Trying to understand example of non-rational number

[_nash]

In a video https://www.youtube.com/watch?v=isbt-7DQBy0 the instructor gives an example of a non-rational number, with the example of the square root of 2 = a/b.



He says:

If the square root of 2 would be a rational number, it could be expressed as a ratio of two integers, like a/b

If we square both sides of the equation, we get

2 = a^2/b^2 This can be rewritten as
2b^2 = a^2

So he reaches the conclusion that a is even, because it's 2 times some integer b.
He proceeds to say that a^2 = 2(2k^2) - this is the red text in the image. I don't understand how that step is happening, why is a^2 equal to 2(2k^2) ?

 

[HallsofIvy]

First show that "a number is even if and only if a^2 is even". The "only if" part is easy- if a is even then a= 2k for some integer k so a^2= (2k)^2= 4k^2= 2(2k). That last step is true because 4= 2 times 2!

Since a^2 is 2 times an integer (2k), a^2 is even.

To prove the "if" part, if a^2 is even then a is even, use "proof by contradiction". If a is not even then it is odd and can be written in the form 2k+ 1, two times an integer plus 1. Then a^2= (2k+1)^2= 4k^2+ 4k+ 1= 2(2k^2+ 2k)+ 1. That is two times an integer plus 1 so is odd, not even.

 

[JeffM]

In a video https://www.youtube.com/watch?v=isbt-7DQBy0 the instructor gives an example of a non-rational number, with the example of the square root of 2 = a/b.

View attachment 7777

He says:

If the square root of 2 would be a rational number, it could be expressed as a ratio of two integers, like a/b

If we square both sides of the equation, we get

2 = a^2/b^2 This can be rewritten as
2b^2 = a^2

So he reaches the conclusion that a is even, because it's 2 times some integer b.
He proceeds to say that a^2 = 2(2k^2) - this is the red text in the image. I don't understand how that step is happening, why is a^2 equal to 2(2k^2) ?

The video is sloppy with respect to variables and order . When he gets to the red part, he has not introduced k as a variable. All he has shown is that

\(\displaystyle a^2 = 2b^2\). Not that \(\displaystyle b^2 = 2k^2\).

He needs a lemma and should start there. I like Hall's proof of the needed lemma.

\(\displaystyle a \in \mathbb Z\ and\ a^2\ is\ even \implies a^2\ is\ not\ odd.\)

\(\displaystyle ASSUME\ a\ is\ odd \implies \exists\ n \in \mathbb Z^+\ such\ that\ a = 2n - 1 \implies\)

\(\displaystyle a^2 = 4n^2 - 4n + 1 = 2(2n^2 - 2n) + 1 \implies a^2\ is\ odd.\)

CONTRADICTION. \(\displaystyle \therefore a^2\ even \implies a\ even.\)

Now follow the video and ignore the red part.