Question

Prove that $2 - 5\sqrt{3}$ is an irrational number given that $\sqrt{3}$ is irrational.

Hint:

For questions involving the proof of irrationality of a combination of numbers (like $a \pm b\sqrt{c}$), always assume the entire expression is a rational variable $r$.
Isolating the radical term on one side will automatically establish the proof by contradiction with minimal steps!

Answer 1

Step 1: Assume the opposite, using a p/q form.
Suppose, for contradiction, that $2 - 5\sqrt{3}$ is rational. Then it can be written as $\frac{p}{q}$, where $p, q$ are integers, $q \neq 0$, and $p,q$ have no common factor other than 1.
Step 2: Isolate the radical term. \[ 2 - 5\sqrt{3} = \frac{p}{q} \] \[ 5\sqrt{3} = 2 - \frac{p}{q} = \frac{2q-p}{q} \] \[ \sqrt{3} = \frac{2q-p}{5q} \]
Step 3: Examine the right-hand side.
Since $p$ and $q$ are integers, $2q-p$ and $5q$ are also integers, so $\frac{2q-p}{5q}$ is a rational number. This forces $\sqrt{3}$ to be rational too.
Step 4: Reach the contradiction and conclude.
This directly contradicts the given fact that $\sqrt{3}$ is irrational. So our assumption was wrong, and $2-5\sqrt{3}$ must be irrational.
Hence, $2 - 5\sqrt{3}$ is proved to be an irrational number.