<meta http-equiv="refresh" content="1; url=/nojavascript/">
You are reading an older version of this FlexBook® textbook: CK-12 Algebra I - Second Edition Go to the latest version.

# 2.1: Integers and Rational Numbers

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Graph and compare integers.
• Classify and order rational numbers.
• Find opposites of numbers.
• Find absolute values.
• Compare fractions to determine which is bigger.

## Graph and Compare Integers

Integers are the counting numbers (1, 2, 3...), the negative opposites of the counting numbers (-1, -2, -3...), and zero. There are an infinite number of integers and examples are 0, 3, 76, -2, -11, and 995.

Example 1

Compare the numbers 2 and -5

First, we will plot the two numbers on a number line.

We can compare integers by noting which is the greatest and which is the least. The greatest number is farthest to the right, and the least is farthest to the left.

In the diagram above, we can see that 2 is farther to the right on the number line than -5, so we say that 2 is greater than -5. We use the symbol “>” to mean “greater than”.

Solution $2 > -5$

Example 2

A frog is sitting perfectly on top of number 7 on a number line. The frog jumps randomly to the left or right, but always jumps a distance of exactly 2. Describe the set of numbers that the frog may land on, and list all the possibilities for the frog’s position after exactly 5 jumps.

Solution

We will graph the frog’s position, and also indicate what a jump of 2 looks like. We see that one possibility is that the frog lands on 5. Another possibility is that it lands on 9. It is clear that the frog will always land on an odd number.

After one jump the frog could be on either the 9 or the 5 (but not on the 7). After two jumps the frog could be on 11, 7 or 3. By counting the number of times the frog jumps to the right or left, we may determine where the frog lands. After five jumps, there are many possible locations for the frog. There is a systematic way to determine the possible locations by how many times the frog jumped right, and by how many times the frog jumped left.

$& \text{RRRRR} = 5\ \text{jumps right} && \text{location} = 7 + (5 \cdot 2) = 17 \\& \text{RRRRL} = 4\ \text{jumps right, 1 jump left} && \text{location} = 7 + (3 \cdot 2) = 13 \\& \text{RRRLL} = 3\ \text{jumps right, 2 jumps left} && \text{location} = 7 + (1 \cdot 2) = 9 \\& \text{RRLLL} = 2\ \text{jumps right, 3 jumps left} && \text{location} = 7 - (1 \cdot 2) = 5 \\& \text{RLLLL} = 1\ \text{jump right, 3 jumps left} && \text{location} =7 - (3 \cdot 2) = 1 \\& \text{LLLLL} = 5\ \text{jumps left} && \text{location} = 7 - (5 \cdot 2) = -3$

These are the possible locations of the frog after exactly five jumps. Notice that the order does not matter: three jumps right, one left and one right is the same as four jumps to the right and one to the left.

## Classifying Rational Numbers

When we divide an integer by another integer (not zero) we get what we call a rational number. It is called this because it is the ratio of one number to another. For example, if we divide one integer $a$ by a second integer $b$ the rational number we get is $\left( \frac{a}{b} \right )$, provided that $b$ is not zero. When we write a rational number like this, the top number is called the numerator. The bottom number is called the denominator. You can think of the rational number as a fraction of a cake. If you cut the cake into $b$ slices, your share is $a$ of those slices.

For example, when we see the rational number $\left (\frac{1}{2}\right )$, we imagine cutting the cake into two parts. Our share is one of those parts. Visually, the rational number $\left (\frac{1}{2}\right )$ looks like this.

With the rational number $\left (\frac{3}{4}\right )$, we cut the cake into four parts and our share is three of those parts. Visually, the rational number $\left (\frac{3}{4}\right )$ looks like this.

The rational number $\left (\frac{9}{10}\right )$ represents nine slices of a cake that has been cut into ten pieces. Visually, the rational number $\left (\frac{9}{10}\right )$ looks like this.

Proper fractions are rational numbers where the numerator (the number on the top) is less than the denominator (the number on the bottom). A proper fraction represents a number less than one. With a proper fraction you always end up with less than a whole cake!

Improper fractions are rational numbers where the numerator is greater than the denominator. Improper fractions can be rewritten as a mixed number – an integer plus a proper fraction. An improper fraction represents a number greater than one.

Equivalent fractions are two fractions that give the same numerical value when evaluated. For example, look at a visual representation of the rational number $\left (\frac{2}{4}\right )$.

You can see that the shaded region is identical in size to that of the rational number one-half $\left (\frac{1}{2}\right )$. We can write out the prime factors of both the numerator and the denominator and cancel matching factors that appear in both the numerator and denominator.

$\left (\frac{2} {4}\right ) = \left (\frac{\cancel{2} \cdot 1} {\cancel{2} \cdot 2 \cdot 1}\right ) && \text{We then re-multiply the remaining factors.} && \left ( \frac{2}{4} \right )=\left ( \frac{1}{2} \right )$

This process is called reducing the fraction, or writing the fraction in lowest terms. Reducing a fraction does not change the value of the fraction. It just simplifies the way we write it. When we have canceled all common factors, we have a fraction in its simplest form.

Example 3

Classify and simplify the following rational numbers

a) $\left ( \frac{3}{7} \right )$

b) $\left ( \frac{9}{3} \right )$

c) $\left ( \frac{50}{60} \right )$

a) 3 and 7 are both prime – there is no simpler form for this rational number so...

Solution

$\frac{3}{7}$ is already in its simplest form.

b) $9 = 3\cdot 3$ and 3 is prime. We rewrite the fraction as: $\left (\frac{9} {3}\right ) = \left (\frac{\cancel{3} \cdot 3 \cdot 1} {\cancel{3} \cdot 1} \right ).$ $9 > 3$ so...

Solution

$\frac{9}{3}$ is an improper fraction and simplifies to $\frac{3}{1}$ or simply 3.

c) $50 = 5 \cdot 5 \cdot 2$ and $60 = 5 \cdot 3 \cdot 2 \cdot 2$. We rewrite the fraction thus: $\frac{50} {60} = \left (\frac{\cancel{5} \cdot 5 \cdot \cancel{2} \cdot 1} {\cancel{5} \cdot 5 \cdot \cancel{2} \cdot 2 \cdot 1} \right ).$ $50 <60$ so...

Solution

$\frac{50}{60}$ is a proper fraction and simplifies to $\frac{5}{6}$.

## Order Rational Numbers

Ordering rational numbers is simply a case of arranging numbers in order of increasing value. We write the numbers with the least (most negative) first and the greatest (most positive) last.

Example 4

Put the following fractions in order from least to greatest: $\frac {1}{2}, \frac {3}{4}, \frac {2}{3}$

Let’s draw out a representation of each fraction.

We can see visually that the largest number is $\frac{3}{4}$ and the smallest is $\frac{1}{2}$:

Solution

$\frac {1}{2} < \frac {2}{3} < \frac {3}{4}$

With simple fractions, it is easy to order them. Think of the example above. We know that one-half is greater than one quarter, and we know that two thirds is bigger than one-half. With more complex fractions, however we need to find a better way to compare.

Example 5

Which is greater, $\frac {3}{7}$ or $\frac {4}{9}$?

In order to determine this we need to find a way to rewrite the fractions so that we can better compare them. We know that we can write equivalent fractions for both of these. If we make the denominators in our equivalent fractions the same, then we can compare them directly. We are looking for the lowest common multiple of each of the denominators. This is called finding the lowest common denominator (LCD).

The lowest common multiple of 7 and 9 is 63. Our fraction will be represented by a shape divided into 63 sections. This time we will use a rectangle cut into 9 by $7 = 63$ pieces:

7 divides into 63 nine times so:

$\left ( \frac{3}{7} \right )= \frac {9}{9}\left ( \frac{3}{7} \right )=\left ( \frac{27}{63} \right )$

Note that multiplying by $\frac{9}{9}$ is the same as multiplying by 1. Therefore, $\frac{27}{63}$ is an equivalent fraction to $\frac{3}{7}$. Here it is shown visually.

9 divides into 63 seven times so:

$\left ( \frac{4}{9} \right )=\frac {7}{7}\left ( \frac{4}{9} \right )=\left ( \frac{28}{63} \right )$

$\frac{28}{63}$ is an equivalent fraction to $\frac{4}{9}$. Here it is shown visually.

By writing the fractions over a common denominator of 63, you can easily compare them. Here we take the 28 shaded boxes out of 63 (from our image of $\frac{4}{9}$ above) and arrange them in a way that makes it easy to compare with our representation of $\frac{3}{7}$. Notice there is one little square “left over”.

Solution

Since $\frac {28}{63}$ is greater than $\frac {27}{63}$, $\frac {4}{9}$ is greater than $\frac {3}{7}$.

Remember

To compare rational numbers re-write them with a common denominator.

## Find the Opposites of Numbers

Every number has an opposite. On the number line, a number and its opposite are opposite each other. In other words, they are the same distance from zero, but they are on opposite sides of the number line.

By definition, the opposite of zero is zero.

Example 6

Find the value of each of the following.

a) $3 + (-3)$

b) $5 + (-5)$

c) $(-11.5) + (11.5)$

d) $\frac{3}{7}+\frac{-3}{7}$

Each of the pairs of numbers in the above example are opposites. The opposite of 3 is (-3), the opposite of 5 is (-5), the opposite of (-11.5) is 11.5 and the opposite of $\frac{3}{7}$ is $\frac{3}{7}$.

Solution

The value of each and every sum in this problem is 0.

Example 7

Find the opposite of each of the following:

a) 19.6

b) $-\frac{4}{9}$

c) $x$

d) $xy^2$

e) $(x-3)$

Since we know that opposite numbers are on opposite sides of zero, we can simply multiply each expression by -1. This changes the sign of the number to its opposite.

a) Solution

The opposite of 19.6 is -19.6.

b) Solution

The opposite of is $-\frac{4}{9}$ is $\frac{4}{9}$.

c) Solution

The opposite of $x$ is $-x$.

d) Solution

The opposite of $xy^2$ is $-xy^2$.

e) Solution

The opposite of $(x-3)$ is $-(x-3)=3-x$.

Note: With the last example you must multiply the entire expression by -1. A common mistake in this example is to assume that the opposite of $(x-3)$ is $(x+3)$. DO NOT MAKE THIS MISTAKE!

## Find absolute values

When we talk about absolute value, we are talking about distances on the number line. For example, the number 7 is 7 units away from zero. The number -7 is also 7 units away from zero. The absolute value of a number is the distance it is from zero, so the absolute value of 7 and the absolute value of -7 are both 7.

We write the absolute value of -7 like this $|-7|$

We read the expression $|x|$ like this “the absolute value of $x$.”

• Treat absolute value expressions like parentheses. If there is an operation inside the absolute value symbols evaluate that operation first.
• The absolute value of a number or an expression is always positive or zero. It cannot be negative. With absolute value, we are only interested in how far a number is from zero, and not the direction.

Example 8

Evaluate the following absolute value expressions.

a) $|5+4|$

b) $3-|4-9|$

c) $|-5-11|$

d) $-|7-22|$

Remember to treat any expressions inside the absolute value sign as if they were inside parentheses, and evaluate them first.

Solution

a) $|5+4| &= |9|\\ &= 9$

b) $3-|4-9| &= 3-|-5|\\ &= 3-5\\ &= -2$

c) $|-5-11| &= |-16|\\ &= 16$

d) $-|7-22| &= -|-15|\\ &= -(15)\\&= -15$

## Lesson Summary

• Integers (or whole numbers) are the counting numbers (1, 2, 3...), the negative counting numbers (-1, -2, -3...), and zero.
• A rational number is the ratio of one integer to another, like $\frac{a}{b}$ or $\frac{3}{5}$. The top number is called the numerator and the bottom number (which can not be zero) is called the denominator.
• Proper fractions are rational numbers where the numerator is less than the denominator.
• Improper fractions are rational numbers where the numerator is greater than the denominator.
• Equivalent fractions are two fractions that give the same numerical value when evaluated.
• To reduce a fraction (write it in simplest form) write out all prime factors of the numerator and denominator, cancel common factors, then recombine.
• To compare two fractions it helps to write them with a common denominator: the same integer on the bottom of each fraction.
• The absolute value of a number is the distance it is from zero on the number line. The absolute value of a number or expression will always be positive or zero.
• Two numbers are opposites if they are the same distance from zero on the number line and on opposite sides of zero. The opposite of an expression can be found by multiplying the entire expression by -1.

## Review Questions

1. The tick-marks on the number line represent evenly spaced integers. Find the values of $a, b, c, d$ and $e$.
2. Determine what fraction of the whole each shaded region represents.
3. Place the following sets of rational numbers in order, from least to greatest.
1. $\frac {1}{2},\frac {1}{3},\frac {1}{4}$
2. $\frac {11}{12},\frac {12}{11},\frac {13}{10}$
3. $\frac {39}{60},\frac {49}{80},\frac {59}{100}$
4. $\frac {7}{11},\frac {8}{13},\frac {12}{19}$
4. Find the simplest form of the following rational numbers.
1. $\frac {22}{44}$
2. $\frac {9}{27}$
3. $\frac {12}{18}$
4. $\frac {315}{420}$
5. Find the opposite of each of the following.
1. 1.001
2. $(5-11)$
3. $(x+ y)$
4. $(x-y)$
6. Simplify the following absolute value expressions.
1. $11-|-4|$
2. $|4-9|-|-5|$
3. $|-5-11|$
4. $7-|22-15-19|$
5. $-|-7|$
6. $|-2-88|-|88+2|$

1. $a = -3; b = 3; c = 9; d = 12; e = 15$
2. $a =\frac{1}{3}; b = \frac{7}{12}; c = \frac{22}{35}$
1. $\frac {1}{4} < \frac {1}{3} < \frac {1}{2}$
2. $\frac {11}{12} < \frac {12}{11} < \frac {13}{10}$
3. $\frac {59}{100} < \frac {49}{80} < \frac {39}{60}$
4. $\frac {8}{13} < \frac {12}{19} < \frac {7}{11}$
1. $\frac {1}{2}$
2. $\frac {1}{3}$
3. $\frac {2}{3}$
4. $\frac {3}{4}$
1. -1.001
2. $6 -(x + y)$
3. $(y-x)$
1. 7
2. 0
3. 16
4. -5
5. -7
6. 0

## Date Created:

Feb 22, 2012

Sep 28, 2014
Files can only be attached to the latest version of None