Question

Prove that \(\sqrt{3}\) is an irrational number.

Hint:

In these proofs, the "Fundamental Theorem of Arithmetic" is your best friend: if a prime \(p\) divides \(a^2\), then \(p\) must divide \(a\).

Answer 1

To Prove: √3 is an irrational number.

Proof (By Contradiction):

Step 1: Assume the opposite
Suppose √3 is a rational number.
Then it can be written in the form:
√3 = p/q
where p and q are integers having no common factor, and q ≠ 0.

Step 2: Square both sides
3 = p² / q²
Multiply both sides by q²:
p² = 3q²

Step 3: Interpret the equation
Since p² is divisible by 3, p must also be divisible by 3.
Let p = 3k.

Step 4: Substitute back
p² = (3k)² = 9k²
Substitute into the earlier equation:
9k² = 3q²
Divide both sides by 3:
3k² = q²

Step 5: Interpretation
This means q² is divisible by 3, so q must also be divisible by 3.

This contradicts our assumption
We assumed p and q have no common factor, but both are divisible by 3.

Final Conclusion:
Therefore, √3 is an irrational number.