For a real number, the absolute value of \(x\), denoted \( \lvert x \rvert \), is defined as

\[ \left|x \right| = \begin{cases} x, & \text{if } x \geq 0 \\ -x, & \text{if } x < 0. \end{cases} \]

Here are some properties of the absolute value:

Non-negative: \( \vert x \rvert \geq 0 \) for all real values.

\( \lvert x \rvert = \sqrt{ x^2 }\) and \(\lvert x \rvert^2 = x^2 \) for all real values.

Multiplicative: \( \lvert xy \rvert = \lvert x \rvert \cdot \lvert y \rvert \).

Sub-additive: \( \lvert x+y \rvert \leq \lvert x \rvert + \lvert y \rvert \).

Symmetry: \( \lvert -x \rvert = \lvert x \rvert \).

Distance to origin: \( \lvert x \rvert \) measures the geometric distance of the real number \(x\) to the origin \(0\). Since distance is always positive, the absolute value of a number is always positive. Thinking of absolute value as the distance from zero is also helpful when considering complex numbers \( z = a + bi \). This distance from \( z \) to the origin is given by the distance formula: \[ \text{distance}_{(a,b)\leftrightarrow(0,0)} = \sqrt{(a-0)^2 + (b-0)^2} \] In fact, this is also the definition of the absolute value for a complex number \( z = a+bi \):

\[ \left| z \right| = \sqrt{a^2 + b^2} \]

## Worked Examples

## 1. Find all real values \(x\) satisfying \( \lvert x - 3 \rvert = 5 \).

Solution 1: If \( x-3 \geq 0 \), then we need to solve \( x-3 = 5 \), which gives \( x = 8 \). This satisfies the original condition of \( x - 3 \geq 0 \), hence is a valid solution. If \( x-3 < 0 \), then we need to solve \( -(x-3) = 5 \), which gives \( -x+3 = 5 \) or \( x = -2 \). This satisfies the original condition of \( x-3 < 0 \), hence is a valid solution.

Solution 2: Using property 2, we obtain \( (x-3)^2 = \lvert x-3\rvert ^2 = 5^2\), or \( x^2 - 6x + 9 = 25 \), which reduces to \( 0 = x^2 -6x - 16 = (x-8)(x+2) \). This has solutions \(x=8, -2 \). We can verify that both of these are solutions.

## 2. How many real values \(x\) satisfy \( \lvert x^2 -1 \rvert = \lvert x-2 \rvert \)?

Solution: Let's approach this problem by considering the different regions. We have \( x^2 -1 \geq 0 \Leftrightarrow x\leq -1, x \geq 1 \). Also, \( x -2 \geq 0 \Leftrightarrow x \geq 2 \).

If \(x \leq -1 \) then we have \( x^2 -1 = 2 - x \), or \( x^2 + x - 3 = 0 \). This has roots \( \frac{-1 \pm \sqrt{ 1^2 - 4 \times 1 \times (-3) }} {2} \), of which only the choice of \( -\) satisfies \( x < -1 \). There is one solution in this case.

If \( -1 \leq x \leq 1 \), then we have \( 1 - x^2 = 2 - x \), or \( x^2 -x + 1 = 0 \). This has roots \( \frac { 1 \pm \sqrt{ 1^2 - 4 \times 1 \times 1 } } {2} \) which are not real. Hence, there are no solutions in this case.

If \( 1 \leq x \leq 2 \), then we have \( x^2 -1 = 2 - x \), or \( x^2 + x - 3 = 0\). We check that the root \( \frac{ -1 + \sqrt{13} } { 2} \) is within this domain. Hence, there is one solution in this case.

If \( 2 \leq x \), then we have \( x^2-1 = x - 2 \), or \( x^2 - x + 1 = 0 \). This has no real root.

Hence, there are two real values \(x\) which satisfy \( \vert x^2 -1 \rvert = \lvert x-2 \rvert \).

## 3. Prove the sub-additive property:

\[ \lvert x+y \rvert \leq \lvert x \rvert + \lvert y \rvert.\]

Solution 1: We apply the triangle inequality to the triangle with vertices \( O = 0 \), \( A = x \), \( B = -y \) on the real number line. Then \( AB \leq OA + OB \), implying \( \lvert x - (-y) \rvert \leq \lvert x \rvert + \lvert y \rvert \).

Solution 2:

If \( x, y \geq 0 \), then \( \lvert x+y \rvert = x+y, \lvert x \rvert + \lvert y \rvert = x + y \), so \( \lvert x+y \rvert \leq \lvert x \rvert + \lvert y \rvert \).

If \( x, y \leq 0 \), then \( \lvert x+y \rvert = -x -y, \lvert x \rvert = -x, \lvert y \rvert = - y \), so \( \lvert x+y \rvert \leq \lvert x \rvert + \lvert y \rvert \).

If \( x \geq -y \geq 0 \), then \( \lvert x+y \rvert = x+y, \lvert x \rvert = x, \lvert y \rvert = -y \) so \( \lvert x+y \rvert = x+y \leq x -y = \lvert x \rvert + \lvert y \rvert\). Likewise, if \( y \geq -x \geq 0 \).

If \( x \leq -y \leq 0 \), then \( \lvert x+y \rvert = -x - y , \lvert x \rvert = -x, \lvert y \rvert = y\), so \( \lvert x+y \rvert = -x - y \leq -x + y = \lvert x \rvert + \lvert y \rvert \). Likewise, if \( y \leq x \leq 0 \).

For more problems, see the Technique Trainer.

## Comments

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TopNewestTypo in #1, Sol'n 2: \( (x−3)^2=|x−2∣^2 \). I believe it should be \( |x-3|^2 \). – Jonathan Wong · 3 years, 10 months ago

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– Calvin Lin Staff · 3 years, 10 months ago

Updated. Thanks!Log in to reply

An alternative proof of the sub-additive property:

Recall that if \(a,b\geq0\), then \(a\leq b\) is equivalent to \(a^2\leq b^2\).

Let \(a=|x+y|,b=|x|+|y|\). We now need to show that \(a^2\leq b^2\), i.e. \[x^2+y^2+2xy \leq x^2+y^2+2|x||y|.\]This is true since \(xy\leq|xy|=|x||y|\). – Ang Yan Sheng · 3 years, 10 months ago

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How is the result in the fist line of Solution 2 for Example 1 made? Apologies, I'm a little confused. – Skyler Hill · 3 years, 10 months ago

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– Sotiri Komissopoulos · 3 years, 10 months ago

If you are referring to the "\((x - 3)^2 = |x - 2|^2\)," it is simply a typo and should read "\(|x - 3|^2\)."Log in to reply

– Skyler Hill · 3 years, 10 months ago

Yes, I was just trying to confirm that it was a typo. Thank you!Log in to reply

Typo in #2: In the first line of math text of the solution, it should read \(x\leq -1, x\geq 1\) – Bob Krueger · 3 years, 10 months ago

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– Calvin Lin Staff · 3 years, 10 months ago

Updated. Thanks!Log in to reply

– Vamsi Krishna Appili · 3 years, 10 months ago

Welcome!Log in to reply

– Tim Ye · 3 years, 10 months ago

You didn't even say anything,Log in to reply

Hi,calvin l. I read this article on high school in my country, When you learn it? – Hamid Hassani · 3 years, 10 months ago

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– Calvin Lin Staff · 3 years, 10 months ago

I'm not certain what you mean. Absolute value is generally taught in high school.Log in to reply

awesum post......i was really in need of clarification over this topic....thank u Calvin Sir. .... – Riya Gupta · 3 years, 10 months ago

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Nice post Master Calvin, I am waiting it ready to post in the blog.. :D – Andrias Yuwantoko · 3 years, 10 months ago

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