**Proof by Contradiction: Showing that \(\sqrt{2}\) is Not a Rational Number**To demonstrate that \(\sqrt{2}\) is not a rational number, we'll use a proof by contradiction. Assume that \(\sqrt{2}\) is rational, which means it can be expressed as a fraction \(\frac{a}{b}\) where \(a\) and \(b\) are integers with no common factors other than 1 (i.e., the fraction is in its simplest form) and \(b \neq 0\).Start by expressing \(\sqrt{2}\) as:\[\sqrt{2} = \frac{a}{b}\]Squaring both sides, we have:\[2 = \frac{a^2}{b^2}\]This implies:\[a^2 = 2b^2\]From this equation, we can see that \(a^2\) is even because it equals \(2\) times an integer (\(b^2\)). Since \(a^2\) is even, \(a\) must also be even. Let \(a = 2k\) for some integer \(k\).Substituting \(a = 2k\) into the equation \(a^2 = 2b^2\), we get:\[(2k)^2 = 2b^2\]\[4k^2 = 2b^2\]\[2k^2 = b^2\]This equation shows that \(b^2\) is even, which means \(b\) is also even.Since both \(a\) and \(b\) are even, they have a common factor of \(2\). This contradicts our initial assumption that \(\frac{a}{b}\) was in its simplest form. Therefore, \(\sqrt{2}\) cannot be expressed as a fraction of two integers, proving that it is not a rational number.*Note: You may use the theorem stated in question 14 to further justify this proof or explore related concepts.*

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Description: **Proof by Contradiction: Showing that \(\sqrt{2}\) is Not a Rational Number**To demonstrate that \(\sqrt{2}\) is not a rational number, we'll use a proof by contradiction. Assume that \(\sqrt{2}\) is rational, which means it can be expressed as a f

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Answered: Use proof by contradiction to show that…

Math

Advanced Math

Use proof by contradiction to show that 2 is not a rational number. Note that you may use the theorem stated in question 14.

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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