Can someone give me a counter example to disprove this statement(:

"The irrational numbers are closed under multiplication"
Thank youu (:

This probably seems hard because you're thinking of the harder sort of irrational numbers, like pi or e, the base of natural logarithms, that don't satisfy any polynomial equations with integer (or even rational coefficients). Instead remember that if p is any prime, then the square root of p is irrational. (If not, if sqrt(p) = a/b, one can clear the denominator b*sqrt(p) = a, then square both sides p*b^2 = a^2. Count the number of factors of p in the prime factorization of each side: on the rhs, the number must be even, twice the number of times p occurs in the prime factorization of a, on the lhs, the number must be odd, twice the number of times p occurs in the prime factorization of b, plus one more. This is absurd, so our assumption that sqrt(p) is rational must have been wrong.)

So pick a prime and you've got what you want by taking its square root.