Lecture 3:
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 \(\sim\! B\Rightarrow \sim\! A\).
- If \(0\leq a\leq\varepsilon\) for all \(\varepsilon>0\), then \(a=0\).
- \((\forall\,\varepsilon>0, 0\leq a\leq\varepsilon)\Rightarrow (a=0).\)
- \(\sim\! (a=0)\Rightarrow\sim\! (\forall\,\varepsilon>0, 0\leq a\leq\varepsilon)\)
- \((a\neq 0)\Rightarrow (\exists\,\varepsilon>0, a<0\text{ or }a> \varepsilon)\)
- If \(a\neq 0\), then there exists \(\varepsilon>0\) such that either \(a<0\) or \(a> \varepsilon\).
- If \(a\neq 0\), then either \(a<0\) or there exists \(\varepsilon>0\) such that \(a> \varepsilon\).
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 \(nx\geq 1\).
End Lecture 3:
Read Section 2.1 from the text.
Lecture 3
By John Jasper
Lecture 3
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