prove 7√5 irrational

Assume that √5 is rational, so it has of the form √5=a/b, where a and b are coprimes (both numbers have only common factor as1) √5 =a/b square both sides then 5=a^2/b^2 ie a^2 =5b^2 let it be eq1 a^2 /5=b^2 now a^2is divisible by 5 therefore a is divisible by 5 now take a=5c from eq1 we get 25c^2 =5b^2 ,5c^2 =b^2 Now c^2 =b^2/5 now b^2 is divisible by 5 by theorem 1.3 a and b have a common factor as 5 so our assumption is wrong √5 is irrational Now we have to prove 7√5 is irrational we already proved √5 is irrational Now assume 7√5 is irrational therefore 7√5 =a/b we know √5 is irrational so 7√5=a/b √5=a/7b where √5 is irrational and a/7b is rational So it is contradiction to our assumption therefore our assumption is wrong 7√5 is irrational

Assume that √5 is rational, so it has of the form √5=a/b, where a and b are coprimes (both numbers have only common factor as1) √5 =a/b square both sides then 5=a^2/b^2 ie a^2 =5b^2 let it be eq1 a^2 /5=b^2 now a^2is divisible by 5 therefore a is divisible by 5 now take a=5c from eq1 we get 25c^2 =5b^2 ,5c^2 =b^2 Now c^2 =b^2/5 now b^2 is divisible by 5 by theorem 1.3 a and b have a common factor as 5 so our assumption is wrong √5 is irrational Now we have to prove 7√5 is irrational we already proved √5 is irrational Now assume 7√5 is irrational therefore 7√5 =a/b we know √5 is irrational so 7√5=a/b √5=a/7b where √5 is irrational and a/7b is rational So it is contradiction to our assumption therefore our assumption is wrong 7√5 is irrational