Let's see that the nth root of 2 is irrational
With n greater or equal than 3. This is very important!
It's not because we have seen the sqrt of 2 case before in another video.
It's because the tool we are about to use, only allow us to use n greater or equal than 3.
We will use the usual strategy.
Suppose that the nth root of two is rational and lets look for the absurd.
We can write it as a/b with a and b integers.
So, we pass the root to the other side.
Then 2 equals a^n/b^n
We pass the b^n to the other side.
2b^n equals a^n
And we write it as b^n+b^n=a^n
Remember, a and b are Integers.
Positive integers. (Naturals)
And, now we are going to use Fermat's last Theorem.
This Theorem was conjectured by Pierre de Fermat in 1637
and proven by Andrew Wiles in 1995.
And it says:
If n is equal or greater than 3 (this is very important for our proof!)
then no three positive integers x, y, and z satisfy the equation x^n + y^n = z^n
That is to say that this equality can not hold.
We don't have b and a
(in this case, b is twice in the formula.)
that comply with the equation that we have there.
So this does not hold
this does not hold
this does not hold
this does not hold
And this too.
There we have the absurd.
The nth root of two is irrational.
With n greater or equal than 3.
We forgot to say that a and b are coprime.
(the fraction is simplified.)
This does not change the proof.
So we have an absurd
but the reason for this absurd
the reason of this n greater or equal than 3
the reason why we can't do the n=2 case
It's because Fermat's last theorem isn't strong enough.
To prove it.
That is to say that as Fermat's last thorem  leaves out the n=2 case
Implies that we can't use it to prove that square root of two is irrational.
