Question

Given that \(\sqrt{5}\) is an irrational number, prove that \(3 + 2\sqrt{5}\) is also an irrational number.

Hint:

In contradiction proofs, always aim to isolate the known irrational part (like \(\sqrt{5}\)) on one side of the equation.

Answer 1

Step 1: Assume the contrary

Assume that 3 + 2√5 is a rational number.
That means we assume 3 + 2√5 = r, where r is a rational number.

Step 2: Rearranging the equation

3 + 2√5 = r
2√5 = r − 3

Since r is rational and 3 is also rational,
(r − 3) is rational.

So, 2√5 = rational number.

Step 3: Divide both sides by 2

√5 = (r − 3) / 2

Since (r − 3) is rational and 2 is rational,
(r − 3) / 2 is rational.

This implies √5 is rational.

Step 4: Contradiction

But it is given that √5 is irrational.
This contradicts our assumption.

Step 5: Conclusion

Therefore, our assumption is wrong.
Hence, 3 + 2√5 is an irrational number.