Algebra and Trigonometry | Adaptive Books

It is frequently desirable to measure how large a quantity is, regardless of its sign. In such cases, we use merely the absolute value of the quantity.

Definition 1.1 . The absolute value (or modulus) of a real number \(x\) , is a nonnegative real number denoted by \(|x|\) , and defined as follows \[|x|=\begin{cases} x & \text{if }x>0\\ 0 & \text{if }x=0\\ -x & \text{if }x<0 \end{cases}\]

Geometrically the absolute value of a number \(x\) is its distance from 0 regardless of the direction.

Example 1.2 . Find \(|2|,|-2|,|0|,|3-\sqrt{3}|,|\sqrt{3}-3|\)

Solution

Because \(2,0,3-\sqrt{3}\) are nonnegative [ \(\sqrt{3}\approx1.73\) , so \(3-\sqrt{3}\approx1.27\) ] , we have \[|2|=2,\ |0|=0,\ |3-\sqrt{3}|=3-\sqrt{3}\] but \(-2\) and \(\sqrt{3}-3\) are negative, so we have \[|-2|=-(-2)=2,\quad|\sqrt{3}-3|=-(\sqrt{3}-3)=3-\sqrt{3}.\]

In computer languages and mathematical packages, the absolute value of \(x\) is often denoted by abs(x) .
We have by definition: \[-|x|\leq x\leq|x|\] because if \(x>0\) , then \(|x|=x\) and we have the sign of equality on the right and the sign of inequality on the left (a positive number is larger than a negative one). If \(x<0\) , then \(|x|=-x\) , and we have the sign of equality on the left and the sign of inequality on the right.
[ Note that \(a\leq b\) means \(aor \(a=b\) ]

From the above definition, it follows that for every real numbers \(a\) and \(b\) we have:

\(|a|\geq0\)
\(|a|=0\) if and only if \(a=0\)
\(||a||=|a|\) (because \(|a|\ge0\) )
\(|-a|=|a|\) .
\(|ab|=|a|\ |b|\) .
\(\left|\dfrac{a}{b}\right|=\dfrac{|a|}{|b|}\) (provided \(b\neq0\) )
If \(r>0\)

\[\bbox[5px,border:1px solid black;background-color:#f2f2f2]{|x|\le r\qquad\text{is equivalent to}\qquad-r\le x\le r}\tag{i}\] \[\bbox[5px,border:1px solid black;background-color:#f2f2f2]{r\le|x|\qquad\text{is equivalent to}\qquad x\le-r\quad\text{or }\quad r\le x}\tag{ii}\] \[\bbox[5px,border:1px solid black;background-color:#f2f2f2]{|x|=r\qquad\text{is equivalent to}\qquad x=r\quad\text{or}\quad x=-r}\tag{iii}\]

Geometric Interpretation

Let \(r>0\) and \[|x|\leq r.\] The above relationship implies that \(x\) is nearer to 0 than \(r\) , and you can see from the following figure that this is possible if and only if \(x\) lies between \(-r\) and \(r\) : \[-r\leq x\leq r.\]

You can show the significance of the other two statements geometrically.

\(|a+b|\leq|a|+|b|\) (known as triangle inequality )

If \(a\) and \(b\) are either both positive, both negative, or at least one of them is zero, then \(|a+b|=|a|+|b|\) . Otherwise, when \(a\) and \(b\) have opposite signs, \(|a+b|<|a|+|b|\) .

Proof

The proof of the triangle inequality is as follows: we know \[-|a|\leq a\leq|a|\] and \[-|b|\leq b\leq|b|.\] Adding two inequalities we obtain \[-(|a|+|b|)\leq a+b\leq|a|+|b|.\] Let \(c=|a|+|b|\) and \(x=a+b\) . Because \(-c\leq x\leq c\) , it follows from (1) that \(|x|\leq c\) or \[|x|=|a+b|\leq|a|+|b|.\] This is what we were trying to prove.

Example 1.3 . Show that \[\bbox[5px,border:1px solid black;background-color:#f2f2f2]{|a| - |b| \leq |a+b| \leq |a| + |b|.}\]

Solution

This result follows from the triangle inequality and the fact that \(|-b| = |b|\) (Property 4). Using the triangle inequality, we have: \[|a| = |(a+b) - b| \leq |a+b| + |-b| = |a+b| + |b|.\] By subtracting \(|b|\) from both sides of the inequality, we obtain: \[|a| - |b| \leq |a+b|.\] Combining the above inequality with the standard triangle inequality \(|a+b| \leq |a| + |b|\) , we conclude that: \[|a| - |b| \leq |a+b| \leq |a| + |b|.\]

Example 1.4 . Prove that for all real number \(a\) and \(b\) , we have \[|a-b|\leq|a|+|b|\]

Solution

This follows directly from the triangle inequality and the fact that \(|-b| = |b|\) . Applying the triangle inequality to \(a + (-b)\) , we get: \[|a-b| = |a + (-b)| \leq |a| + |-b| = |a| + |b|.\] Thus, \(|a-b| \leq |a| + |b|\) holds for all real numbers \(a\) and \(b\) .

Example 1.5 . Prove that for all real numbers \(a\) and \(b\) , \[|a| - |b| \leq |a-b|.\]

Solution

We start by adding and subtracting \(b\) and then applying the triangle inequality. Specifically: \[|a| = |(a-b) + b| \leq |a-b| + |b|.\] Subtracting \(|b|\) from both sides gives: \[|a| - |b| \leq |a-b|.\]

Thus, the inequality \(|a| - |b| \leq |a-b|\) holds for all real numbers \(a\) and \(b\) .

It follows from (4) that

\(|a-b|=|b-a|\)

Distance Between Points on the Real Line

Look at the following figure. The distance between \(-2\) and \(10\) is \(12\) units. The distance can be calculated as the difference \(10 - (-2)\) , which involves subtracting the smaller value from the larger one. However, since the absolute value of \(10 - (-2)\) equals \(|12| = 12\) , and similarly, \(|-2 - 10| = |-12| = 12\) , we can use the absolute value function to find the distance between two points without worrying about which number is larger or smaller.

Definition 1.2 . If \(P\) and \(Q\) are two points located on a number line with coordinates \(a\) and \(b\) , the distance between \(P\) and \(Q\) , denoted by \(d(P, Q)\) , is calculated as: \[d(P, Q) = |b - a|\]

Since \(|b - a|\) is the same as \(|a - b|\) , it follows that the distance from \(P\) to \(Q\) is equal to the distance from \(Q\) to \(P\) : \[d(P,Q)=d(Q,P).\]

It follows from Equation (ii) that for any real number \(a\) and any positive number \(r\) , \[|x-a|is equivalent to \[-ror \[a-rThis means that the distance of \(x\) from \(a\) is less than \(r\) if and only if \(x\) is between \(a-r\) and \(a+r\) .

Example 1.6 . Find the set \(I\) , if it consists of all points whose distance from the point 2 is less than 0.6.

Solution

As discussed above \[\begin{align} I & =\{x|\ |x-2|<0.6\}\\ & =\{x|\ -0.6This set is graphed in the following figure.

The set of all points whose distance from 2 is less than 0.6

Neighborhoods

Let \(I\) be the set of all points whose distance from a fixed point \(a\) is less than a number \(\delta>0\) . Then

\[\begin{align} I & =\{x|\ |x-a|<\delta\}\\ & =\{x|\ -\delta

Such a set is called a neighborhood (or more precisely the \(\delta\) -neighborhood) of \(a\) and \(\delta\) is called the radius of the neighborhood. The \(\delta\) -neighborhood of \(a\) is shown in the following figure.

The \(\delta\) -neighborhood of \(a\) is the set of all points whose distance from \(a\) is less than \(\delta>0\) . The radius of this neighborhood is \(\delta\) .

Now let’s consider the set of all points such that \[0<|x-a|<\delta\] or \[J=\{x|\ 0<|x-a|<\delta\}\] Here we have two inequalities \[0<|x-a|\quad\text{and}\quad|x-a|<\delta\] Recall that the absolute value is always nonnegative (that is \(0\le|t-a|\) for all \(t\) ) , so \[0<|x-a|\] means \[|x-a|\neq0\] or equivalently \[x\neq a\] [ Recall that \(|t|=0\) if and only if \(t=0\) ] . Therefore, \[\begin{align} J & =\{x|\quad0<|x-a|<\delta\}\\ & =\{x|\ |x-a|<\delta\quad\text{and}\quad x\neq a\} \end{align}\] That is \(J\) is the \(\delta\) -neighborhood of \(a\) with the midpoint \(a\) removed. The set \(J\) is called the deleted neighborhood or punctured neighborhood of \(a\) . The deleted \(\delta\) -neighborhood of \(a\) is shown in the following figure.

The deleted \(\delta\) -neighrbood of \(a\) is the \(\delta\) -neighborhood of \(a\) minus the midpoint \(a\) .