Absolute value¶
Exercises · Further topics · with worked solutions · Lecture notes (PDF) · Chapter A.1: Absolute value
Exercise 1
Using the definition of absolute value, compute:
Solution
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a) Since \(-3<0\), \(2-7=-5<0\), \(-4<0\) and \(\frac{1}{2} \ge 0\):
\[ |-3| + |2-7| - |-4| \; \left|\frac{1}{2}\right| = 3 + 5 - 4 \cdot \frac{1}{2} = 6 \] -
b) Since \(\pi > 3\), we have \(3 - \pi < 0\) and then \(|3-\pi| = -(3-\pi) = \pi - 3\).
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c) Since \(\frac{1}{3} - \frac{1}{2} = -\frac{1}{6} < 0\), we have \(\left| \frac{1}{3} - \frac{1}{2}\right| = \frac{1}{6}\).
Exercise 2
Prove that for every real number \(b\) we have:
Solution
We distinguish the two cases of the definition of absolute value.
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If \(b \ge 0\), then \(|b| = b \ge 0\); moreover \(|b|=0\) if and only if \(b=0\); we have \(-|b| = -b \le 0 \le b = |b|\); finally \(|b|^2 = b^2\).
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If \(b < 0\), then \(|b| = -b > 0\) (in particular \(|b| \neq 0\)); we have \(-|b| = b < 0 < -b = |b|\); finally \(|b|^2 = (-b)^2 = b^2\).
In both cases a), b) and c) hold.
Exercise 3
Find all real numbers \(x\) such that:
Solution
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a) From \(|a| < \varepsilon \Longleftrightarrow -\varepsilon < a < \varepsilon\) with \(a = x-2\) and \(\varepsilon = 3\):
\[ -3 < x - 2 < 3 \Longleftrightarrow -1 < x < 5 \] -
b) From \(|a| \le \varepsilon \Longleftrightarrow -\varepsilon \le a \le \varepsilon\) with \(a = 2\:x+1\) and \(\varepsilon = 5\):
\[ -5 \le 2\:x + 1 \le 5 \Longleftrightarrow -6 \le 2\:x \le 4 \Longleftrightarrow -3 \le x \le 2 \] -
c) The inequality \(|x-4| \ge 1\) is false exactly when \(|x - 4| < 1\), that is when \(-1 < x-4 < 1\), i.e., \(3 < x < 5\). Then the solutions are:
\[ x \le 3 {\rm ~~~or~~~} x \ge 5 \]
Exercise 4
Find all real numbers \(x\) such that \(|x-1| = |x+3|\).
Solution
Two real numbers have the same absolute value if and only if they are equal or opposite. Then:
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\(x - 1 = x + 3\) gives \(-1 = 3\), which is impossible;
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\(x - 1 = -(x+3)\) gives \(2\:x = -2\), that is \(x = -1\).
The only solution is \(x=-1\). Check: \(|-1-1| = 2 = |-1+3|\).
Exercise 5
Find all real numbers \(x\) such that \(|x| + |x-2| = 4\).
Solution
The signs of \(x\) and \(x-2\) change at \(x=0\) and \(x=2\), so we distinguish three cases:
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\(x < 0\): \(|x| = -x\) and \(|x-2| = 2-x\), then \(-2\:x + 2 = 4\), i.e., \(x = -1\), which satisfies \(x<0\);
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\(0 \le x < 2\): \(|x| = x\) and \(|x-2| = 2-x\), then \(x + 2 - x = 2 = 4\), which is impossible;
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\(x \ge 2\): \(|x| = x\) and \(|x-2| = x-2\), then \(2\:x - 2 = 4\), i.e., \(x = 3\), which satisfies \(x \ge 2\).
The solutions are \(x=-1\) and \(x=3\).
Exercise 6
Prove that for every pair of real numbers \(b\) and \(c\):
Solution
Both \(|b+c|\) and \(|b|+|c|\) are non-negative, and two non-negative numbers are equal if and only if their squares are equal. Using Exercise Exercise 2 c) and \(|b\:c| = |b|\:|c|\), we have:
Then \(|b+c| = |b| + |c|\) if and only if \(b\:c = |b\:c|\), that is (by definition of absolute value) if and only if \(b\:c \ge 0\).
Exercise 7
Verify the triangle inequality \(|b+c| \le |b| + |c|\) and the reverse triangle inequality \(\big|\,|b| - |c|\,\big| \le |b - c|\) for \(b = 5\) and \(c = -3\).
Solution
We have \(|b| = 5\) and \(|c| = 3\).
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Triangle inequality: \(|b+c| = |5-3| = 2 \le 8 = |b| + |c|\).
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Reverse triangle inequality: \(\big|\,|b| - |c|\,\big| = |5 - 3| = 2 \le 8 = |5-(-3)| = |b-c|\).
The triangle inequality holds with the strict inequality, in agreement with Exercise Exercise 6, since \(b\:c = -15 < 0\).
Exercise 8
Let \(x\) be a real number such that \(|x - 3| \le 1\). Prove that:
and show that the bound in b) cannot be improved.
Solution
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a) From the form \(|g| \le |g-h| + |h|\) of the triangle inequality, with \(g = x\) and \(h = 3\):
\[ |x| \le |x-3| + |3| \le 1 + 3 = 4 \] -
b) We have \(x^2 - 9 = (x-3)\:(x+3)\) and then \(|x^2 - 9| = |x-3| \; |x+3|\). From the triangle inequality:
\[ |x+3| = |(x-3) + 6| \le |x-3| + 6 \le 7 \]and therefore \(|x^2-9| = |x-3|\; |x+3| \le 1 \cdot 7 = 7\).
The bound cannot be improved: for \(x=4\) (which satisfies \(|4-3| = 1\)) we have \(|16 - 9| = 7\).