If \(s_1\), \(s_2\), \(s_3\), \(\ldots\) and \(t_1\), \(t_2\), \(t_3\), \(\ldots\) are sequences of positive numbers, we write
\[
(s_n)\le (t_n)
\]
to mean
"there exists a positive integer \(m\) such that \(s_n \le t_n\) whenever \(n\ge m\)".
Determine whether each of the following statements is true or false. In the case of a true statement, you should give a proof which includes an explicit determination of an appropriate \(m\); in the case of a false statement, you should give a counterexample.
\((1000n) \le (n^2)\,\).
If it is not the case that \((s_n)\le (t_n)\), then it is the case that \((t_n)\le (s_n)\,\).
If \((s_n)\le (t_n)\) and \((t_n) \le (u_n)\), then \((s_n)\le (u_n)\,\).
\((n^2)\le (2^n)\,\).
Solution:
If \(m = 1000\), then \(n \geq m \Rightarrow n^2 \geq 1000n \Rightarrow (1000n) \leq (n^2)\)
This is false. Let \(s_i = 1,2,1,2,\cdots\) and \(t_i = 2,1,2,1,\cdots\).
Suppose that for \(n \geq m_1, s_n \le t_n\) and for \(n \geq m_2, s_t \le u_n\), then for \(n \geq m = \max(m_1, m_2), s_n \leq t_n \leq u_n \Rightarrow s_n \leq u_n \Rightarrow (s_n) \leq (u_n)\)
Let \(m = 6\), then if \(n \geq m, 2^n \geq 1 + n + \frac{n(n-1)}{2} + \frac{n(n-1)}{2} + n + 1 = n^2 + n + 2 \geq n^2\), so \((2^n) \geq (n^2)\)
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.