To prove that the sum of \( \frac{3}{4} \) and \( \sqrt{10} \) is irrational, we start by recalling the definitions of rational and irrational numbers:
- A rational number can be expressed as a fraction \( \frac{a}{b} \) where \( a \) and \( b \) are integers and \( b \neq 0 \).
- An irrational number cannot be expressed as a fraction of integers.
Step 1: Identify the numbers involved.
- \( \frac{3}{4} \) is a rational number because it can be written as a fraction of integers (3 and 4).
- \( \sqrt{10} \) is an irrational number because it cannot be expressed as a fraction of integers; its decimal form is non-repeating and non-terminating.
Step 2: Add the two numbers. Let \( x = \frac{3}{4} + \sqrt{10} \).
Step 3: Assume for contradiction that \( x \) is rational. If \( x \) is rational, we can write: \[ x = \frac{3}{4} + \sqrt{10} \] Rearranging gives: \[ \sqrt{10} = x - \frac{3}{4} \]
Since \( \frac{3}{4} \) is rational, \( x - \frac{3}{4} \) would also be rational if \( x \) is rational. Therefore, if \( x \) is rational, then \( \sqrt{10} \) must also be rational.
Step 4: Conclude based on the nature of \( \sqrt{10} \). However, we know that \( \sqrt{10} \) is irrational. This creates a contradiction because we cannot have an irrational number being equal to a rational number.
Conclusion: Since our assumption that \( \frac{3}{4} + \sqrt{10} \) is rational led us to a contradiction, we conclude that: \[ \frac{3}{4} + \sqrt{10} \text{ is irrational.} \]
In one sentence: The sum \( \frac{3}{4} + \sqrt{10} \) is irrational because the sum of a rational number and an irrational number is always irrational.