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# 2.5: Square Roots and Real Numbers

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Find square roots.
• Approximate square roots.
• Identify irrational numbers.
• Classify real numbers.
• Graph and order real numbers.

## Find Square Roots

The square root of a number is a number which, when multiplied by itself gives the original number. In algebraic terms, the square root of $x$ is a number, $b$, such that $b^2=x$.

Note: There are two possibilities for a numerical value for $b$. The positive number that satisfies the equation $b^2=x$ is called the principal square root. Since $(-b)\cdot (-b)=+b^2=x, -b$ is also a valid solution.

The square root of a number, $x$, is written as $\sqrt {x}$ or sometimes as $\sqrt[2]{x}$ . For example, $2^2=4$, so the square root of 4, $\sqrt{4}= \pm 2$ .

Some numbers, like 4, have integer square roots. Numbers with integer square roots are called perfect squares. The first five perfect squares (1, 4, 9, 16, 25) are shown below.

You can determine whether a number is a perfect square by looking at its prime factors. If every number in the factor tree appears an even number of times, the number is a perfect square. Further, to find the square root of that number, simply take one of each pair of factors and multiply them together.

Example 1

Find the principal square root of each of these perfect squares.

a) 121

b) 225

c) 324

d) 576

a) $121 = 11 \times 11$

Solution

$\sqrt {121}=11$

b) $225 = (5 \times 5) \times (3 \times 3)$

Solution

$\sqrt {225}=5\times 3=15$

c) $324 = (2 \times 2) \times (3 \times 3) \times (3 \times 3)$

Solution

$\sqrt {324}=2\times 3\times 3=18$

d) $576 = (2 \times 2) \times (2 \times 2) \times (2 \times 2) \times (3 \times 3)$

Solution

$\sqrt {576}=2\times 2\times 2\times 3=24$

When we have an odd number of prime factors, we leave any unpaired factors under a radical sign. Any answer that contains both whole numbers and irreducible radicals should be written $A\sqrt {b}$.

Example 2

Find the principal square root of the following numbers.

a) 8

b) 48

c) 75

d) 216

a) $8 = (2 \times 2) \times 2$

Solution

$\sqrt {8}=2\times \sqrt {2}=2\sqrt {2}$

b) $48 = (2 \times 2) \times (2 \times 2) \times 3$

Solution

$\sqrt {48}=2\times 2\times\sqrt{3}=4\sqrt{3}$

c) $75 = (5 \times 5) \times 3$

Solution

$\sqrt {75}=5\times \sqrt{3}=5\sqrt{3}$

d) $216 = (2 \times 2) \times 2 \times (3 \times 3) \times 3$

Solution

$\sqrt {216}=2\times \sqrt{2}\times 3\times \sqrt{3}=6\sqrt{2}\sqrt{3}=6\sqrt{6}$

Note that in the last example we collected the whole numbers and multiplied them first, then we collect unpaired primes under a single radical symbol. Here are the four rules that govern how we treat square roots.

$\sqrt{a}\times\sqrt{b} & =\sqrt{ab} & \sqrt{a}\div\sqrt{b}& =\sqrt{\frac {a}{b}} \\A\sqrt{a}\times B\sqrt{b}& =AB\sqrt{ab} & A\sqrt{a}\div B\sqrt{b} &=\frac {A}{B}\sqrt{\frac {a}{b}}$

Example 3

Simplify the following square root problems

a) $\sqrt{8}\times\sqrt{2}$

b) $3\sqrt{4}\times 4\sqrt{3}$

c) $\sqrt{12}\div \sqrt{3}$

d) $12\sqrt{10}\div 6\sqrt{5}$

a) $\sqrt {8}\times \sqrt{2}=16$

Solution

$\sqrt {8}\times \sqrt {2}=4$

b) $3\sqrt {4}\times 4\sqrt{3}=12\sqrt {12}=12\sqrt {(2\times 2)\times 3}=12\times 2\sqrt {3}$

Solution

$3\sqrt {4}\times 4\sqrt{3}=24\sqrt{3}$

c) $\sqrt {12}\div \sqrt{3}=\sqrt{\frac {12}{3}}=\sqrt{4}$

Solution

$\sqrt {12}\div \sqrt{3}=2$

d) $12\sqrt {10}\div 6\sqrt{5}=\frac {12}{6}\sqrt{\frac {10}{5}}$

Solution

$12\sqrt {10}\div 6\sqrt{5}=2\sqrt{2}$

## Approximate Square Roots

When we have perfect squares, we can write an exact numerical solution for the principal square root. When we have one or more unpaired primes in the factor tree of a number, however, we do not get integer values for the square root and we have seen that we leave a radical in the answer. Terms like $\sqrt {2}$, $\sqrt {3}$ and $\sqrt {7}$ (square roots of prime numbers) cannot be written as rational numbers. That is to say, they cannot be expressed as the ratio of two integers. We call them irrational numbers. In decimal form, they have an unending, seemingly random, string of numbers after the decimal point.

To find approximate values for square roots, we use the $\sqrt{\;\;}$ or $\sqrt{x}$ button on a calculator. When the number we are finding the square root of is a perfect square, or the square of a rational number, we will get an exact answer. When the number is a non-perfect square, the decimals will appear random and we will have an irrational number as our answer. We call this an approximate answer. Even though we may have an answer to eight or nine decimal places, it still represents an approximation of the real answer which has an infinite number of non-repeating decimals.

Example 4

Use a calculator to find the following square roots. Round your answer to three decimal places.

a) $\sqrt {99}$

b) $\sqrt{5}$

c) $\sqrt{0.5}$

d) $\sqrt{1.75}$

a) The calculator returns 9.949874371.

Solution

$\sqrt {8}\approx 9.950$

b) The calculator returns 2.236067977.

Solution

$\sqrt {5}\approx 2.236$

c) The calculator returns 0.7071067812.

Solution

$\sqrt {0.5}\approx 0.707$

d) The calculator returns 1.322875656.

Solution

$\sqrt {1.75}\approx 1.323$

## Identify Irrational Numbers

Any square root that cannot be simplified to a form without a square root is irrational, but not all square roots are irrational. For example, $\sqrt {49}$ reduces to 7 and so $\sqrt {49}$ is rational, but $\sqrt {50}$ cannot be reduced further than $\sqrt {25}\times \sqrt {2}=5\sqrt{2}$. The fact that we cannot remove the factor of square root of 2 makes $\sqrt {50}$ irrational.

Example 5

Identify which of the following are rational numbers and which are irrational numbers.

a) 23.7

b) 2.8956

c) $\pi$

d) $\sqrt {6}$

e) $3.\overline{27}$

a) $23.7=23\frac{7}{10}$. This is clearly a rational number.

b) $2.8956=2\frac{8956}{10000}$. Again, this is a rational number.

c) $\pi = 3.141592654\ldots$ The decimals appear random, and from the definition of $\pi$ we know they do not repeat. This is an irrational number.

d) $\sqrt {6} = 2.44949489743\ldots$ Again the decimals appear to be random. We also know that $\sqrt {6}=\sqrt{2}\times \sqrt {3}$. Square roots of primes are irrational. $\sqrt {6}$ is an irrational number.

e) $3.\overline{27} = 3.2727272727\ldots$ Although these decimals are recurring they are certainly not unpredictable. This is a rational number (in actual fact, $3.\overline {27}=\frac {36}{11}$)

## Classify Real Numbers

We can now see how real numbers fall into one of several categories.

If a real number can be expressed as a rational number, it falls into one of two categories. If the denominator of its simplest form is one, then it is an integer. If not, it is a fraction (this term is used here to also include decimals, as $3.27 = 3\frac{27}{100}$.

If the number cannot be expressed as the ratio of two integers (i.e. as a fraction), it is irrational.

Example 6

Classify the following real numbers.

a) 0

b) -1

c) $\frac {\pi}{3}$

d) $\frac {\sqrt {2}}{3}$

e) $\frac {\sqrt {36}}{9}$

a) Solution

Zero is an integer.

b) Solution

-1 is an integer.

c) Although $\frac {\pi}{3}$ is written as a fraction, the numerator $(\pi)$ is irrational.

Solution

$\frac {\pi}{3}$ is an irrational number.

d) $\frac {\sqrt {2}}{3}$ cannot be simplified to remove the square root.

Solution

$\frac {\sqrt {2}}{3}$ is an irrational number.

e) $\frac {\sqrt {36}}{9}$ can be simplified to $\frac {\sqrt {36}}{3}=\frac {6}{9}=\frac {2}{3}$

Solution

$\frac {\sqrt {36}}{9}$ is a rational number.

## Graph and Order Real Numbers

We have already talked about plotting integers on the number line. It gives a visual representation of which number is bigger, smaller, etc. It would therefore be helpful to plot non-integer rational numbers (fractions) on the number line also. There are two ways to graph rational numbers on the number line. You can convert them to a mixed number (graphing is one of the few instances in algebra when mixed numbers are preferred to improper fractions), or you can convert them to decimal form.

Example 7

Plot the following rational numbers on the number line.

a) $\frac {2}{3}$

b) $-\frac {3}{7}$

c) $\frac {17}{3}$

d) $\frac {57}{16}$

If we divide the intervals on the number line into the number on the denominator, we can look at the fraction’s numerator to determine how many of these sub-intervals we need to include.

a) $\frac {2}{3}$ falls between 0 and 1. We divide the interval into three units, and include two sub-intervals.

b) $-\frac {3}{7}$ falls between 0 and -1. We divide the interval into seven units, and move left from zero by three sub-intervals.

c) $\frac {17}{5}$ as a mixed number is $2\frac {2}{5}$ and falls between 3 and 4. We divide the interval into five units, and move over two sub-intervals.

d) $\frac {57}{16}$ as a mixed number is $3\frac {9}{16}$ and falls between 3 and 4. We need to make sixteen sub-divisions.

Example 8

Plot the following numbers, in the correct order, on a number line.

a) $\pi$

b) $\frac {22}{7}$

c) 3.14

d) $\sqrt {10}$

We will use a calculator to find decimal expansions for each of these, and use a number line divided into 1000 sub-divisions. When we have two extremely close numbers, we will ensure that we place them in the correct order by looking at the expansion to the $3^{rd}$ decimal place and writing as a fraction of 1000.

a) $\pi=3.14159 \ldots \approx 3\frac {142}{1000}$

b) $\frac {22}{7}=3.14288 \ldots \approx 3\frac {143}{1000}$

c) $3.14\approx 3\frac {140}{1000}$

d) $\sqrt {10}=3.16227 \ldots \approx 3\frac {162}{1000}$

Solution

## Lesson Summary

• The square root of a number is a number which gives the original number when multiplied by itself. In algebraic terms, the square root of $x$ is a number, $b$, such that $b^2=x$, or $b=\sqrt {x}$
• There are two possibilities for a numerical value for $b$. A positive value called the principal square root and a negative value (the opposite of the positive value).
• A perfect square is a number with an integer square root.
• Here are some mathematical properties of square roots.

$\sqrt{a}\times\sqrt{b} & =\sqrt{ab} & \sqrt{a}\div\sqrt{b}& =\sqrt{\frac {a}{b}} \\A\sqrt{a}\times B\sqrt{b} & =AB\sqrt{ab} & A\sqrt{a}\div B\sqrt{b} & =\frac {A}{B}\sqrt{\frac {a}{b}}$

• Square roots of prime numbers are irrational numbers. They cannot be written as rational numbers (the ratio of two integers). In decimal form, they have an unending, seemingly random, string of numbers after the decimal point.
• Computing a square root on a calculator will produce an approximate solution since there are a finite number of digits after the decimal point.

## Review Questions

1. Find the following square roots exactly without using a calculator, giving your answer in the simplest form.
1. $\sqrt {25}$
2. $\sqrt {24}$
3. $\sqrt {20}$
4. $\sqrt {200}$
5. $\sqrt {2000}$
6. $\sqrt {\frac {1}{4}}$
7. $\sqrt {\frac {9}{4}}$
8. $\sqrt {0.16}$
9. $\sqrt {0.1}$
10. $\sqrt {0.01}$
2. Use a calculator to find the following square roots. Round to two decimal places.
1. $\sqrt {13}$
2. $\sqrt {99}$
3. $\sqrt {123}$
4. $\sqrt {2}$
5. $\sqrt {2000}$
6. $\sqrt {0.25}$
7. $\sqrt {1.35}$
8. $\sqrt {0.37}$
9. $\sqrt {0.7}$
10. $\sqrt {0.01}$
3. Classify the following numbers as an integer, a rational number or an irrational number.
1. $\sqrt {0.25}$
2. $\sqrt {1.35}$
3. $\sqrt {20}$
4. $\sqrt {25}$
5. $\sqrt {100}$
4. Place the following numbers in numerical order, from lowest to highest. $\frac {\sqrt {6}}{2} && \frac {61}{50} && \sqrt {1.5} && \frac {16}{13}$
5. Use the marked points on the number line and identify each proper fraction.

1. 5
2. $2\sqrt {6}$
3. $2\sqrt {5}$
4. $10\sqrt {2}$
5. $20\sqrt {5}$
6. $\frac {1}{2}$
7. $\frac {3}{2}$
8. 0.4
9. $\frac {1}{\sqrt {10}}$ or $\frac {\sqrt {10}}{10}$
10. 0.1
1. 3.61
2. 1.16
3. 9.95
4. 11.09
5. 44.72
6. 0.5
7. 1.16
8. 0.61
9. 0.84
10. 0.1
1. rational
2. irrational
3. irrational
4. integer
5. integer
1. ${\;\;}$ $\frac {61}{50} \qquad \frac {\sqrt {6}} {2} \qquad \frac {16}{13} \qquad \sqrt {1.6}$
1. $\frac {2}{3}$
2. $\frac {9}{5}$
3. $\frac {9}{4}$
4. $\frac {34}{9}$

Feb 22, 2012

Sep 28, 2014