To prove that the sum \( \frac{1}{2} + \sqrt{2} \) is irrational, we will use a proof by contradiction.
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Assume that \( \frac{1}{2} + \sqrt{2} \) is rational. This means we can express it as: \[ \frac{1}{2} + \sqrt{2} = \frac{a}{b} \] for some integers \( a \) and \( b \) where \( b \neq 0 \).
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Rearranging the equation: \[ \sqrt{2} = \frac{a}{b} - \frac{1}{2} \] To combine the two fractions, we get: \[ \sqrt{2} = \frac{2a - b}{2b} \]
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Next, we square both sides: \[ 2 = \left( \frac{2a - b}{2b} \right)^2 \] This simplifies to: \[ 2 = \frac{(2a - b)^2}{(2b)^2} \] Multiplying both sides by \( (2b)^2 \) gives: \[ 2(2b)^2 = (2a - b)^2 \] which simplifies to: \[ 8b^2 = (2a - b)^2 \]
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Expanding the right side: \[ 8b^2 = 4a^2 - 4ab + b^2 \] Rearranging this, we have: \[ 4a^2 - 4ab + b^2 - 8b^2 = 0 \] Simplifying leads to: \[ 4a^2 - 4ab - 7b^2 = 0 \]
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This is a quadratic in \( a \): The discriminant \( D \) of the quadratic equation \( Ax^2 + Bx + C = 0 \) (where \( A = 4 \), \( B = -4b \), and \( C = -7b^2 \)) is given by: \[ D = B^2 - 4AC = (-4b)^2 - 4 \cdot 4 \cdot (-7b^2) = 16b^2 + 112b^2 = 128b^2 \]
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Since \( D \) is positive, we can find two distinct solutions for \( a \). Note that since the discriminant \( D \) is positive and is a perfect square, the solutions for \( a \) will be rational.
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Referring Back to the Sum: If \( \frac{1}{2} + \sqrt{2} \) is rational, then \( \sqrt{2} \) must also be rational (since it can be expressed in terms of \( a \) and \( b \)). However, it is known that \( \sqrt{2} \) is irrational.
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Conclusion of Proof: This leads to a contradiction since both \( \sqrt{2} \) being rational and its known irrationality cannot coexist.
Thus, \( \frac{1}{2} + \sqrt{2} \) must be irrational.
In one simple sentence: The sum \( \frac{1}{2} + \sqrt{2} \) is irrational because it combines the rational number \( \frac{1}{2} \) with the irrational number \( \sqrt{2} \), resulting in an irrational number (the sum of a rational and an irrational number is always irrational).