Rational Numbers
6.2 Addition and Subtraction with Rational Numbers
Adding Rational Numbers
Adding rational numbers is very similar to adding integers, with the exception that we must first find a common denominator. It doesn't make sense to add $\frac{1}{2}$ and $\frac{1}{4}$, because they are measures of different things. It would be like asking, "what is 4 oranges plus 3 apples?". In comparison, $\frac{1}{5}$ and $\frac{2}{5}$ are both measures of the same sized pieces. We have $1$ fifth of something, and $2$ fifths of that same something. We can add them directly. $1$ fifth plus $2$ fifths is $3$ fifths.
$$\frac{1}{5}+\frac{2}{5}=\frac{3}{5}$$
Definition: Addition with Common Denominators
If fractions have same denominator,
$$\frac ab+\frac cb=\frac{a+c}{b}.$$
MyOpenMath: Adding with common denominators
If the fractions do not have a common denominator, we can find one using our fundamental law of fractions. Then we can add th emore finely divided pieces. For $\frac{1}{2}$ and $\frac{1}{4}$,
$$\frac{1}{2}\cdot \frac{2}{2}+\frac{1}{4}$$ $$\frac{2}{4}+\frac{1}{4}$$ $$\frac{3}{4}$$
The area model lends itself nicely to visualizing this. Follow this Geogebra link and try it for yourself.
Example: Add with unlike denominators
Find $\frac{2}{15}+\frac{4}{21}$.
Show Solution
We can find any denominator, but the least common denominator of 15 and 21 is the lease common multiple of the two numbers. Lets factor everyting and see whats missing.
$$\frac{2}{15}=\frac{2}{3\cdot 5},\qquad \frac{4}{21}=\frac{4}{3\cdot 7}$$
We see that the left fraction needs a $7$ to match the right, and the right needs a $5$ to match the left.
$$\frac{2}{15}+\frac{4}{21}=\frac{2}{3\cdot 5}\cdot \frac{7}{7}+\frac{4}{3\cdot 7}\cdot \frac{5}{5}=\frac{14}{105}+\frac{20}{105}=\frac{34}{105}$$
Our number theory topics on GCD and LCM can really reduce the workload here. Only adding what is neccessary to each number can greatly simplify the arithmetic.
MyOpenMath: Adding with uncommon denominators
Example: Symbolic addition
Add $\frac{3}{x}+\frac{4}{y}$.
Show Solution
Common denominator is $xy$:
$$\frac{3}{x}=\frac{3y}{xy},\qquad \frac{4}{y}=\frac{4x}{xy}.$$
So
$$\frac{3}{x}+\frac{4}{y}=\frac{3y+4x}{xy},\quad x\ne0,\ y\ne0.$$
Here's some more practice finding least common denominators.
MyOpenMath: Finding a least common denominator symbolically
Again, our goal here is to find a least common denominator to reduce our arithmetic later on. Its the difference between this
and this
They're equivalent rational numbers, but one is fully reduced while the other isn't.
Mixed Numbers and Improper Fractions
A mixed number is a number that is made up of an integer and a fraction together. On the number line, these numbers are represented by arrows that don't end on an integer, but a rational number between two integers. For example, $3\frac{2}{5}$ is a mixed number, it is the number 3 and $\frac{2}{5}$ together. or
$$3\frac{2}{5}=3 + \frac{2}{5}=\frac{15}{5}+\frac{2}{5}=\frac{17}{5}$$
Mixed numbers are often used when measuring things quickly. They're easy to understand, albiet a little confusing in the notation. If we need $2\frac{3}{5}$ cups of water, we can go ahead and measure the $2$ cups easily, and then worry about the $\frac{3}{5}$. Visualized on our numberline modely, its as easy as going to $2$ first, and then counting out the remaining fifths.
To be clear, we often imply multiplication when we write two numbers side by side without a symbol. $ab$ means $a\times b$ in many cases. But with mixed numbers, the implied operation is addition. It will be clear because we don't normally write an integer beside a rational number without some operation between them unless we mean it to be a mixed number.
Definition: Improper Fractions
Fractions whose numerator is larger than their denominator are called improper fractions and can be turned into mixed numbers.
Example: Convert improper fractions
Covert each of the improper fractions below to a mixed number.
- $\frac{18}{5}$
- $\frac{11}{3}$
- $-\frac{23}{6}$
Show Solution
- Since our denominator is $5$, we're looking to remove as many $5$'s from the numerator as possible. $$\frac{18}{5}=\frac{15+3}{5}=\frac{15}{3}+\frac{3}{5}=5+\frac{3}{5}=5\frac{3}{5}$$
- Similarly, $$\frac{11}{3}=\frac{9+2}{3}=\frac{9}{3}+\frac{2}{3}=3\frac{2}{3}$$
- And finally, the negative sign. Its important to remember that it belongs to the whole number, not just any pieces we break off. $$-\frac{23}{6}=-\left(\frac{18}{6}+\frac{5}{6}\right)=-\left(3+\frac{5}{6}\right)=-3\frac{5}{6}$$
MyOpenMath: Converting Improper Fractions
Example: Convert mixed numbers
Convert $4\frac{1}{3}$, $-5\frac{2}{7}$, and $-3\frac{1}{4}$.
Show Solution
- $4\frac{1}3=4+\frac{1}{3}=\frac{12}{3}+\frac{1}{3}=\frac{13}{3}$
- $-5\frac27=-\left(5+\frac{2}{7}\right)=-\left(\frac{35}{7}+\frac{2}{7}\right)=-\frac{37}{7}$
- $-3\frac14=-\left(3+\frac14\right)=-\frac{13}{4}$
MyOpenMath: And the other direction...
The key thing to remember with improperfracion conversions is that the denominator is deciding how many pieces make up the whole. It is the number that dictates the division of the numerator.
Subtraction and Least Common Denominator
We can borrow our integer definition of subtraction to greatly simplify subtracting rational numbers. We can think of subtraction as added the opposite of the number in question.
But we need to be clear of what the opposite of a rationial number is. Take $a,b\in \mathbb{W}$ with $b\neq 0$.
$\frac{a}{b}$ is a rational number, and its opposite would be, $-\frac{a}{b}$, which we will interpret as $\frac{-a}{b}$. We have our unit divided into $b$ pieces, and we have $-a$ of those pieces.
For simplicity, anytime we have some expression with a negative denominator, we can simplify it by making the numerator negative.
$$-\frac{2}{3}=\frac{-2}{3}=\frac{2}{-3}$$
These are all the same, but we'll prefer $\frac{-2}{3}$ for our physical models. We can remove $2$ of the $\frac{1}{3}$ pieces, and this is easier to model.
Definition: Rational Subtraction via Addition
$$\frac ab-\frac cd=\frac ab+\left(-\frac cd\right).$$
Example: Subtract fractions
Find each difference.
- $\frac{5}{8}-\frac{1}{4}$
- $5\frac{1}{3}-2\frac{3}{4}$
- $\frac{7}{9}-\frac{14}{15}$
Show Solution
- First we'll rewrite it as addition. $$\frac{5}{8}-\frac{1}{4}=\frac{5}{8}+\frac{-1}{4}$$ Now we have to make sure they have the same denominator so we can add them. $$\frac{5}{8}+\frac{-1}{4}\cdot \frac{2}{2}=\frac{5}{8}+\frac{-2}{8}$$ And now that we're comparing apples to apples, if we have five and we add the opposite of two: $$\frac{5+(-2)}{8}=\frac{3}{8}$$
- First, let's convert the mixed numbers so we're working in one format. $$5\frac{1}{3}=\frac{16}{3},\qquad 2\frac{3}{4}=\frac{11}{4}$$ Now rewrite subtraction as addition of the opposite: $$\frac{16}{3}-\frac{11}{4}=\frac{16}{3}+\frac{-11}{4}$$ The least common denominator of 3 and 4 is 12, so we'll rewrite both fractions. $$\frac{16}{3}\cdot\frac{4}{4}+\frac{-11}{4}\cdot\frac{3}{3}=\frac{64}{12}+\frac{-33}{12}$$ Now add the numerators: $$\frac{64+(-33)}{12}=\frac{31}{12}=2\frac{7}{12}$$
- Again, we'll rewrite as addition first. $$\frac{7}{9}-\frac{14}{15}=\frac{7}{9}+\frac{-14}{15}$$ The least common denominator of 9 and 15 is 45. $$\frac{7}{9}\cdot\frac{5}{5}+\frac{-14}{15}\cdot\frac{3}{3}=\frac{35}{45}+\frac{-42}{45}$$ So we add the numerators: $$\frac{35+(-42)}{45}=\frac{-7}{45}$$
MyOpenMath: Subtraction Using the Area Model
And for good measure, lets confirm that our numberline model for subtraction still works.
MyOpenMath: Subtraction Using the Numberline Model
Properties of Addition with Rational Numbers
As with integers, addition of rational numbers have several properties of note.
Fact: Properties of Rational Number Addition
For integers $a,b,c,d$ and $c,d\neq 0$ we have:
- Closure: $\frac{a}{b}+\frac{c}{d}\in\mathbb{Q}$
- Commutative: $\frac{a}{b}+\frac{c}{d}=\frac{c}{d}+\frac{a}{b}$
- Associative: With $\frac{e}{f}$ being another rational number, $$\left(\frac{a}{b}+\frac{c}{d}\right)+\frac{e}{f}=\frac{a}{b}+\left(\frac{c}{d}+\frac{e}{f}\right)$$
- Identity: $\frac{a}{b}+0=\frac{a}{b}$
- Additive inverse: for each $\frac{a}{b}$, there exists $-\frac{a}{b}$ with $\frac{a}{b}+(-\frac{a}{b})=0$
Estimation with Rational Numbers
Estimating is a great skill to build, because it can help us weed out some obvious mistakes in our work. Estimation also helps us make practical decisions in our everyday lives. For instance, when adding $\frac{7}{8}+\frac{11}{10}$, we might estimate it to be 2. $\frac{7}{8}$ is just under 1, and $\frac{11}{10}$ is just over 1, so the sum should be close to 2. We can make reasonable estimates based on how close a rational number is to an integer.
Example: Is the result reasonable?
A student says $\frac{3}{4}+\frac{1}{2}=\frac{4}{6}$. Is that reasonable?
Show Solution
No. It looks like the student tried to add straight across the fractions which is a very common mistake. We can quickly see that both $\frac{3}{4}$ is close to $\frac{1}{2}$ because $\frac{2}{4}=\frac{1}{2}$, its just a tad bit more. And $\frac{1}{2}$ is exactly a half. So we have two numbers that are at least one half in size, which means our number should at least be one. But $\frac{4}{6}$ is less than one.
$\frac46=\frac23$, is far too small.
Correct exact sum is
$$\frac34+\frac12=\frac34+\frac24=\frac54=1\frac14$$
which a reasonable estimate for it would be a little more than $1$.
Example: Estimate and classify
Estimate $\frac{27}{13}+\frac{10}{9}$ and decide high or low if you round to whole numbers.
Show Solution
Find the nearest whole number to each fraction:
$$\frac{27}{13}\approx \frac{26}{13}=2,\qquad \frac{10}{9}\approx\frac{9}{9}=1.$$
The first number is close to $2$ while the second is close to $1$, so we would estimate $2+1=3$.
Exact value is a bit higher, because we rounded down both times. $3$ is a low estimate.
When numbers are very close to $\frac{1}{2}$ it can be tough to decide which way to round. As a rule of thumb, at exactly $\frac{1}{2}$ we'll round up. So something like $$\frac{1}{2}+5\frac{2}{8}\approx 1 + 5 = 6$$
In these cases its a bit harder to see if our estimate is over or uinder. We rounded $\frac{1}{2}$ up, which would put us over. But we rounded $\frac{2}{8}$ down, which lowers our estimate. In this case $\frac{1}{2}>\frac{2}{8}$, we went up more than down, so our estimate is still over.