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2008 Paper 1 Q1
D: 1500.0 B: 1484.0
proof by contradiction irrational numbers counterexample logical reasoning real numbers rationality and irrationality

What does it mean to say that a number \(x\) is irrational? Prove by contradiction statements A and B below, where \(p\) and \(q\) are real numbers.

  • A: If \(pq\) is irrational, then at least one of \(p\) and \(q\) is irrational.
  • B: If \(p+q\) is irrational, then at least one of \(p\) and \(q\) is irrational.
Disprove by means of a counterexample statement C below, where \(p\) and \(q\) are real numbers.
  • C: If \(p\) and \(q\) are irrational, then \(p+q\) is irrational.
If the numbers \(\e\), \(\pi\), \(\pi^2\), \(\e^2\) and \(\e\pi\) are irrational, prove that at most one of the numbers \(\pi+\e\), \(\pi -\e\), \(\pi^2-\e^2\), \(\pi^2+\e^2\) is rational.

Solution:

  • A: Suppose for sake of contradiction that neither \(p\) nor \(q\) is irrational, then \(pq\) is the product of two rational numbers, ie is also rational. Therefore \(pq\) is rational. Contradiction.
  • B: Suppose for the sake of contradiction both \(p\) and \(q\) are rational, but then \(p+q\) is also rational, contradicting \(p+q\) is irrational.
  • C: Note that \(\sqrt{2}\) and \(-\sqrt{2}\) are both irrational, but \(\sqrt{2}+(-\sqrt{2}) = 0\) which is rational.
Since \((\pi + \e) + (\pi - \e) = 2\pi\) is irrational, at most one of \(\pi+\e\) and \(\pi - \e\) can be rational. Since \((\pi+\e)(\pi-\e) = \pi^2 - \e^2\) is is the product of a (non-zero) rational and an irrational, \(\pi^2 - \e^2\) cannot be rational. Therefore for two of these numbers to be irrational, we need \(\pi^2 + \e^2\) to be rational. But then squaring whichever of \(\pi \pm \e\) is rational and subtracting \(\pi^2+\e^2\) we obtain \(\pm 2\pi \e\) which is irrational. But the product and sum of rationals is irrational. Therefore it cannot be the case that more than one of these numbers is rational.

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2003 Paper 2 Q3
D: 1600.0 B: 1541.7
proof by contradiction proof by induction irrational numbers cube root sequences convergence integer limit

Prove that the cube root of any irrational number is an irrational number. Let \(\displaystyle u_n = {5\vphantom{\dot A}}^{1/{(3^n)}}\,\). Given that \(\sqrt[3]5\) is an irrational number, prove by induction that \(u_n\) is an irrational number for every positive integer \(n\). Hence, or otherwise, give an example of an infinite sequence of irrational numbers which converges to a given integer \(m\,\). [An irrational number is a number that cannot be expressed as the ratio of two integers.]

Solution: Claim: \(x \in \mathbb{R}\setminus \mathbb{Q} \Rightarrow x^{1/3} \in \mathbb{R} \setminus\mathbb{Q}\) Proof: We will prove the contrapositive, since \(x^{1/3} = p/q\) but then \(x = p^3/q^3 \in \mathbb{Q}\), therefore we're done. Claim: \(u_n = 5^{1/(3^n)}\) is irrational for \(n \geq 1\) Proof: We are assuming the base case, but then \(u_{n+1} = \sqrt[3]{u_n}\) which is clearly irrational by our first lemma, so we're done. Note that \(u_n \to 1\) and so \((m-1)+u_n \to m\) for any integer \(m\).

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