Quiz 5
Discrete Mathematics · 3 problems · solutions hidden, click to reveal
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Problems
State the definition of a real number \(x\) being rational.
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Answer: \(x\) is rational if there exist integers \(a, b\) with \(b \neq 0\) such that \(x = a/b\).
A real number \(x\) is rational iff \(x = a/b\) for some integers \(a, b\) with \(b \neq 0\).
Disprove: the product of two irrational numbers is irrational.
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Answer: Counterexample: \(\sqrt{2}\cdot\sqrt{2} = 2\), which is rational.
\(\sqrt{2}\) is irrational, yet \(\sqrt{2}\cdot\sqrt{2} = 2\) is rational. A single counterexample disproves the claim.
Use proof by contradiction to prove that the product of a nonzero rational number and an irrational number is irrational.
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Let \(x\) be a nonzero rational and \(y\) irrational. Toward a contradiction, suppose \(xy\) is rational. Then \(x = a/b\) and \(xy = c/d\) for integers \(a,b,c,d\), with \(a \neq 0\) since \(x \neq 0\). Then
\(y = \dfrac{xy}{x} = \dfrac{c/d}{a/b} = \dfrac{bc}{ad}\),
which is rational - contradicting that \(y\) is irrational. Therefore \(xy\) is irrational.
\(y = \dfrac{xy}{x} = \dfrac{c/d}{a/b} = \dfrac{bc}{ad}\),
which is rational - contradicting that \(y\) is irrational. Therefore \(xy\) is irrational.