Question

Prove that $14 - 2\sqrt{3}$ is an irrational number, given that $\sqrt{3}$ is irrational.

Hint:

For any combination of rational and irrational numbers like $a - b\sqrt{c}$, rearranging to isolate the radical $\sqrt{c}$ on one side will automatically establish the proof by contradiction!

Answer 1

Step 1: Set up the proof by contradiction, isolating the radical differently.
Suppose $14 - 2\sqrt3$ were rational. Since $14$ is rational, the difference of two rationals, $14 - (14-2\sqrt3) = 2\sqrt3$, would also have to be rational.
Step 2: Divide out the constant multiplier.
If $2\sqrt3$ is rational, then dividing it by the nonzero rational number $2$ must also give a rational number:
\[ \sqrt3 = \frac{2\sqrt3}{2} \]
Step 3: Spot the contradiction.
This forces $\sqrt3$ to be rational, but it is a well known fact (given to us) that $\sqrt3$ is irrational.
Step 4: Conclude.
Our assumption was wrong, so $14 - 2\sqrt3$ cannot be rational, it must be irrational.