Question:

Given: \(\sqrt{32}+\sqrt{5}\).

To Prove: The number is irrational.

Step 1 – Assume the contrary
Assume that \(\sqrt{32}+\sqrt{5}\) is rational. Then there exist integers \(p,q\) (with \(q
eq0\) and \(\gcd(p,q)=1\)) such that
\[\sqrt{32}+\sqrt{5}=\frac{p}{q}.\]

Step 2 – Isolate one radical
\[\sqrt{5}=\frac{p}{q}-\sqrt{32}.\]
Since \(\sqrt{32}=4\sqrt{2}\) and \(4\sqrt{2}\) is irrational, the right‑hand side is the difference of a rational number and an irrational number; therefore it is irrational.

Step 3 – Square both sides
Square the equality in Step 1:
\[\left(\sqrt{32}+\sqrt{5}\right)^2 = \left(\frac{p}{q}\right)^2\]
\[32+5+2\sqrt{32\cdot5}=\frac{p^2}{q^2}\]
\[37+2\sqrt{160}=\frac{p^2}{q^2}.\]
Hence
\[2\sqrt{160}=\frac{p^2}{q^2}-37.\]
The right‑hand side is rational (difference of two rationals), so \(\sqrt{160}\) must be rational.

Step 4 – Reduce \(\sqrt{160}\)
\[\sqrt{160}=\sqrt{16\times10}=4\sqrt{10}.\]
Thus \(4\sqrt{10}\) is rational, which implies \(\sqrt{10}\) is rational.

Step 5 – Contradiction
If \(\sqrt{10}\) were rational, then 10 would be a perfect square, which is false. Hence our assumption that \(\sqrt{32}+\sqrt{5}\) is rational leads to a contradiction.

Conclusion
Therefore \(\sqrt{32}+\sqrt{5}\) is irrational.