How do you prove that square root of 2 is irrational?

Answer 1

It is proved by contradiction (reductio ad absurdum), a powerful type of proof in mathematics.

Assume that the square root of 2 is rational and is equal to a/b where a and b are integers.

Squaring both sides gives:

2=a2/b2

a2=2b2

Since a2 is even, it implies that a is even

So, replacing a by 2k where k is an integer, we have:

(2k)2=2b2

b2=2k2

Since b2 is even, it implies that b is even, which is in contradiction to our first statement: a/b is in lowest terms.

Thus, the square root of 2 is irrational. (Q.E.D)

Note: It can be proved that if a2 is even, then a is also even.

Proof:

Odd numbers are of the form 2n+1, where n is an integer.

When an odd number is squared, we have:

(2n+1)2 = 4n2+4n+1 = 2(2n2+2n) +1 = 2y+1

where y = 2n2+2n

y is also an integer; so, 2y+1 is an odd number, which in turn means that the square of any odd number is also odd.

Therefore, if the square of a number is even, the number cannot be odd; it has to be even. (Q.E.D)