Question 461858
Right away, we know statement 1 is false; square root is not commutative. For example, let a = 9, b = 16, then *[tex \sqrt{9} + \sqrt{16} = \sqrt{9 + 16}] which pretty much says that 7 = 5.


Statement 2 is false; multiplying by sqrt(3) will only turn the denominator into another radical expression. Multiplying by 3 - sqrt(3) will rationalize the denominator.