Question

prove that underroot 3 is irrational ...

Answer 1

suppose 3–√3 is rational, then 3–√=ab3=ab for some (a,b)(a,b) suppose we have a/ba/b in simplest form.

3–√a2=ab=3b23=aba2=3b2

if b is even, then a is also even in which case a/b is not in simplest form.
if b is odd then a is also odd. Therefore:

ab(2n+1)24n2+4n+12n2+2n2(n2+n)=2n+1=2m+1=3(2m+1)2=12m2+12m+3=6m2+6m+1=2(3m2+3m)+1a=2n+1b=2m+1(2n+1)2=3(2m+1)24n2+4n+1=12m2+12m+32n2+2n=6m2+6m+12(n2+n)=2(3m2+3m)+1

Since (n^2+n) is an integer, the left hand side is even. Since (3m^2+3m) is an integer, the right hand side is odd and we have found a contradiction, therefore our hypothesis is false.