If d 1 is a squarefree natural number ie if p d is prime t

If d > 1 is a square-free natural number (i.e. if p \\ d is prime, then p2 \\ d), then prove that squareroot d is irrational. If d notelement {0, plusminus 1} is a cube-free integer, then prove that d^1/3 irrational

Solution

1)

Assume sqrt{d} is rational

So, sqrt{d}=p/q , p, q are integers, gcd(p,q)=1

Squaring gives

d=p^2/q^2

d is a natural number and gcd(p,q)=1

So, q=1

So, d=p^2

Hence, d is not square free which is a contradiction

Hence, sqrt{d} is irrational

2)

Assume d^{1/3} is rational

So, d^{1/3}=p/q

d=p^3/q^3

d is integer so q=1

Hence, d=p^3 , where p is integer

So, d is not cube free

Hence a contradiction

So, d^{1/3} is irrational