Answer:
Step-by-step explanation:
Hello,
Basically, we need to prove that [tex]\sqrt{2}[/tex] is irrational
Let s assume that [tex]\sqrt{2}[/tex] is rational
it means that we can find p and q (q different from 0) two integers with no common factors other than 1
so that
[tex]\sqrt{2}=\dfrac{p}{q}[/tex]
And then we can write that
[tex]2=\dfrac{p^2}{q^2}\\<=> p^2=2q^2[/tex]
So [tex]p^2[/tex] is even so it means that p is even
so [tex]p^2[/tex] is divisible by 2*2=4
as [tex]p^2=2q^2[/tex] it means that [tex]q^2[/tex] is even, meaning q is even
wait, p and q are then even !? but by definition they have no common factors. This is not possible.
so our assumption that [tex]\sqrt{2}[/tex] is rational is false
So it means that this is irrational
and then [tex]3-\sqrt{2}[/tex] is irrational too
Hope this helps
