###### In this section, you will...

review definitions surrounding integers.

add, subtract, multiply, divide, and reduce fractions

use the order of operations to put expressions into simplest form

do computations with percents

review radicals

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Algebra is often described as the generalization of arithmetic. The systematic use of variables, letters used to represent numbers, allows us to communicate and solve a wide variety of real-world problems. For this reason, we begin by reviewing real numbers and their operations.

review definitions surrounding integers.

add, subtract, multiply, divide, and reduce fractions

use the order of operations to put expressions into simplest form

do computations with percents

review radicals

The main topics of this section are also presented in the following videos:

We call the set of numbers \(\{\dots,-3,-2,-1,0,1,2,3,\dots\}\) the integers.

A prime number is an integer greater than \(1\) that is divisible only by \(1\) and itself. The smallest prime number is \(2\) and the rest are necessarily odd:

\begin{equation*}
2,3,5,7,11,13,17,19,23,29,\dots\text{.}
\end{equation*}

Any integer greater than \(1\) that is not prime can be uniquely written as a product of primes; we call such a number composite. Note that the number 1 is neither prime nor composite.

When a composite number, like \(42\text{,}\) is written as a product, \(42=2\cdot 21\text{,}\) we say that \(2\cdot 21\) is a factorization of \(42\) and that \(2\) and \(21\) are factors. Note that factors divide the number evenly.

When a factorization of a number consists only of prime numbers, we call it the prime factorization of that number. Since \(21\) is not prime, \(42=2\cdot21\) is not the prime factorization of \(42\text{.}\) However, we can continue to write factors as products until only a product of primes remains.

Therefore the prime factorization of \(42\) is \(2\cdot 3\cdot 7\text{.}\)

Determine the prime factorization of \(210\text{.}\)

Solution

We begin by writing \(210\) as a product with \(10\) as a factor:

\begin{equation*}
\begin{aligned}
210\amp=10\cdot21
\end{aligned}
\end{equation*}

We then continue by factoring the factors, breaking apart both \(10\) and \(21\) until only a product of primes remains.

\begin{equation*}
\begin{aligned}
210 \amp= 10\cdot 21 \\
\amp= 2\cdot 5\cdot 3\cdot 7
\end{aligned}
\end{equation*}

Since the prime factorization is unique, it does not matter how we choose to initially factor the number - the end result will be the same (although you may write your answer in a different order).

A fraction is a number written as a quotient, or ratio \(\frac{a}{b}\text{,}\) of two integers \(a\) and \(b\) where \(b\neq 0\text{.}\) The integer above the fraction bar is called the numerator and the integer below is called the denominator. Two equal ratios expressed using different numerators and denominators are called equivalent fractions. For example,

\begin{equation*}
\frac{50}{100}=\frac{1}{2}\text{.}
\end{equation*}

One way that we can find equivalent fractions is to identify a common factor of the numerator and denominator, which is simply a factor that two or more numbers share. For example, consider the following factorizations of \(50\) and \(100\text{:}\)

\begin{equation*}
\begin{aligned}
50\amp = 2\cdot 25 \text{ and }\\
100\amp =4\cdot 25\text{.}
\end{aligned}
\end{equation*}

Here we see that 25 is a common factor of 50 and 100. Making use of the fact that \(\frac{25}{25}=1\text{,}\) we see that

\begin{equation*}
\frac{50}{100}=\frac{2\cdot 25}{4\cdot 25}=\frac{2}{4}\cdot 1=\frac{2}{4}\text{.}
\end{equation*}

Dividing \(\frac{25}{25}\) and replacing this factor with a \(1\) is called cancelling. Together, these basic steps for finding equivalent fractions define the process of reducing a fraction. Since factors divide their product evenly, we achieve the same result by dividing both the numerator and denominator by \(25\) as follows:

\begin{equation*}
\frac{50\alert{\div 25}}{100\alert{\div 25}}=\frac{2}{4}\text{.}
\end{equation*}

Finding equivalent fractions where the numerator and denominator are relatively prime, or have no common factor other than \(1\text{,}\) is called reducing to lowest terms. This can be done by dividing the numerator and denominator by the greatest common factor (GCF). The GCF is the largest number that divides a set of numbers evenly. One way to find the GCF of 50 and 100 is to list all the factors of each and identify the largest number that appears in both lists. Remember, each number is also a factor of itself.

\begin{equation*}
\{{\bf 1,2,5,10,25,50}\} \alert{\text{ Factors of }50}
\end{equation*}

\begin{equation*}
\{{\bf1,2, } 4,{\bf 5,10, }\, 20,{\bf 25,50, }\, 100\} \alert{\text{ Factors of }100}
\end{equation*}

Another way to find the GCF of two numbers is to consider their prime factorizations. The GCF will be the product of all the prime factors that they have in common. For example, we can compute the GCF of 108 and 72 as follows:

\begin{equation*}
\begin{aligned}
108\amp = \alert{2}\cdot\alert{2}\cdot\alert{3}\cdot\alert{3}\cdot 3\\
72\amp =\alert{2}\cdot\alert{2}\cdot 2\cdot\alert{3}\cdot\alert{3}
\end{aligned}
\end{equation*}

\begin{equation*}
\text{GCF}(108,72)=\alert{2}\cdot\alert{2}\cdot\alert{3}\cdot\alert{3}=36
\end{equation*}

In this case, the product of the common prime factors is \(36\text{.}\)

Find the greatest common factor of the numerator and denominator. This can be done by listing all the factors of each and identifying the largest number that appears in both lists, or by finding the prime factorizations of both and taking the product of the prime factors they have in common.

Reduce the fraction by dividing both the numerator and the denominator by their GCF.

Simplify \(\frac{172}{64}\) by reducing it to lowest terms.

Solution

First, we find the GCF of 172 and 64 by considering their prime factorizations:

\begin{equation*}
\begin{aligned}
172 \amp= \alert{2} \cdot \alert{2} \cdot 43 \\
64 \amp= \alert{2} \cdot \alert{2} \cdot 2 \cdot 2 \cdot 2 \cdot 2
\end{aligned}
\end{equation*}

The product of the common prime factors is \(2\cdot2=4\text{,}\) so the GCF of 172 and 64 is 4. We can now reduce the fraction:

\begin{equation*}
\frac{172}{64}=\frac{172\alert{\div 4}}{64\alert{\div 4}}=\frac{43}{16}\text{.}
\end{equation*}

Simplify \(\frac{108}{72}\) by reducing it to lowest terms.

Solution

We have already found that the GCF of 108 and 72 is 36. We can then reduce the fraction:

\begin{equation*}
\frac{108}{72}=\frac{108\alert{\div 36}}{72\alert{\div 36}}=\frac{3}{2}\text{.}
\end{equation*}

In the previous example, we can convert the improper fraction \(\frac{3}{2}\) to a mixed number \(1\frac{1}{2}\text{;}\) however, it is important to note that converting to a mixed number is not part of the reducing process. We consider improper fractions, such as \(\frac{3}{2}\text{,}\) to be reduced to lowest terms. In algebra it is often preferable to work with improper fractions, although in some applications, mixed numbers are more appropriate. It is also worth noting that the mixed number notation \(1\frac{1}{2}\) is actually a confusing notation. What we really mean is \(1+\frac{1}{2}\)

Next we will explore some interesting properties of division. Division is a restatement of multiplication; for example,

\begin{equation*}
\frac{12}{6}=2\, \, \text{ since }\, \, 6\cdot 2=12\text{.}
\end{equation*}

In this case, the dividend 12 is evenly divided by the divisor 6 to obtain the quotient 2. It is true in general that if we multiply the divisor by the quotient we obtain the dividend. Now consider the case where the dividend is zero and the divisor is nonzero:

\begin{equation*}
\frac{0}{6}=0\, \, \text{ since }\, \, 6\cdot 0=0\text{.}
\end{equation*}

This demonstrates that zero divided by any nonzero real number must be zero. Now consider a nonzero number divided by zero:

\begin{equation*}
\frac{12}{0}=\alert{?}\, \, \text{ or }\, \, 0\, \cdot\alert{?}=12\text{.}
\end{equation*}

Zero times anything is zero and we conclude that there is no real number to satisfy the equation \(0\cdot\text{?}=12\text{.}\) Thus, the quotient \(\frac{12}{0}\) is undefined. Try it on a calculator, what does it say? For our purposes, we will simply write undefined. To summarize, given any real number \(a\neq 0\text{,}\) then zero divided by \(a\) is zero and \(a\) divided by zero is undefined.

\begin{equation*}
0\div a=\frac{0}{a}=0\, \, \alert{\text{ zero}} \text{ and }\, \, a\div 0=\frac{a}{0}\, \, \alert{\text{ undefined}}
\end{equation*}

We are left to consider the case where the dividend and divisor are both zero.

\begin{equation*}
\frac{0}{0}=\alert{?}\, \, \text{ or }\, \, 0\, \cdot\alert{?}=0
\end{equation*}

Here, any real number seems to work. For example, \(0\cdot 5=0\) and also, \(0\cdot 3=0.\) Therefore, the quotient is uncertain or indeterminate.

\begin{equation*}
0\div 0=\frac{0}{0} \, \, \alert{\text{ indeterminate}}
\end{equation*}

In this course, we state that \(0\div 0=\frac{0}{0}\) is undefined.

Notice that in the definition of fractions we allowed for \(a\) and \(b\) to be negative. Remember that for any number \(c\text{,}\) we have \(-c=(-1)\cdot c\text{.}\) This tells us that

\begin{equation*}
\frac{-a}{b}=\frac{(-1)\cdot a}{b}=-\frac{a}{b}
\end{equation*}

and also

\begin{equation*}
\frac{a}{-b}=\frac{a}{(-1)\cdot b}=-\frac{a}{b}\text{.}
\end{equation*}

Thus as a rule, we write the negative sign out in front of a fraction if we have it:

\begin{equation*}
\frac{-a}{b}=\frac{a}{-b}=-\frac{a}{b}.
\end{equation*}

Numerical calculations often involve more than one operation. So that everyone agrees on how such expressions should be evaluated, we follow the order of operations.

Simplify any expressions within grouping symbols (parentheses, brackets, square root bars, or fraction bars). Start with the innermost grouping symbols and work outward.

Evaluate all exponents and roots.

Perform multiplications and divisions in order from left to right.

Perform additions and subtractions in order from left to right.

You may have heard of the acronym PEMDAS to remember the order of operations easily. This acronym stands for Parentheses, Exponents, Multiplication, Division, Addition, and Subtraction. Notice the differences between PEMDAS and the Order of Operations described in the box above. For example, PEMDAS seems to indicate that multiplication is always done before division and that addition is done before subtraction, while in the Order of Operations, multiplication and division have equal weight and are performed in order from left to right (and similarly for addition and subtraction). PEMDAS can help you *remember* the Order of Operations, and we agree to always use the Order of Operations when simplifying expressions - it is important that we all use the same order when performing operations, as using different orders can lead to drastically different answers.

Simplify \(-10-(-10)-5+3^2+\frac{15}{5}\) by reducing to lowest terms.

Solution

Follow order of operations:

\begin{equation*}
\begin{aligned}
-10-(-10)-5+3^2+\frac{15}{5}\amp = -10-(-10)-5+3^2+3 \alert{\text{ Simplify fraction}}\\
\amp = -10-(-10)-5+9+3 \alert{\text{ Evaluate power}}\\
\amp=-10+10-5+9+3 \alert{\text{ Perform multiplication}}\\
\amp =0-5+9+3\\
\amp = 7.
\end{aligned}
\end{equation*}

We can use parentheses to override the multiplication before addition or subtraction rule. Compare the two expressions below.

\begin{align*}
\amp\text{The sum of 4 times 6 and 10} \amp\amp 4 \cdot 6 + 10\\
\amp\text{4 times the sum of 6 and 10} \amp\amp 4(6 + 10)
\end{align*}

In the first expression, we perform the multiplication \(4\cdot 6\) first, but in the second expression we perform the addition \(6 + 10\) first, because it is enclosed in parentheses.

The location (or absence) of parentheses can drastically alter the meaning of an expression. In the following example, note how the location of the parentheses changes the value of the expression.

\(\begin{aligned}5-3\cdot 4^2 \amp = 5-3\cdot 16\\ \amp = 5-48=-43 \end{aligned}\)

\(\begin{aligned}5-(3\cdot 4)^2 \amp = 5-12^2\\ \amp= 5-144 = -139 \end{aligned}\)

\(\begin{aligned}(5-3\cdot 4)^2 \amp = (5-12)^2\\ \amp = (-7)^2 = 49 \end{aligned}\)

\(\begin{aligned}(5-3)\cdot 4^2 \amp = 2\cdot 16\\ \amp = 32 \end{aligned}\)

In the expression \(5 - 12^2\text{,}\) which appears in Example5, the exponent \(2\) applies only to \(12\text{,}\) not to \(-12\text{.}\) Thus, \(5 - 12^2 \ne 5 + 144\text{.}\)

When dealing with exponents, it's important to pay attention to what the exponent is being applied to. For example,

\begin{equation*}
\begin{aligned}
-1^2 \amp= -(1\cdot1) = -1 \\
(-1)^2 \amp= -1\cdot-1 = 1
\end{aligned}
\end{equation*}

The properties of exponents will be explored in more detail in the next chapter.

The order of operations mentions other grouping devices besides parentheses - fraction bars and square root bars. Notice how the placement of the fraction bar affects the expressions in the next example.

\(\begin{aligned} \frac{1+2}{3\cdot 4} \amp = \frac{3}{12} \end{aligned}\)

\(\begin{aligned} 1+\frac{2}{3\cdot 4} \amp = 1+ \frac{2}{12} \end{aligned}\)

\(\begin{aligned} 1+\frac{2}{3}\cdot 4 \amp = 1+ \frac{8}{3} \end{aligned}\)

\(\frac{1+2}{3}\cdot 4 = \frac{3}{3}\cdot 4= 1\cdot 4=4\)

As we saw above in part (d) of Example7, we were able to achieve a "simple" answer by noticing that \(\frac{3}{3}=1\text{.}\) We can use this sort of notion to simplify all our answers above. A number is in simplest form if all order of operations have been performed and the number is either a reduced fraction or a whole number.

Adding or subtracting fractions requires a common denominator. Assume the common denominator \(c\) is a nonzero integer and we have

\begin{equation*}
\frac{a}{c}+\frac{b}{c}=\frac{a+b}{c}\text{ and }\frac{a}{c}-\frac{b}{c}=\frac{a-b}{c}\text{.}
\end{equation*}

Simplify \(\frac{2}{9}-\frac{1}{15}+\frac{8}{45}\text{.}\)

Solution

In order to add or subtract fractions, we need them to have a common denominator. To keep the numbers as simple as possible, it is helpful to find the least common denominator (LCD). To find the LCD, we first determine the least common multiple (LCM) of the denominators. We begin by listing the multiples of each given denominator.

\begin{equation*}
\{9,18,27,36,{\bf 45,}\, 54,63,72,81,{\bf{90}},\dots\} \alert{\text{ Multiples of }9}
\end{equation*}

\begin{equation*}
\{15,30,{\bf 45,}\, 60,75,{\bf{90}},\dots\} \alert{\text{ Multiples of }15}
\end{equation*}

\begin{equation*}
\{{\bf 45,90,135,}\dots\} \alert{\text{ Multiples of }45}
\end{equation*}

Here we see that the

\begin{equation*}
\text{LCM}(9,15,45)=45\text{.}
\end{equation*}

Multiply the numerator and the denominator of each fraction by values that result in equivalent fractions with the determined common denominator.

\begin{equation*}
\begin{aligned}
\frac{2}{9}-\frac{1}{15}+\frac{8}{45}\amp =\frac{2}{9}\cdot\alert{\frac{5}{5}}-\frac{1}{15}\cdot\alert{\frac{3}{3}}+\frac{8}{45}\\
\amp =\frac{10}{45}-\frac{3}{45}+\frac{8}{45}
\end{aligned}
\end{equation*}

Once we have equivalent fractions, with a common denominator, we can perform the operations on the numerators and write the result over the common denominator.

\begin{equation*}
\begin{aligned}
\amp =\frac{10-3+8}{45}\\
\amp =\frac{15}{45}
\end{aligned}
\end{equation*}

Then we reduce if necessary:

\begin{equation*}
\begin{aligned}
\amp =\frac{15\alert{\div 15}}{45\alert{\div 15}}\\
\amp =\frac{1}{3}\text{.}
\end{aligned}
\end{equation*}

Finding the LCM using lists of multiples, as described in the previous example, is often very cumbersome. For example, try making a list of multiples for 12 and 81. We can streamline the process of finding the LCM by using prime factors.

\begin{equation*}
\begin{aligned}
12\amp = 2^2\cdot 3\\
81\amp =3^4
\end{aligned}
\end{equation*}

The least common multiple is the product of each prime factor raised to the highest power. In this case,

\begin{equation*}
\text{LCM}(12,81)=2^2\cdot 3^4=324\text{.}
\end{equation*}

Given two fractions you want to add:

\begin{equation*}
\frac{a}{b} + \frac{c}{d}.
\end{equation*}

If you use the least common multiple of \(b\) and \(d\) as the common denominator, then the solution is already in its simplest form.

Find the simplest form of each of the answers in Example7.

Solution

\(\begin{aligned} \frac{1+2}{3\cdot 4} \amp = \frac{3}{12}=\frac{3}{3\cdot 4}=\frac{1}{4} \end{aligned}\)

\(\begin{aligned} 1+\frac{2}{3\cdot 4} \amp = 1+ \frac{2}{12}=\frac{12}{12}+\frac{2}{12}=\frac{14}{12}=\frac{2\cdot 7}{2\cdot 6}=\frac{7}{6} \end{aligned}\)

\(\begin{aligned} 1+\frac{2}{3}\cdot 4 \amp = 1+ \frac{8}{3}=\frac{3}{3}+\frac{8}{3}=\frac{11}{3} \end{aligned}\)

It is already in simplest form.

Unlike addition and subtraction, multiplying or dividing fractions does not require a common denominator. To multiply two fractions, we simply multiply the numerators and multiply the denominators:

\begin{equation*}
\frac{a}{b}\cdot\frac{c}{d}=\frac{a\cdot c}{b\cdot d}
\end{equation*}

In order to divide fractions, we need to define the reciprocal of a number. Two real numbers whose product is \(1\) are called reciprocals. Therefore, the reciprocal of \(\frac{a}{b}\) is \(\frac{b}{a}\) because \(\frac{a}{b}\cdot\frac{b}{a}=\frac{ab}{ab}=1\text{.}\) For example,

\begin{equation*}
\frac{2}{3}\cdot\frac{3}{2}=\frac{6}{6}=1\text{.}
\end{equation*}

Since their product is \(1,\frac{2}{3}\) and \(\frac{3}{2}\) are reciprocals. Some other reciprocals are

\begin{equation*}
\frac{5}{8}\, \, \text{ and } \, \, \frac{8}{5},\, \, \, \, 7\, \, \text{ and }\, \, \frac{1}{7},\, \, \, \, -\frac{4}{5}\, \, \text{ and }\, -\frac{5}{4}\text{.}
\end{equation*}

This definition is important because dividing fractions requires that you multiply the dividend by the reciprocal of the divisor.

\begin{equation*}
\frac{a}{b}\div\alert{\frac{c}{d}}=\frac{\frac{a}{b}}{\frac{c}{d}}\cdot\frac{\alert{\frac{d}{c}}}{\alert{\frac{d}{c}}}=\frac{\frac{a}{b}\cdot\frac{d}{c}}{1}=\frac{a}{b}\cdot\alert{\frac{d}{c}}
\end{equation*}

Simplify \(\frac{5}{4}\div\frac{3}{5}\cdot\frac{1}{2}\text{.}\)

Solution

Perform the multiplication and division left to right.

\begin{equation*}
\begin{aligned}
\frac{5}{4}\div\alert{\frac{3}{5}}\cdot\frac{1}{2}\amp =\frac{5}{4}\cdot\alert{\frac{5}{3}}\cdot\frac{1}{2}\\
\amp =\frac{5\cdot 5\cdot 1}{4\cdot 3\cdot 2}\\
\amp =\frac{25}{24}
\end{aligned}
\end{equation*}

In algebra, it is often preferable to work with improper fractions. In this case, we leave the answer expressed as an improper fraction.

Simplify \(\frac{(3^2-1)2+\frac{1}{2}}{6\div3\cdot2}\text{.}\)

Solution

Recall that fraction bars are a grouping operation; this means that we need to simplify everything in the numerator and everything in the denominator separately first. In the numerator, we simplify everything in parentheses before moving on to multiplication and addition. In the denominator, we perform multiplication and division from left to right.

\begin{equation*}
\begin{aligned}
\frac{(3^2-1)2+\frac{1}{2}}{6\div3\cdot2} \amp= \frac{(8)2+\frac{1}{2}}{6\div3\cdot2} \amp\amp\text{Simplify what's in parentheses.}\\
\amp= \frac{16+\frac{1}{2}}{6\div3\cdot2} \amp\amp\text{Multiply in the numerator.} \\
\amp= \frac{\frac{32}{2}+\frac{1}{2}}{6\div3\cdot2} \amp\amp\text{Rewrite }16\text{ as a fraction.} \\
\amp= \frac{\frac{33}{2}}{6\div3\cdot2} \amp\amp\text{Add in the numerator.} \\
\amp= \frac{\frac{33}{2}}{2\cdot2}\amp\amp\text{Divide in the denominator.} \\
\amp= \frac{\frac{33}{2}}{4} \amp\amp\text{Multiply in the denominator.} \\
\amp= \frac{33}{2}\cdot\frac{1}{4} \amp\amp\text{Divide by multiplying by the reciprocal.} \\
\amp= \frac{33}{8}
\end{aligned}
\end{equation*}

In the previous example, we could have change the fraction \(\frac{33}{2}\) to the decimal \(16.5\) to obtain an answer of

\begin{equation*}
\frac{16.5}{4}
\end{equation*}

Although this may look like a simplified fraction, it is not! To be a simplified fraction, the numerator and denominator must both be whole numbers.

Percents are used to express a number as a fraction out of \(100\text{.}\) For example, the number \(\frac{1}{2}\) is equal to \(50\)% since \(\frac{1}{2}=\frac{50}{100}\text{.}\) When working with percents, we will often have to change a percentage to a decimal value or vice versa:

Suppose \(a\) represents a partial amount of the whole amount \(b\text{.}\) Then we have

\begin{equation*}
\frac{\text{partial amount }a}{\text{whole amount }b} = \text{decimal value }
\end{equation*}

The decimal value is then converted to a percent by multiplying by \(100\) and adding the percent symbol %.

For example, say we have 15 red marbles in a bag of 75. If we want to know the percentage of the marbles that are red, we set up the following:

\begin{equation*}
\begin{aligned}
\frac{\text{partial amount }a}{\text{whole amount } b}\amp = \frac{15}{75} \amp\text{Let }a=15,\, b=75\\
\amp =\frac{15}{75}\cdot\frac{\frac{4}{3}}{\frac{4}{3}}\amp\text{to make the denominator }100\\
\amp=\frac{20}{100} \\
\amp=0.2\amp\text{Write as a decimal}\\
\amp=20\text{%}\amp\text{Convert to a percent}
\end{aligned}
\end{equation*}

Lara's choir consists of 25 singers and 18 are women. What is the percent of women in the choir?

Solution

\begin{equation*}
\begin{aligned}
\frac{\text{partial amount }a}{\text{whole amount } b}\amp = \frac{18}{25} \amp\text{Let }a=32,\, b=50\\
\amp =\frac{18}{25}\left(\alert{\frac{4}{4}}\right)\\
\amp =\frac{72}{100}\\
\amp =0.72 \amp\text{Write as a decimal}\\
\amp =72\text{%}\amp\text{Convert to a percent}
\end{aligned}
\end{equation*}

Lara's choir is \(72\)% women.

Areeba's mathematics grade includes 300 points for exams. If the entire amount of points she can earn is 750, what percentage of her grade comes from exams?

Solution

The partial amount is the \(300\) points for exams, and the whole amount is the \(750\) points overall.

\begin{equation*}
\begin{aligned}
\frac{\text{partial amount }a}{\text{whole amount }b}=\frac{300}{750}=0.4=40\text{%}
\end{aligned}
\end{equation*}

So \(40\)% of Areeba's grade comes from exams.

Percents are often used to represent the change in some quantity. When representing change, percents give you the ability to see how big the change is relative to the quantity that is changing.

\(\text{percent change (as a decimal)} =\frac{\text{amount of change}}{\text{original amount}}\)

Your favorite store is having a sale on jeans. Their original price is \($40\text{,}\) but today, their sale price is \($28\text{.}\) What is the percent discount on the jeans?

Solution

The amount of change in the price of the jeans is \(28-40 = -12\) dollars. So the percent change is \(\frac{-12}{40} = -.3\text{.}\) We convert this to a percent, or \(-30\text{%}\text{.}\) The negative tells us that the price has decreased. So the percent discount on the jeans was \(30\text{%}\text{.}\)

The population of Mongolia in 2007 was estimated to be \(2.62\) million. In 2010, it was estimated to be \(2.76\) million. What was the percent increase in the population of Mongolia over this three year period?

Solution

The amount of change in population was \(2.76 - 2.62 = 0.14\) million. So the percent change was \(\frac{0.14}{2.62} = 0.053\) (approximately). We convert this to a percent, or \(5.3\)%. So the population of Mongolia increased by \(5.3\)% between 2007 and 2010.

The following section is another way we operate on numbers. This is good to keep in the back of your mind when we see more stuff about Exponents in future sections.

You may already be familiar with square roots. Every nonnegative number has two square roots, defined as follows:

\begin{equation*}
s ~\text{ is a square root of }~ n ~\text{ if }~s^2 = n.
\end{equation*}

For example, 4 is a square root of 16, because \(4^2=16\text{.}\) Since \((-4)^2=16\text{,}\) we can say that \(-4\) is a square root of 16 as well. We use the radical sign \(\sqrt{ }\) to denote the square root of a number. By convention, we will use the positive answer when calculating a square root if there are two choices. There are several other kinds of roots, one of which is called the cube root, denoted by \(\sqrt[3]{n}\text{.}\) We define the cube root as follows.

\begin{equation*}
b ~\text{is a cube root of}~n~\text{if}~ b ~\text{cubed (raised to the power of }3~\text{) equals}~n.
\end{equation*}

In symbols, we write

\begin{equation*}
b=\sqrt[3]{n} ~\text{ if } ~ b^3=n.
\end{equation*}

Although we cannot take the square root of a *negative number*, we can take the cube root of *any* real number. For example,

\begin{equation*}
\sqrt[3]{64}=4 ~\text{ because }~ 4^3=64
\end{equation*}

and

\begin{equation*}
\sqrt[3]{-27}=-3 ~\text{ because }~ (-3)^3=-27.
\end{equation*}

In the order of operations, simplifying radicals and powers comes after parentheses but before products and quotients.

Simplify each expression.

- \(3\sqrt[3]{-8}\)
- \(2-\sqrt[3]{-125}\)
- \(\displaystyle{\frac{6-\sqrt[3]{-27}}{2}}\)

Solution

- \(3\sqrt[3]{-8}=3(-2)=-6\)
- \(2-\sqrt[3]{-125}=2-(-5)=7\)
- \(\displaystyle{\frac{6-\sqrt[3]{-27}}{2}=\frac{6-(-3)}{2}=\frac{9}{2}}\)

It will not always be the case that the number inside a square root is a perfect square. If not, we use the following two properties to simplify the expression. Let \(\sqrt[n]{a}\) and \(\sqrt[n]{b}\) be real numbers where \(b\neq 0\text{.}\)

\begin{gather*}
\text{Product Rule for Radicals: }\, \sqrt[n]{a\cdot b}=\sqrt[n]{a}\cdot\sqrt[n]{b}\\
\text{Quotient Rule for Radicals: }\, \sqrt[n]{\frac{a}{b}}=\frac{\sqrt[n]{a}}{\sqrt[n]{b}}
\end{gather*}

A simplified radical is one where the inside number does not consist of any factors that can be written as perfect powers. Given a square root, the idea is to identify the largest square factor of the inside number and then apply the property shown above. As an example, to simplify \(\sqrt{12}\text{,}\) notice that \(12\) is not a perfect square. However, \(12\) does have a perfect square factor, \(12=4\cdot 3\text{.}\) Apply the property as follows:

\begin{equation*}
\begin{aligned}
\sqrt{12}\amp=\sqrt{4\cdot 3}\\
\amp=\sqrt{4}\cdot\sqrt{3}\\
\amp=2\cdot\sqrt{3}\text{.}
\end{aligned}
\end{equation*}

The number \(2\sqrt{3}\) is a simplified radical.

Simplify the expression \(\sqrt{135}\text{.}\)

Solution

Begin by finding the largest perfect square factor of 135. \begin{aligned} 135 \amp=3\cdot 3\cdot 3\cdot 5\\ \amp=3^2\cdot 3\cdot 5\\ \amp=9\cdot 15. \end{aligned} Therefore we have \begin{aligned} \sqrt{135} \amp=\sqrt{9\cdot 15}\\ \amp=\sqrt{9}\cdot\sqrt{15}\\ \amp=3\cdot\sqrt{15}. \end{aligned}

Simplify the expression \(\sqrt{\frac{108}{169}}\text{.}\)

Solution

Begin by finding the prime factorizations of both 108 and 169. This will enable us to easily determine the largest perfect square factors. \begin{aligned} 108 \amp=2\cdot 2\cdot 3\cdot 3\cdot 3=2^2\cdot 3^2\cdot 3,\\ 169\amp=13\cdot 13=13^2. \end{aligned} Therefore we have \begin{aligned} \sqrt{\frac{108}{169}} \amp=\sqrt{\frac{2^2\cdot 3^2\cdot 3}{13^2}}\\ \amp=\frac{\sqrt{2^2\cdot 3^2\cdot 3}}{\sqrt{13^2}}\\ \amp=\frac{\sqrt{2^2}\sqrt{3^2}\sqrt{3}}{\sqrt{13^2}}\\ \amp=\frac{2\cdot 3\cdot \sqrt{3}}{13}\\ \amp=\frac{6\cdot{\sqrt{3}}}{13}. \end{aligned}

Using order of operations, simplify the expression \(2(\sqrt{(2+4)\cdot 3}-\sqrt{2})\text{.}\)

Solution

\(\begin{aligned} 2(\sqrt{(\alert{2+4})\cdot 3}-\sqrt{2}) \amp = 2(\sqrt{\alert{6\cdot 3}}-\sqrt{2}) \\ \amp = 2(\alert{\sqrt{18}}-\sqrt{2}) \\ \amp = 2(\sqrt{9}\sqrt{2}-\sqrt{2}) \\ \amp = 2(3\sqrt{2}-\sqrt{2}) \\ \amp = 2(2\sqrt{2}) \\ \amp = 4\sqrt{2} \end{aligned}\)

A cube root is simplified if it does not contain any factors that can be written as perfect cubes. The idea is to identify the largest cube factor of the radicand and then apply the product or quotient rule for radicals. As an example, to simplify \(\sqrt[3]{80}\text{,}\) notice that 80 is not a perfect cube. However, \(80=8\cdot 10\text{,}\) so we can write

\begin{equation*}
\begin{aligned}
\sqrt[3]{80} \amp=\sqrt[3]{8\cdot 10}\\
\amp=\sqrt[3]{8}\cdot\sqrt[3]{10}\\
\amp=2\cdot\sqrt[3]{10}.
\end{aligned}
\end{equation*}