Question
Prove that 3 + √5 is an irrational number.
✓
Answer
Let us assume that 3 + √5 is a rational number.
So, we can find co-prime integers ‘a’ and ‘b’ (b ≠ 0) such that
$
\begin{aligned}
& 3+\sqrt{5}=\frac{a}{b} \\
\therefore \quad & \sqrt{5}=\frac{a}{b}-3
\end{aligned}
$
Since, $a$ and $b$ are integers, $\frac{a}{b}-3$ is a rational
number and so √5 is a rational number.
But this contradicts the fact that √5 is an irrational number.
∴ Our assumption that 3 – √5 is a rational number is wrong.
3 + √5 is an irrational number.
So, we can find co-prime integers ‘a’ and ‘b’ (b ≠ 0) such that
$
\begin{aligned}
& 3+\sqrt{5}=\frac{a}{b} \\
\therefore \quad & \sqrt{5}=\frac{a}{b}-3
\end{aligned}
$
Since, $a$ and $b$ are integers, $\frac{a}{b}-3$ is a rational
number and so √5 is a rational number.
But this contradicts the fact that √5 is an irrational number.
∴ Our assumption that 3 – √5 is a rational number is wrong.
3 + √5 is an irrational number.
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