Show That 3−√5 Is Irrational

Video Explanation

Watch the video below to understand the complete solution step by step:

Solution

Question: Show that the number 3−√5 is irrational.

Step 1: Assume the Contrary

Assume that 3−√5 is rational. Then it can be written as a fraction of two integers in lowest terms:

3−√5 = p/q, where p and q are integers with no common factors and q ≠ 0.

Step 2: Isolate √5

Rearrange the equation to isolate √5:

√5 = 3 − (p/q)

Now square both sides:

5 = [3 − (p/q)]²

Expand the right side:

5 = 9 − 6(p/q) + (p²/q²)

Multiply both sides by q²:

5q² = 9q² − 6pq + p²

Rearranging:

0 = 9q² − 6pq + p² − 5q²

0 = 4q² − 6pq + p²

0 = p² − 6pq + 4q²

Step 3: Contradiction

Consider the equation:

p² − 6pq + 4q² = 0

This is a quadratic in p with discriminant:

D = (−6q)² − 4(1)(4q²) = 36q² − 16q² = 20q²

For p to be an integer, the discriminant must be a perfect square. But 20q² is not a perfect square unless q=0, which is impossible (q ≠ 0). Thus we reach a contradiction.

Final Answer

∴ The number 3−√5 cannot be expressed as a ratio of two integers, so 3−√5 is irrational.

Conclusion

By assuming 3−√5 is rational and reaching a contradiction, we conclude that 3−√5 is irrational.