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

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:

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

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

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

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

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

Answer 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.