How can I prove that the square root of 8 minus the square root of 3 is an irrational number.?

Question

I am supposed to use thefact that the square root of 3 is an irrational number. I know I need to use a proof by contradiction, but I am stuck after that.

Answer 1

Suppose √8 - √3 is rational. Then its square is rational, so we have that (√8 - √3)² = 8 - 2√24 + 3 is rational. Then subtracting 11 from it will also yield a rational number, so we have that -2√24 is rational. Since dividing a rational number by -2 yields a rational number, we have that √24 is rational. However, 24 is not a perfect square, so its square root cannot be rational -- a contradiction. Therefore, √8 - √3 is not rational.

Answer 2

Assume sqrt(8) - sqrt(3) = m/.n, in other words, that it IS rational. m and n are both integers. Square to get
8 - 2*sqrt(24) + 3 = m^2/n^2
Simplify to get 11 - (m^2/n^2) = 4*sqrt(3)
Left side is rational, right side is not, therefore a contradiction with the only assumption.

Answer 3

See http://en.wikipedia.org/wiki/Irrational_number#Example_proofs
for an example proof using the sqrt of 2.

Answer 4

if it is a rational number it can be expressed as a fraction. approximating the upper and lower limits will show that it cannot be done