prove that 3/2root5 is irrational

Question

anonymous · Answer

please give urgently

parthkohli · Answer

Suppose that $\large \frac{3}{2\sqrt5}$ is rational. Then$$\dfrac{3}{2\sqrt5} = \dfrac{p}{q} \tag{p,q are coprime integers, q is not 0 }$$$$\dfrac{9}{10} = \dfrac{p^2}{q^2}$$$$9q^2 = 10p^2$$$$p^2 = \dfrac{9}{10} q^2$$So $q^2$ is divisible by 10, meaning that $q$ is a also divisible by 10. Let $q = 10k$.$$p^2 = \dfrac{9}{10}100k^2$$$$p^2 = 90k^2 $$$$\dfrac{p^2}{10} = 9k^2$$So $p^2$ is divisible by 10, meaning that $p$ is also divisible by 10. Hence, p and q are both divisible by 10. But this is a contradiction to our original assumption that p,q are coprime. Hence proved.

anonymous · Answer

@ParthKohli  I think 2 root 5 whole squared is 4x5 = 20?