Real numbers · Definition
Absolute value
Before you start you need to know
Almost every definition in analysis says “is close to”; the absolute value turns that phrase into an inequality we can manipulate.
Definition
The absolute value \(|x|\) of a real number \(x\) is its distance from zero on the real line; \(|x-a|\) measures the distance between \(x\) and \(a\).
It is defined by cases: \(|x|=x\) if \(x\ge0\) and \(|x|=-x\) if \(x<0\). The basic properties follow: \(|x|\ge0\), \(|x|=0\) only if \(x=0\), \(|xy|=|x||y|\) and \(-|x|\le x\le|x|\). A frequent toolbox: \(|x|<c\) is equivalent to \(-c<x<c\), and \(|x-a|<\varepsilon\) is equivalent to \(a-\varepsilon<x<a+\varepsilon\): the points at distance less than \(\varepsilon\) from \(a\).
The central property is the triangle inequality, \(|x+y|\le|x|+|y|\). It is proved by adding \(-|x|\le x\le|x|\) and \(-|y|\le y\le|y|\). A useful consequence is the reverse version, \(\big||x|-|y|\big|\le|x-y|\), which gives lower bounds.
For example, if \(|x|=3\) and \(|y|=4\), then \(|x+y|\) can be anywhere between \(1\) and \(7\); the maximum occurs when both have the same sign. The typical mistake is writing \(|x+y|=|x|+|y|\) or \(|x-y|=|x|-|y|\); both fail in general.
Another constant use: if \(|x|<\varepsilon\) for every \(\varepsilon>0\), then \(x=0\). This argument appears when proving that a limit is unique. The absolute value gives the distance \(d(x,y)=|x-y|\) that will be used to define convergent sequences.
Think about it
Prove that \(|x-z|\le|x-y|+|y-z|\).
Write \(x-z=(x-y)+(y-z)\) and apply the triangle inequality to those two terms.
For which \(x\) does \(|x-3|=|x+1|\) hold?
Only for \(x=1\), the point equidistant from \(3\) and \(-1\).
Exercises 10
- Compute \(|-7.5|\).
- Compute \(|3-8|+|-2|\).
- Solve \(|x-2|=5\).
- The inequality \(|x-4|<2\) is equivalent to:
- Which equality holds for all reals \(x,y\)?
- Solve \(|2x-1|\le5\).
- If \(|x|=3\) and \(|y|=4\), what is the largest possible value of \(|x+y|\)?
- If \(|x|=5\) and \(|y|=2\), what is the smallest possible value of \(|x-y|\)?
- If \(|x|<\varepsilon\) for every \(\varepsilon>0\), then:
- The solution set of \(|x-1|+|x+1|=2\) is:
What comes next
References
- Stephen Abbott, Understanding Analysis, 2nd edition, Springer, 2015. §1.2
- Walter Rudin, Principles of Mathematical Analysis, 3rd edition, McGraw-Hill, 1976. Cap. 1, 1.33
- Robert G. Bartle, Donald R. Sherbert, Introduction to Real Analysis, 4th edition, Wiley, 2011. §2.2