To prove that the sum \( \frac{3}{4} + \sqrt{10} \) is irrational, we will use proof by contradiction.
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Assume that \( \frac{3}{4} + \sqrt{10} \) is rational. This means that we can express it as the fraction of two integers, say \( \frac{a}{b} \) where \( a \) and \( b \) are integers, and \( b \neq 0 \).
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Rearranging the equation gives: \[ \sqrt{10} = \frac{a}{b} - \frac{3}{4} \] To combine the fractions on the right, we find a common denominator which is \( 4b \): \[ \sqrt{10} = \frac{4a - 3b}{4b} \]
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This implies that \( \sqrt{10} \) is a rational number, since the right-hand side is a fraction of two integers (where \( 4b \neq 0 \), because \( b \neq 0 \)).
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However, we know that \( \sqrt{10} \) is an irrational number. This is because there are no two integers \( p \) and \( q \) such that \( \left(\frac{p}{q}\right)^2 = 10 \).
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Since our assumption that \( \frac{3}{4} + \sqrt{10} \) is rational leads to the conclusion that an irrational number is rational, we have a contradiction. Thus, we conclude that: \[ \frac{3}{4} + \sqrt{10} \text{ is irrational.} \]
Explanation: The sum \( \frac{3}{4} + \sqrt{10} \) is irrational because it is the sum of a rational number \( \frac{3}{4} \) and an irrational number \( \sqrt{10} \), and the sum of a rational number and an irrational number is always irrational.