Absolute value
Definition. The absolute value of a real number \(a\), denoted by \(|a|\), is defined by
\[|a|: = \begin{cases} a & \text{if }a>0,\\ 0 & \text{if } a=0,\\ -a & \text{if }a<0.\end{cases}\]
Theorem 2.2.2 (a) \(|ab|=|a||b|\) for all \(a,b\in\R\).
Proof. Let \(a,b\in\R\) be given. If either \(a=0\) or \(b=0\), then \(|ab|=|0|=0\) and \(|a||b|=0\). Now we consider each possible sign of \(a\) and \(b\) as a separate case.
If \(a>0\) and \(b>0\), then \(ab>0\) and hence \(|ab|=ab=|a||b|\).
If \(a>0\) and \(b<0\), then \(ab<0\) and hence \(|ab| = -(ab)=a(-b)=|a||b|\).
If \(a<0\) and \(b>0\), then \(ab<0\) and hence \(|ab|=-(ab)=(-a)b=|a||b|\).
If \(a<0\) and \(b<0\), then \(ab>0\) and hence \(|ab|=ab=(-a)(-b)=|a||b|.\)
\(\Box\)
Prove or provided a counterexample to the following:
Proposition. If \(x\in\R\), then \(|-x|=x\).
This statement is false. Indeed, if \(x=-3\), then \[|-x| = |-(-3)|=|3|=3\neq x.\]
Theorem 2.2.2 (a) \(|ab|=|a||b|\) for all \(a,b\in\R\).
Theorem 2.2.2 (b) \(|a|^2=a^2\) for all \(a\in\R\).
Proof. Let \(a\in\R\). Since \(a^2\geq 0\) we see that \(a^2=|a^2|\). By Theorem 2.2.2(a) we see that \(|a^2|=|aa|=|a||a|=|a|^2.\) \(\Box\)
Theorem 2.2.2(c). If \(c\geq 0\), then \(|a|\leq c\) if and only if \(-c\leq a\leq c\).
Proof. First, we will prove the forward direction. Thus, we suppose that \(|a|\leq c\). There are two cases to consider.
Case 1. Suppose \(a\geq 0\). In this case we have \(a=|a|\leq c\). Since \(c\geq 0\) and \(a\geq 0\) we have \(-c\leq 0\leq a\).
Case 2. Suppose \(a<0\). In this case we have \(|a|=-a>0\geq -c\). This implies \(a\leq c\). For the other half of the inequality we note that \(-a=|a|\leq c\). This implies \(-c\leq a\).
For the other direction we suppose that \(-c\leq a\leq c\). This implies \(a\leq c\) and \(-a\leq c\). Hence, no matter the sign of \(a\) (or if it is zero) we have \(|a|\leq c\). \(\Box\)
Theorem 2.2.2(d). \(-|a|\leq a\leq |a|\) for all \(a\in\R\).
Proof. Apply Theorem 2.2.2(c) with \(c=|a|\). \(\Box\)
Inequalities involving absolute values are very common in real analysis, and Theorem 2.2.2 (c) gives us a very nice way to deal with them.
Example. Describe the following set using interval notation:
\[A=\left\{x\in\R : |2x-1|<\frac{1}{2}\right\}.\]
Example: \(\frac{1}{2}\in A\) since \(|2\cdot\frac{1}{2}-1|=0<\frac{1}{2}\).
Assume that \(x\in A\), that is, \(x\) is a number which satisfies the inequality \(|2x-1|<\frac{1}{2}\). By Theorem 2.2.2 (c) we see that this is equivalent to
\[-\tfrac{1}{2}<2x-1<\tfrac{1}{2}.\]
Adding \(1\) and multiplying by \(\tfrac{1}{2}\) we see that
\[\tfrac{1}{4}<x<\tfrac{3}{4}.\]
Therefore, \(x\in (\tfrac{1}{4},\tfrac{3}{4})\). Hence, we see that \(A \subset \left(\frac{1}{4},\frac{3}{4}\right)\)
Inequalities involving absolute values are very common in real analysis, and Theorem 2.2.2 (c) gives us a very nice way to deal with them.
Next, we assume \(x\in \left(\frac{1}{4},\frac{3}{4}\right)\), that is, \(x\) is a number such that \[\tfrac{1}{4}<x<\tfrac{3}{4}.\]
Multiplying by \(2\) and adding \(-1\) we see \[-\tfrac{1}{2}<2x-1<\tfrac{1}{2}.\]
By Theorem 2.2.2 (c) this implies \[|2x-1|<\tfrac{1}{2}.\]
Therefore \(x\in A\). This shows that \(\left(\tfrac{1}{4},\tfrac{3}{4}\right)\subset A.\) Therefore \[\textstyle{A = (\frac{1}{4},\frac{3}{4}) = \{x\in\R: \frac{1}{4}<x<\frac{3}{4}\}}\]
Example. Describe the following set using interval notation:
\[A=\left\{x\in\R : |2x-1|<\frac{1}{2}\right\}.\]
Practice problem:
Describe the following set using interval notation:
\[A=\left\{x\in\R : |5x+1|<\frac{3}{4}\right\}.\]
You don't need to prove your answer, but you should know how!
Read Section 2.2 from the text.
\(|5x+1|<\frac{3}{4}\quad\Leftrightarrow\quad -\frac{3}{4}<5x+1<\frac{3}{4}\quad\Leftrightarrow\quad -\frac{7}{20}<x<-\frac{1}{20}\)
\[A=\left(-\frac{7}{20},-\frac{1}{20}\right)\]