Question

Prove that $\sqrt{5}$ is an irrational number.

Hint:

Proof by contradiction is often used to prove irrationality.

Answer 1

To Prove:
\( \sqrt{5} \) is an irrational number

Proof by Contradiction:
Assume, to the contrary, that \( \sqrt{5} \) is a rational number.
Then, it can be expressed as a fraction of two integers in lowest terms:
\[ \sqrt{5} = \frac{a}{b}, \quad \text{where } a \text{ and } b \text{ are integers, } b \ne 0, \text{ and } \gcd(a, b) = 1 \]
Step 1: Square both sides
\[ \sqrt{5} = \frac{a}{b} \Rightarrow 5 = \frac{a^2}{b^2} \Rightarrow a^2 = 5b^2 \]
Step 2: Analyze the result
The equation \( a^2 = 5b^2 \) implies that \( a^2 \) is divisible by 5.
Hence, \( a \) must also be divisible by 5.
Let \( a = 5k \) for some integer \( k \).

Step 3: Substitute back
\[ a^2 = (5k)^2 = 25k^2 \Rightarrow 25k^2 = 5b^2 \Rightarrow 5k^2 = b^2 \]
This implies that \( b^2 \) is divisible by 5 ⟹ so \( b \) is divisible by 5.

Step 4: Contradiction
Now, both \( a \) and \( b \) are divisible by 5.
This contradicts our original assumption that \( \gcd(a, b) = 1 \), i.e., the fraction is in lowest terms.

Conclusion:
Our assumption that \( \sqrt{5} \) is rational leads to a contradiction.
Hence, \( \boxed{\sqrt{5} \text{ is an irrational number}} \).