Question:

Given: \(\sqrt{5}-\sqrt{3}\)

To Prove: \(\sqrt{5}-\sqrt{3}\) is irrational.

Step 1: Assume the contrary, i.e., \(\sqrt{5}-\sqrt{3}=\dfrac{p}{q}\) where \(p,q\in\mathbb{Z},\;q
eq0\) and \(\gcd(p,q)=1\).

Step 2: Rearrange the equation
\[\sqrt{5}=\sqrt{3}+\dfrac{p}{q}.\]

Step 3: Square both sides
\[5 = \left(\sqrt{3}+\dfrac{p}{q}\right)^{2}=3+\dfrac{p^{2}}{q^{2}}+2\sqrt{3}\,\dfrac{p}{q}.\]

Step 4: Isolate the term containing \(\sqrt{3}\)
\[2\sqrt{3}\,\dfrac{p}{q}=5-3-\dfrac{p^{2}}{q^{2}}=2-\dfrac{p^{2}}{q^{2}}.\]

Step 5: Multiply by \(q^{2}\) to clear denominators
\[2\sqrt{3}\,p q = 2q^{2}-p^{2}.\]

Step 6: The right‑hand side is an integer; therefore \(2\sqrt{3}\,p q\) must be an integer. Since \(p,q\) are integers, this implies \(\sqrt{3}\) is rational, which is impossible because \(\sqrt{3}\) is known to be irrational (proved earlier in the chapter).

Step 7: The contradiction arises from the assumption that \(\sqrt{5}-\sqrt{3}\) is rational. Hence the assumption is false.

Conclusion: \(\sqrt{5}-\sqrt{3}\) is irrational.