We prove by contradiction method.Let assume that √5 is rational.√5=p/q —eq……1P and q are co prime integer with common factor 1.Squaring both side equation 15=p²/q²:. q²=p²/5 eq…2It means p² have factor 5 so p would also have sector 5.p=5xsquaring both sidep²=25x²Puting this value of p² in equation 2q²=25x²/5q²=5x²q²/5=x²This means q² have factor 5.But our assumption says p and q have common factor 1.So our assumption was wrong√5 is irrational number.
yes √ 5 is an irrational no.
Let\xa05\u200b\xa0be a rational number.then it must be in form of qp\u200b where, q\ue020=0 (\xa0p\xa0and\xa0q\xa0are co-prime)5\u200b=qp\u200b5\u200b×q=pSuaring on both sides,5q2=p2 ————–(1)p2\xa0is divisible by\xa05.So,\xa0p\xa0is divisible by\xa05.p=5cSuaring on both sides,p2=25c2 ————–(2)Put\xa0p2\xa0in eqn.(1)5q2=25(c)2q2=5c2So,\xa0q\xa0is divisible by\xa05..Thus\xa0p\xa0and\xa0q\xa0have a common factor of\xa05.So, there is a contradiction as per our assumption.We have assumed\xa0p\xa0and\xa0q\xa0are co-prime but here they a common factor of\xa05.The above statement contradicts our assumption.Therefore,\xa05\u200b\xa0is an irrational number.
√5 =p/q
Let us assume that√5 is rational