To prove that \( \frac{3}{4} + \sqrt{10} \) is irrational, we will assume for contradiction that \( \frac{3}{4} + \sqrt{10} \) is rational. A rational number can be expressed as the ratio of two integers.
Let \( \frac{3}{4} + \sqrt{10} = r \), where \( r \) is rational. Rearranging gives us:
\[ \sqrt{10} = r - \frac{3}{4} \]
Thus, we can express \( r - \frac{3}{4} \) as a single fraction:
\[ \sqrt{10} = \frac{4r - 3}{4} \]
Since \( r \) is rational, \( 4r \) is also rational, and therefore \( 4r - 3 \) is rational. Therefore, the right side of the equation is rational (as the ratio of two rational numbers).
However, this leads to the conclusion that \( \sqrt{10} \) is rational because it can be expressed as a fraction of two integers, which contradicts the known fact that \( \sqrt{10} \) is irrational (as 10 is not a perfect square).
Since our assumption that \( \frac{3}{4} + \sqrt{10} \) is rational leads to a contradiction, we conclude that \( \frac{3}{4} + \sqrt{10} \) must be irrational.
Explanation: The sum of a rational number (\( \frac{3}{4} \)) and an irrational number (\( \sqrt{10} \)) is always irrational; thus, \( \frac{3}{4} + \sqrt{10} \) is irrational.