3 number questions...?

Question

1) Prove that 3^(1/2) is a real number.
2) Prove that 3^(1/2), 5^(1/2), 6^(1/2) are irrational. Evaluate [2^(1/2)+3^(1/2)]^2 and deduce that 2^(1/2)+3^(1/2) is irrational.
3) Prove that p^(1/2) is irrational for any prime number p. Hence prove that n^(1/2) is irrational unless n is a perfect square. Hence show that n^(1/2)+m^(1/2) is rational if and only if both n^(1/2)+m^(1/2) are rational.

Answer 1

Use this theorem: Suppose
c_nx^n + c_(n-1)x^(n-1) + ... + c_1x + c_0 = 0 is a polynomial equation with integer coefficients. If this equation has a rational root a/b (where a and b are integers), then a divides c_0 and b divides c_n.

For 2) consider the equation x^2 - 3 = 0. Clearly +/-3^(1/2) are solutions. But if a/b is a rational root, then a divides 3 and b divides 1. Thus, the only possible rational roots are +/-1 and +/- 3, and none of these works. Now the discriminant of the quadratic is 12, so the equation has two real and distinct roots (which we know are +/-3^(1/2)). Real roots which are not rational are irrational.

Use similar arguments for the other parts of the problem.

Answer 2

Another way :

Suppose that 3^(1/2) is rationnal : so, we can write 3^(1/2) = a/b with a and b integer numbers and relatively prime.

then a²/b² = 3 or a² = 3b²

then 3 divides a² , so 3 divides a
we can write a = 3a' and a² = 9a'²
3b² = 9a'²
b² = 3a'2
3 divides b² , 3 divides b
then a and b are not relatively prime : it's impossible
so we cannot write 3^(1/2) = a/b

It's the same way for 5, 6, ... (but not 4 : no contradiction because if 4 divides a², then 2 (and not 4) divides a) and for a prime number p.