To prove that the sum \( \frac{3}{4} + \sqrt{10} \) is irrational, we can use a proof by contradiction.
Step 1: Assume the Contrary
Assume that \( \frac{3}{4} + \sqrt{10} \) is rational. This means we can express it as a fraction of two integers: \[ \frac{3}{4} + \sqrt{10} = r \] where \( r \) is a rational number.
Step 2: Isolate \( \sqrt{10} \)
Rearranging the equation gives: \[ \sqrt{10} = r - \frac{3}{4} \]
Step 3: Simplify the Right Side
Since \( r \) is rational, we can also express \( r \) as \( \frac{a}{b} \) for integers \( a \) and \( b \) (where \( b \neq 0 \)). Thus, \[ \sqrt{10} = \frac{a}{b} - \frac{3}{4} \] Finding a common denominator (which is \( 4b \)): \[ \sqrt{10} = \frac{4a - 3b}{4b} \]
Step 4: Conclude Rationality of \( \sqrt{10} \)
Since both \( 4a \) and \( 3b \) are integers, \( 4a - 3b \) is also an integer. Consequently, \( \sqrt{10} \) can be expressed as the ratio of two integers: \[ \sqrt{10} = \frac{k}{m} \text{ where } k = 4a - 3b \text{ and } m = 4b \] This implies that \( \sqrt{10} \) is rational.
Step 5: Contradiction
However, it is a well-known fact that \( \sqrt{10} \) is irrational. Since our assumption led to a contradiction, our initial assumption must be incorrect.
Conclusion
Thus, \( \frac{3}{4} + \sqrt{10} \) must be irrational.
Explanation
The sum of a rational number (\( \frac{3}{4} \)) and an irrational number (\( \sqrt{10} \)) is always irrational because the addition of a rational number cannot eliminate the irrationality of an irrational number. Therefore, \( \frac{3}{4} + \sqrt{10} \) is irrational.
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