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 \( z = a+bi \):

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

## 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.

No vote yet

24 votes

×

Problem Loading...

Note Loading...

Set Loading...

## Comments

Sort by:

TopNewestTypo in #1, Sol'n 2: \( (x−3)^2=|x−2∣^2 \). I believe it should be \( |x-3|^2 \).

Log in to reply

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 \(xy\leq|xy|=|x||y|\).

Log in to reply

How is the result in the fist line of Solution 2 for Example 1 made? Apologies, I'm a little confused.

Log in to reply

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

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 \(x\leq -1, x\geq 1\)

Log in to reply

Updated. Thanks!

Log in to reply

Welcome!

Log in to reply

Log in to reply

Hi,calvin l. I read this article on high school in my country, When you learn it?

Log in to reply

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. ....

Log in to reply

Nice post Master Calvin, I am waiting it ready to post in the blog.. :D

Log in to reply