Question

Prove that \(\frac{2 + 3\sqrt{5}}{7}\) is an irrational number, given that \(\sqrt{5}\) is an irrational number.

Hint:

In board exams, clearly state that the combination of integers under standard arithmetic operations (multiplication, subtraction, division) always yields a rational number.

Answer 1

Step 1: Assume the opposite of what we want to prove.
Suppose $\frac{2+3\sqrt{5}}{7}$ is rational, and call this rational value $r$.
Step 2: Isolate the square root term.
\[ 2 + 3\sqrt{5} = 7r \implies 3\sqrt{5} = 7r - 2 \implies \sqrt{5} = \frac{7r-2}{3} \]
Step 3: Spot the contradiction.
Since $r$ is rational, $7r-2$ and hence $\frac{7r-2}{3}$ must also be rational (rational numbers stay rational under addition, subtraction and division by a nonzero integer). This forces $\sqrt{5}$ to be rational, which contradicts the given fact that $\sqrt{5}$ is irrational.
Hence our assumption was wrong, and $\frac{2+3\sqrt{5}}{7}$ is irrational.