More on inequalities
Proposition 1. \(\frac{1}{2}>0\).
Proof. Assume to the contrary, that \(\frac{1}{2}\notin\mathbb{P}\). Since \(\frac{1}{2}\neq 0\), we deduce that \(-\frac{1}{2}\in\mathbb{P}\). Since \(\mathbb{P}\) is closed under addition, we see that \(-1 = (-\frac{1}{2}) + (-\frac{1}{2})\in\mathbb{P}\). We have already shown that \(1>0\), so by the Trichotomy Property this is a contradiction. We conclude that \(\frac{1}{2}\in\mathbb{P}\), that is, \(\frac{1}{2}>0\). \(\Box\)
Can you come up with another proof that uses the fact that \(2\in\mathbb{P}\) and the fact that \(\mathbb{P}\) is closed under multiplication?
Proof. Assume to the contrary, that \(\frac{1}{2}\notin\mathbb{P}\). Since \(\frac{1}{2}\neq 0\), we deduce that \(-\frac{1}{2}\in\mathbb{P}\). We have already shown that \(2\in\mathbb{P}\). Since \(\mathbb{P}\) is closed under multiplication we see that \(-1 = 2\cdot(-\frac{1}{2})\in\mathbb{P}\). We have already shown that \(1>0\), so by the Trichotomy Property this is a contradiction. We conclude that \(\frac{1}{2}\in\mathbb{P}\), that is, \(\frac{1}{2}>0\). \(\Box\)
Proposition 1. \(\frac{1}{2}>0\).
Proposition 2. If \(0\leq a\leq \varepsilon\) for all \(\varepsilon>0\), then \(a=0\).
We will prove the contrapositive of this statement. Recall that given an implication \(A\Rightarrow B\), the contrapositive is the equivalent statement \(\sim\! B\Rightarrow \sim\! A\).
Proposition 2. If \(0\leq a\leq \varepsilon\) for all \(\varepsilon>0\), then \(a=0\).
Proof. We will prove the contrapositive. Hence, we assume \(a\neq 0\). If \(a<0\), then the desired conclusion clearly holds. So we assume \(a>0\). By Proposition 1 we see that \(\frac{a}{2} = a\cdot \frac{1}{2}>0\). Adding \(\frac{a}{2}\) to both sides, we also we see that \(a>\frac{a}{2}\).
Let \(\varepsilon_{0} = \frac{a}{2}\), and note that
\[a>\frac{a}{2} = \varepsilon_{0}.\]
Hence, we have a positive number, \(\varepsilon_{0}\), so that \(a>\varepsilon_{0}\). \(\Box\)
The statement we will prove:
If \(a\neq 0\), then either \(a<0\) or there exists \(\varepsilon>0\) such that \(a> \varepsilon\).
Practice problem:
Proposition. If \(x\geq 0\) and \(nx<1\) for all \(n\in\N\), then \(x=0\).
State the contrapositive of the proposition:
Answer:
If \(x\neq 0\), then either \(x<0\) or there exists \(n\in\N\) such that \(nx\geq 1\).
Read Section 2.1 from the text.