Help understanding this example of a non-rational number

Question

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)

Answer 1

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.

Answer 2

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:

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

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

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

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

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