Question

prove by contradiction that 3√3-2√5 is an irrational number

Answer 1

Hi :) Assume that square root 3 is rational and equal to a/b in lowest terms with a and b integers. Squaring both sides gives 3 = a^2/b^2 so 3b^2 = a^2. If a^2 is a multiple of 3 and a is an integer then a itself must be a multiple of 3. Since a is a multiple of 3 replace it by 3c. Then 3b^2 = (3c)^2 = 9c^2 which reduces to b^2 = 3c^2. Again b^2 a multiple of 3 means b itself is a multiple of 3. We have now shown that a, b are both multiples of 3 contradicting the statement that a/b was in lowest terms, i.e. no common factors. This proves that the original assumption that square root 3 is rational must be false.

I got this bit from a website. Next, by doing the same, we get square root 5 is also irrational. Since both numbers are irrational, multiplying them by any other number and then subtracting them will also get us an irrational number.

Hope this helped. Have a nice day! :)