To prove that the sum \( \frac{1}{2} + \sqrt{2} \) is irrational, we can start by assuming the opposite: that the sum is rational.
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Assume \( \frac{1}{2} + \sqrt{2} \) is rational. This means we can express it as a fraction of two integers: \[ \frac{1}{2} + \sqrt{2} = \frac{a}{b} \] where \( a \) and \( b \) are integers, with \( b \neq 0 \).
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Rearranging the equation gives: \[ \sqrt{2} = \frac{a}{b} - \frac{1}{2} = \frac{2a - b}{2b} \]
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Since \( \frac{2a - b}{2b} \) is a rational number (because it is a fraction of two integers), this would imply that \( \sqrt{2} \) is also rational.
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However, it is known that \( \sqrt{2} \) is irrational. This is a standard result in mathematics, often proved using a contradiction approach (assuming \( \sqrt{2} \) can be expressed as \( \frac{p}{q} \), leading to the conclusion that both \( p \) and \( q \) must be even, contradicting the assumption of their being in simplest form).
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Because our assumption that \( \frac{1}{2} + \sqrt{2} \) is rational leads to the contradiction that \( \sqrt{2} \) is rational, we conclude that our original assumption must be false.
Thus, the sum \( \frac{1}{2} + \sqrt{2} \) is irrational.
In one short simple sentence: The sum \( \frac{1}{2} + \sqrt{2} \) is irrational because it leads to the contradiction that an irrational number \( \sqrt{2} \) can be expressed as a rational number.