Question:

Given: \(\sqrt{32}\) \(=\) \(\sqrt{2^5}\).

To Prove: \(\sqrt{32}\) cannot be expressed as a ratio of two integers with no common factor (i.e., it is irrational).

Step 1: Assume the contrary, that \(\sqrt{32}\) is rational. Then there exist coprime integers \(p\) and \(q\) (\(q
eq 0\)) such that

$$\sqrt{32}=\frac{p}{q}$$

Step 2: Square both sides to eliminate the square root.

$$32 = \frac{p^{2}}{q^{2}} \quad \Rightarrow \quad 32 q^{2}=p^{2}.$$

Step 3: Write 32 as a power of 2: \(32 = 2^{5}\). Hence

$$2^{5} q^{2}=p^{2}.$$

Step 4: From the equation, \(p^{2}\) is divisible by \(2\); therefore \(p\) must be even. Let \(p=2k\) for some integer \(k\).

Step 5: Substitute \(p=2k\) into the equation:

$$2^{5} q^{2} = (2k)^{2}=4k^{2} \quad \Rightarrow \quad 2^{5} q^{2}=4k^{2}.$$

Divide both sides by 4 (i.e., \(2^{2}\)):

$$2^{3} q^{2}=k^{2} \quad \Rightarrow \quad 8 q^{2}=k^{2}.$$

Step 6: The right‑hand side \(k^{2}\) is again divisible by 2, so \(k\) is even. Hence \(k=2m\) for some integer \(m\).

Step 7: Substituting back, we get

$$8 q^{2}= (2m)^{2}=4m^{2} \quad \Rightarrow \quad 2 q^{2}=m^{2}.$$

Thus \(m^{2}\) is even, implying \(m\) is even, and consequently \(q\) is also even.

Step 8: We have shown that both \(p\) and \(q\) are even, which means they have a common factor 2. This contradicts the initial assumption that \(p\) and \(q\) are coprime.

Conclusion: The assumption that \(\sqrt{32}\) is rational leads to a contradiction. Hence \(\sqrt{32}\) is irrational.