Let us denote the sum of \( \frac{1}{2} \) and \( \sqrt{2} \) as \( S = \frac{1}{2} + \sqrt{2} \). We are tasked with proving that \( S \) is irrational.
We know that \( \sqrt{2} \) is an irrational number. To demonstrate that \( S \) is irrational, we will assume the contrary—that \( S \) is rational. If \( S \) is rational, this means that it can be expressed as \( S = \frac{a}{b} \) for some integers \( a \) and \( b \neq 0 \).
From the equation \( S = \frac{1}{2} + \sqrt{2} \), we can rearrange it to isolate \( \sqrt{2} \):
\[ \sqrt{2} = S - \frac{1}{2} \]
Since \( S \) is assumed to be rational and \( \frac{1}{2} \) is rational, the right-hand side \( S - \frac{1}{2} \) would also be rational (as the difference of two rational numbers is rational). Thus, we conclude that \( \sqrt{2} \) must be rational.
This leads to a contradiction, as we know from established mathematics that \( \sqrt{2} \) is irrational. Therefore, our assumption that \( S \) is rational must be false, implying that \( S \) is irrational.
In summary, the sum \( S = \frac{1}{2} + \sqrt{2} \) is irrational because it can be shown that assuming it to be rational leads to a contradiction regarding the irrationality of \( \sqrt{2} \).