first step in that sequence, ${a^2} = 2(2{k^2})$. Where to this 2k come from? 2k symbolizes any even integer? Couldn't i just insert $2k + 1$ for any odd integer and solve it the same? I don't understand how this is supposed to prove that a is even.

(The proof in the video goes on to conclude that ${\sqrt 2}$ can not be a rational number, but what I described above is the part I'm struggling with)

  • $\begingroup$ The argument you're giving here seems to assume that $b$ is even, which is not justified. Just observe that an integer is even iff its square is even. $\endgroup$
    – pwerth
    Mar 7, 2017 at 6:34
  • $\begingroup$ @pwerth I don't really understand what you mean. I'm justr trying to follow what that guy did in the video, which I can't. $\endgroup$
    – Max
    Mar 7, 2017 at 6:37

2 Answers 2


It has to do with unique prime factorization, which is what Dan was doing above.

Suppose $\sqrt{2} = a/b$

It is vital at this point to clarify that $a/b$ is reduced or simplified to lowest possible terms. If $a$ and $b$ have a common factor, it must be divided out, as for example 3/6 = 1/2 dividing by 3/3, or 42/56 = 6/8 = 3/4 dividing first by 7/7 then by 2/2.

In number theory when there are no more common factors the numbers are called mutually prime. In particular they can't both be even, because then they'd have a common factor of 2.

So now $2 = a^2/b^2$ and $2b^2 = a^2$

This means $a^2$ is divisible by 2, an even number.

By the theorem of unique prime factorization, we say any integer $a$ has prime factors $p_1^{r_1} p_2^{r_2} . . p_n^{r_n}$ and if we put the factors in increasing order there is only one way to do this. For example $24 = 2^33^1$ and there is no other way to break it up.

So $ a^2 = (p_1^{r_1} p_2^{r_2} . . p_n^{r_n})^2 = p_1^{2r_1} p_2^{2r_2} . . p_n^{2r_n}$

We said $a^2$ is divisible by 2. 2 is a prime, the smallest prime, so 2 must be $p_1$.

When we squared $a$, we multiplied all the exponents on the primes by 2. The prime factor 2 must appear in $a^2$ at least twice -- that's how squares work.

Let $a = 2k$ so $a^2 = 2k * 2k$ or as your work above put it $a^2 = 2(2k^2$.

Then $2b^2 = 2(2k^2)$ and we can divide by 2 and get $b^2 = 2k^2$

That means $b^2$ has a factor of 2 and by the same argument as for a, $b$ must have a factor of 2.

This contradicts our hypothesis that $a/b$ was reduced to lowest terms.

This is a bit of a long piece of logic but it is vital in number theory and important in the history of math (Pythagoras and his school figured it out) so it is worth learning.

  • $\begingroup$ This is very interesting, thank you for taking the time to write it out. I will take some time and work through that. $\endgroup$
    – Max
    Mar 7, 2017 at 7:33
  • $\begingroup$ Maybe it is worth noting that this approach doesn't really require that the prime factorization is unique, only that it exists. On the other hand, by relying on uniqueness it's possible to dispense with the common factor business (see my post) which I think makes it a bit simpler. $\endgroup$ Mar 7, 2017 at 8:55
  • $\begingroup$ You assumed unique factorization in your lines 2, 3, and 4 when you equated the two factorizations and divided them out. Yes it is true, but without unique factorization you can't prove it's true. $\endgroup$
    – victoria
    Mar 10, 2017 at 3:20

I watched the video, I think you presented his argument accurately and I agree that this part of it does not make any sense. It is true that if $a^2$ is even then it must be a multiple of $4$ but he does not properly justify this -- he is apparently assuming that $b^2$ is even at this point, and then later indirectly using that assumption to prove the same fact, which is circular.

Here is a different way to reach a contradiction:

$2 \cdot b^2 = a^2$

$2 \cdot (2^{b_1} \cdot 3^{b_2} \cdot 5^{b_3} \dots)^2 = (2^{a_1} \cdot 3^{a_2} \cdot 5^{a_3} \dots)^2$

$2^{1+2 \cdot b_1} \cdot (3^{b_2} \cdot 5^{b_3} \dots)^2 = 2^{2 \cdot a_1} \cdot (3^{a_2} \cdot 5^{a_3} \dots)^2$

$1 + 2 \cdot b_1 = 2 \cdot a_1$

$a_1 = b_1 + \frac{1}{2}$

  • $\begingroup$ Ok, thanks for clarifying that for me. I could follow that video easily up until that point. Do you know of any other reference so that I can understand what he's trying to do (proof that e. g. the square root of 2 is not a rational number)? $\endgroup$
    – Max
    Mar 7, 2017 at 6:48
  • $\begingroup$ @Max If you're looking for the exact result that the author of the video slipped up on (i.e. that $a^2$ being even implies $a$ is even), then there are a plethora of resources that show this (there are a variety of answers in math.stackexchange.com/questions/403346/…). Most of the techniques used involve either a proof by contraposition (i.e. using the fact that $a\implies b$ is logically equivalence to $not(b) \implies not(a)$) or by using the fact that if $p$ is a prime number that divides $ab$, then either $p$ divides $a$ or $p$ divides $b$. $\endgroup$
    – benguin
    Mar 7, 2017 at 6:59
  • $\begingroup$ I accept this answer as it helped be getting stuck at the misleading terms shown in the video. I have to say that it's still a bit above my head as why you substitute $2 \cdot b^2$ with $2 \cdot (2^{b_1} \cdot 3^{b_2} \cdot 5^{b_3} \dots)^2$ $\endgroup$
    – Max
    Mar 7, 2017 at 7:10

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .