- Proof that the √5 is irrational:

This will be a proof by contradiction.

- Assume that the √5 is rational (can be expressed in the form , where a and b are integers.
- Then √5 can be written as an irreducible fraction (the fraction is reduced as much as possible)
*a*/*b*such that*a*and*b*are coprime integers and (*a*/*b*)^{2}= 5. - It follows that
*a*^{2}/*b*^{2}= 5 and*a*^{2}= 5*b*^{2}. - Therefore
*a*^{2}is divisible by 5 because it is equal to 5*b*^{2}which is obviously divisible by 5. - It follows that
*a*must be divisible by 5. - Because
*a*is divisible by 5, there exists an integer*k*that fulfills:*a*= 5*k*. - We insert the last equation of (3) in (6): 5
*b*^{2}= (5*k*)^{2}is equivalent to 5*b*^{2}= 25*k*^{2}is equivalent to*b*^{2}= 5*k*^{2}. - Because 5
*k*^{2}is divisible by 5 it follows that*b*^{2}is also divisible by 5 which means that*b*is divisible by 5 because only numbers divisible by 5 have squares divisible by 5. - By (5) and (8)
*a*and*b*are both divisible by 5, which contradicts that*a*/*b*is irreducible as stated in (2).

Therefore, √5 is irrational.