The square root of 2 is irrational. This means that it cannot be expressed as a fraction of two integers. The proof of its irrationality dates back to ancient times and can be demonstrated by contradiction.
Here's a brief outline of the proof:
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Assume \(\sqrt{2}\) is rational. Then it can be expressed as a fraction in the form \(\frac{p}{q}\), where \(p\) and \(q\) are integers with no common factors (i.e., the fraction is in its simplest form).
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This leads to the equation: \(\sqrt{2} = \frac{p}{q}\), which can be squared to yield: \[ 2 = \frac{p^2}{q^2} \]
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Rearranging gives: \[ p^2 = 2q^2 \]
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This equation implies that \(p^2\) is even (since it is equal to 2 times another integer), which means that \(p\) must also be even.
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If \(p\) is even, it can be expressed as \(p = 2k\) for some integer \(k\). Substituting back, we get: \[ (2k)^2 = 2q^2 \implies 4k^2 = 2q^2 \implies 2k^2 = q^2 \]
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This shows that \(q^2\) is also even, and therefore \(q\) must be even as well.
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Since both \(p\) and \(q\) are even, they share a common factor of 2, which contradicts the assumption that \(\frac{p}{q}\) is in simplest form.
Since this leads to a contradiction, the assumption that \(\sqrt{2}\) is rational must be false. Thus, \(\sqrt{2}\) is irrational.