Question

prove the real number \[\sqrt{2}\] is irrational

Answer 1

Assume sqrt(2) is NOT irrational. So assume sqrt(2) is rational

This would mean

sqrt(2) = p/q

where p/q is a fully reduced fraction, and p and q are integers. So p and q have no common factors (other than 1)

Square both sides to get

2 = (p^2)/(q^2)

2q^2 = p^2

p^2 = 2q^2

So this shows us that p^2 is even.

So q^2 must be even for (p^2)/(q^2) to be even.

Answer 2

Taking the square root of each, you can prove (either here or elsewhere) that p and q must be even.


But this is a contradiction. We stated that p/q was fully reduced and p,q had no factors in common. This is contradicted by the fact that p,q are both even (so they at least have 2 in common)

Answer 3

thank you, but why would we have to show that they are even?

Answer 4

so because of this contradiction, the assumption "sqrt(2) is NOT irrational" is false

so the opposite must be true and the statement "sqrt(2) is irrational" is true

Answer 5

i see.... contradiction!

Answer 6

yes you basically go from saying p/q is fully reduced (p,q have nothing in common) to saying p/q can be reduced (since both have 2 in common)....which is a contradiction

Answer 7

wow. u r the best!!
i will have this on a test today for sure, so ty for your help

Answer 8

you would/might have to prove that squaring an even number leads to an even number, so you might have to take that backwards to say taking the square root of an even number leads to an even number

Answer 9

got it :)