Question 485329
Assume that sqrt(2) is a rational number, i.e.


*[tex \LARGE \sqrt{2} = \frac{a}{b}] where a and b are integers and a/b is irreducible. Squaring both sides,


*[tex \LARGE 2 = \frac{a^2}{b^2}]


*[tex \LARGE 2b^2 = a^2] this implies that since the LHS is even, then the RHS is also even, and a is a multiple of 2. We can write 2k instead of a:


*[tex \LARGE 2b^2 = (2k)^2 = 4k^2 \Rightarrow b^2 = 2k^2]


Similarly, we can write b as 2m for some integer m. This contradicts our original statements, because this would imply a = 2k, b = 2m and they are not in simplest form (plus, we can apply this technique infinitely many times -- also not good). Hence we have a contradiction and sqrt(2) is irrational.