Assume that √2 is a rational number, meaning that there exists a pair of integers whose ratio is exactly √2.
If the two integers have a common factor, it can be eliminated using the Euclidean algorithm.
Then √2 can be written as an irreducible fraction
a
/
b
such that a and b are coprime integers (having no common factor) which additionally means that at least one of a or b must be odd .
It follows that
a2
/
b2
= 2 and a2 = 2b2.   ( (
a
/
b
)n =
an
/
bn
 )   ( a2 and b2 are integers)
Therefore, a2 is even because it is equal to 2b2. (2b2 is necessarily even because it is 2 times another whole number and multiples of 2 are even.)
It follows that a must be even (as squares of odd integers are never even).
Because a is even, there exists an integer k that fulfills: a = 2k.
Substituting 2k from step 7 for a in the second equation of step 4: 2b2 = (2k)2 is equivalent to 2b2 = 4k2, which is equivalent to b2 = 2k2.
Because 2k2 is divisible by two and therefore even, and because 2k2 = b2, it follows that b2 is also even which means that b is even.
By steps 5 and 8 a and b are both even, which contradicts that
a
/
b
is irreducible as stated in step 3.
Q.E.D.