Question

Prove that $4 - 2\sqrt{5}$ is an irrational number given that $\sqrt{5}$ 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: Recall two basic facts about rational and irrational numbers.
A nonzero rational number multiplied by an irrational number always gives an irrational number, and a rational number minus an irrational number always gives an irrational number. Both of these are standard, provable facts.
Step 2: Apply the first fact to $2\sqrt{5}$.
Since $\sqrt{5}$ is given to be irrational and $2$ is a nonzero rational number, their product $2\sqrt{5}$ must be irrational.
Step 3: Apply the second fact to $4 - 2\sqrt{5}$.
Now $4$ is rational and $2\sqrt{5}$ is irrational, so their difference $4 - 2\sqrt{5}$ must also be irrational, by the same reasoning as above, without needing to set up a contradiction argument.
Therefore, $4 - 2\sqrt{5}$ is an irrational number, as required.