Definition A.3.2.
The absolute value of a number is the distance between that number and \(0\) on a number line. For the absolute value of \(x\text{,}\) we write \(\abs{x}\text{.}\)
Section A.3 Absolute Value and Square Root
In this section, we will learn the basics of absolute value and square root. These are actions you can do to a given number, often changing the number into something else.
Subsection A.3.1 Introduction to Absolute Value
Let’s look at \(\abs{2}\) and \(\abs{-2}\text{,}\) the absolute value of \(2\) and the absolute value of \(-2\text{.}\)
Since the distance between \(2\) and \(0\) on the number line is \(2\) units, the absolute value of \(2\) is \(2\text{.}\) We write \(\abs{2}=2\text{.}\)
Since the distance between \(-2\) and \(0\) on the number line is also \(2\) units, the absolute value of \(-2\) is also \(2\text{.}\) We write \(\abs{-2}=2\text{.}\)
Fact A.3.4. Absolute Value.
Taking the absolute value of a number results in whatever the “positive version” of that number is. This is because the real meaning of absolute value is its distance from zero.
Checkpoint A.3.5. Calculating Absolute Value.
Try calculating some absolute values.
(a)
\(\abs{57}=\)
(b)
\(\abs{-43}=\)
(c)
\(\abs{\frac{2}{-5}}=\)
Warning A.3.6. Absolute Value Does Not Exactly “Make Everything Positive”.
Students may see an expression like \(\abs{2-5}\) and incorrectly think it is OK to “make everything positive” and write \(2+5\text{.}\) This is incorrect since \(\abs{2-5}\) works out to be \(3\text{,}\) not \(7\text{,}\) as we are actually taking the absolute value of \(-3\) (the equivalent number inside the absolute value).
Subsection A.3.2 Square Root Facts
If you have learned your basic multiplication table, you know:
| \(\times\) | \(1\) | \(2\) | \(3\) | \(4\) | \(5\) | \(6\) | \(7\) | \(8\) | \(9\) |
| \(1\) | \(1\) | \(\lowlight{2}\) | \(\lowlight{3}\) | \(\lowlight{4}\) | \(\lowlight{5}\) | \(\lowlight{6}\) | \(\lowlight{7}\) | \(\lowlight{8}\) | \(\lowlight{9}\) |
| \(2\) | \(\lowlight{2}\) | \(4\) | \(\lowlight{6}\) | \(\lowlight{8}\) | \(\lowlight{10}\) | \(\lowlight{12}\) | \(\lowlight{14}\) | \(\lowlight{16}\) | \(\lowlight{18}\) |
| \(3\) | \(\lowlight{3}\) | \(\lowlight{6}\) | \(9\) | \(\lowlight{12}\) | \(\lowlight{15}\) | \(\lowlight{18}\) | \(\lowlight{21}\) | \(\lowlight{24}\) | \(\lowlight{27}\) |
| \(4\) | \(\lowlight{4}\) | \(\lowlight{8}\) | \(\lowlight{12}\) | \(16\) | \(\lowlight{20}\) | \(\lowlight{24}\) | \(\lowlight{28}\) | \(\lowlight{32}\) | \(\lowlight{36}\) |
| \(5\) | \(\lowlight{5}\) | \(\lowlight{10}\) | \(\lowlight{15}\) | \(\lowlight{20}\) | \(25\) | \(\lowlight{30}\) | \(\lowlight{35}\) | \(\lowlight{40}\) | \(\lowlight{45}\) |
| \(6\) | \(\lowlight{6}\) | \(\lowlight{12}\) | \(\lowlight{18}\) | \(\lowlight{24}\) | \(\lowlight{30}\) | \(36\) | \(\lowlight{42}\) | \(\lowlight{48}\) | \(\lowlight{54}\) |
| \(7\) | \(\lowlight{7}\) | \(\lowlight{14}\) | \(\lowlight{21}\) | \(\lowlight{28}\) | \(\lowlight{35}\) | \(\lowlight{42}\) | \(49\) | \(\lowlight{56}\) | \(\lowlight{63}\) |
| \(8\) | \(\lowlight{8}\) | \(\lowlight{16}\) | \(\lowlight{24}\) | \(\lowlight{32}\) | \(\lowlight{40}\) | \(\lowlight{48}\) | \(\lowlight{56}\) | \(64\) | \(\lowlight{72}\) |
| \(9\) | \(\lowlight{9}\) | \(\lowlight{18}\) | \(\lowlight{27}\) | \(\lowlight{36}\) | \(\lowlight{45}\) | \(\lowlight{54}\) | \(\lowlight{63}\) | \(\lowlight{72}\) | \(81\) |
The numbers along the diagonal are special; they are known as perfect squares. And for working with square roots, it will be helpful if you can memorize these first few perfect square numbers.
“Taking a square root” is the opposite action of squaring a number. For example, when you square \(3\text{,}\) the result is \(9\text{.}\) So when you take the square root of \(9\text{,}\) the result is \(3\text{.}\) Just knowing that \(9\) comes about as \(3^2\) lets us realize that \(3\) is the square root of \(9\text{.}\) This is why memorizing the perfect squares from the multiplication table can be so helpful.
The notation we use for taking a square root is the radical, \(\sqrt{\phantom{x}}\text{.}\) For example, “the square root of \(9\)” is denoted \(\sqrt{9}\text{.}\) And now we know enough to be able to write \(\sqrt{9}=3\text{.}\)
Tossing in a few extra special square roots, it’s advisable to memorize the following:
| \(\sqrt{0}=0\) | \(\sqrt{1}=1\) | \(\sqrt{4}=2\) | \(\sqrt{9}=3\) |
| \(\sqrt{16}=4\) | \(\sqrt{25}=5\) | \(\sqrt{36}=6\) | \(\sqrt{49}=7\) |
| \(\sqrt{64}=8\) | \(\sqrt{81}=9\) | \(\sqrt{100}=10\) | \(\sqrt{121}=11\) |
| \(\sqrt{144}=12\) | \(\sqrt{169}=13\) | \(\sqrt{196}=14\) | \(\sqrt{225}=15\) |
Subsection A.3.3 Calculating Square Roots with a Calculator
Most square roots are actually numbers with decimal places that go on forever. Take \(\sqrt{5}\) as an example:
\begin{align*}
\sqrt{4}\amp=2\amp\sqrt{5}\amp=\mathord{?}\amp\sqrt{9}\amp=3
\end{align*}
Since \(5\) is between \(4\) and \(9\text{,}\) then \(\sqrt{5}\) must be somewhere between \(2\) and \(3\text{.}\) There are no whole numbers between \(2\) and \(3\text{,}\) so \(\sqrt{5}\) must be some number with decimal places. If the decimal places eventually stopped, then squaring it would give you another number with decimal places that stop further out. But squaring it gives you \(5\) with no decimal places. So the only possibility is that \(\sqrt{5}\) is a decimal between \(2\) and \(3\) that goes on forever. With a calculator, we can see:
\begin{equation*}
\sqrt{5}\approx2.236
\end{equation*}
Actually the decimal will not terminate, and that is why we used the \(\approx\) symbol instead of an equal sign. To get \(2.236\) we rounded down slightly from the true value of \(\sqrt{5}\text{.}\) With a calculator, we can check that \(2.236^2=4.999696\text{,}\) a little shy of \(5\text{.}\)
Subsection A.3.4 Square Roots of Fractions
We can calculate the square root of some fractions by hand, such as \(\sqrt{\frac{1}{4}}\text{.}\) The idea is the same: can you think of a number that you would square to get \(\frac{1}{4}\text{?}\) Being familiar with fraction multiplication, we know that \(\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{4}\) and so \(\sqrt{\frac{1}{4}}=\frac{1}{2}\text{.}\)
Checkpoint A.3.8. Square Roots of Fractions.
Try calculating some absolute values.
(a)
\(\sqrt{\dfrac{1}{25}}=\)
(b)
\(\sqrt{\dfrac{4}{9}}=\)
(c)
\(\sqrt{\dfrac{81}{121}}=\)
Subsection A.3.5 Square Root of Negative Numbers
Aside: Imaginary Numbers.
Can we find the square root of a negative number, such as \(\sqrt{-25}\text{?}\) That would mean that there is some number out there that multiplies by itself to make \(-25\text{.}\) Would \(\sqrt{-25}\) be positive or negative? Either way, once you square it (multiply it by itself) the result would be positive. So it couldn’t possibly square to \(-25\text{.}\) So there is no square root of \(-25\) or of any negative number for that matter.
If you are confronted with an expression like \(\sqrt{-25}\text{,}\) or any other square root of a negative number, you can state that “there is no real square root” or that the result “does not exist” (as a real number).
Exercises A.3.6 Exercises
Review and Warmup.
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Absolute Value.
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Evaluate the following.
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\(\displaystyle{ \left\lvert 4 \right\rvert = }\)
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\(\displaystyle{ \left\lvert -5 \right\rvert = }\)
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\(\displaystyle{ \left\lvert 0 \right\rvert = }\)
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\(\displaystyle{ \left\lvert {15+\left(-7\right)} \right\rvert = }\)
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\(\displaystyle{ \left\lvert {-10-\left(-1\right)} \right\rvert = }\)
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Evaluate the following.
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\(\displaystyle{ \left\lvert 5 \right\rvert = }\)
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\(\displaystyle{ \left\lvert -8 \right\rvert = }\)
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\(\displaystyle{ \left\lvert 0 \right\rvert = }\)
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\(\displaystyle{ \left\lvert {20+\left(-10\right)} \right\rvert = }\)
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\(\displaystyle{ \left\lvert {-10-\left(-3\right)} \right\rvert = }\)
Exercise Group.
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Square Roots.
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Exercise Group.
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Exercise Group.
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Exercise Group.
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