# Subtraction of Rational Numbers

## Key Questions

• lets have an eg

$\frac{6}{4}$ +$\frac{10}{4}$

$\frac{6}{4}$ = 1.5

$\frac{10}{4}$ = 2.5

so we add both the decimals to get $4$

Generic explanation given using algebra. Followed by a numeric example.

#### Explanation:

$\textcolor{b l u e}{\text{Initial concept}}$
A fractions construct is such that we have:

$\left(\text{numerator")/("denominator") -> ("count")/("size indicator of what is being counted}\right)$

This may be further considered as:

" count" color(white)("d")ubrace(" of ")color(white)("ddd")1/("size indicator of what is being counted")
$\textcolor{w h i t e}{\text{ddddddd.}} \downarrow$

"count "color(white)("d.")xxcolor(white)("ddd")ubrace(1/("size indicator of what is being counted"))
$\textcolor{w h i t e}{\text{dddddddddddddddddddddddddd.}} \downarrow$

$\text{count "color(white)("d.")xxcolor(white)("dddddddddd")" unit of measurment}$

$\textcolor{red}{\text{To directly add counts the units of measurement must be the same}}$
$\textcolor{g r e e n}{\text{To directly add numerators the denominators must be the same}}$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Answering the question}}$

Let the first rational number be $\frac{a}{b}$
Let the second rational number be $\frac{c}{g}$

Consider the context:
$\frac{a}{b} - \frac{c}{g} \leftarrow \text{ denominators are not the same}$

I opted for $g$ instead of $d$ as its difference in look to $b$ if very obvious. $b \mathmr{and} d$ are easily confused with each other.

Multiply by 1 and you do not change the actual value. However, 1 comes in many forms.

$\textcolor{g r e e n}{\frac{a}{b} - \frac{c}{g} \textcolor{w h i t e}{\text{dddd")->color(white)("dddd}} \left[\frac{a}{b} \textcolor{red}{\times 1}\right] - \left[\frac{c}{g} \textcolor{red}{\times 1}\right]}$

$\textcolor{g r e e n}{\textcolor{w h i t e}{\text{dddddddddd")->color(white)("dddd}} \left[\frac{a}{b} \textcolor{red}{\times \frac{g}{g}}\right] - \left[\frac{c}{g} \textcolor{red}{\times \frac{b}{b}}\right]}$

$\textcolor{g r e e n}{\textcolor{w h i t e}{\text{dddddddddd")->color(white)("dddddd")[(ag)/(bg)]color(white)("d")-color(white)("d}} \left[\frac{b c}{b g}\right]}$

$\textcolor{g r e e n}{\textcolor{w h i t e}{\text{dddddddddd")->color(white)("ddddddddd}} \frac{a g - b c}{b g}}$

So basically you change the denominators to the same value and then directly subtract the suitably adjusted numerators.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$\textcolor{b l u e}{\text{Example}}$

$\frac{3}{5} - \frac{7}{10}$

$\left[\frac{3}{5} \times \frac{2}{2}\right] - \frac{7}{10}$

$\left[\frac{3 \times 2}{5 \times 2}\right] - \frac{7}{10}$

$\textcolor{g r e e n}{\frac{6}{10} \textcolor{red}{- \frac{7}{10}}}$

But $\textcolor{red}{- \frac{7}{10}}$ may be split into $\textcolor{red}{- \frac{6}{10} - \frac{1}{10}}$ giving:

$\textcolor{g r e e n}{\underbrace{\frac{6}{10} \textcolor{red}{- \frac{6}{10}}} \textcolor{red}{- \frac{1}{10}}}$
$\textcolor{w h i t e}{\text{d.d}} \downarrow$
$\textcolor{w h i t e}{\text{ddd")0color(white)("dd}} - \frac{1}{10}$

So $\frac{3}{5} - \frac{7}{10} = - \frac{1}{10}$

• A rational number is a number that can be expressed as a fraction, for instance $\frac{1}{5}$ or $\frac{22}{153}$.

Same denominator

Some subtractions are easy when encountering rational numbers. Here's an example problem: $\frac{5}{2} - \frac{3}{2}$.
What you might instantly notice, is that they both have the same denominator, it's $2$.
Whenever we have this situation, we can just subtract the nominators. The denominators stay the same.
In this case it would be:
$\frac{5}{2} - \frac{3}{2} = \frac{5 - 3}{2} = \frac{2}{2} = 1$

Different denominator

Whenever we have two different denominators, we can't directly subtract them from each other. What we should do in this case, is make them have the same denominator.

How can you legally (without changing the expression) change something's denominator? Easy, you also change the nominator by the same multiplication factor.
To change for example the fraction $\frac{5}{7}$ to have a denominator of $21$, you would have to ask the question: By what number should I multiply $7$, to get $21$? The answer is of course $3$. So how can we make a legal change? We alse multiply the nominator by $3$:
$\frac{5}{7} = \frac{5 \cdot 3}{7 \cdot 3} = \frac{15}{21}$

But how do you know, to what denominator they must be changed? What you're actually asking is: what number is a common multiple of the 2 denominators? This is the LCM: the Lowest Common Multiple. Why would you want that number? It's a multiple of both of the denominators, so you can easily multiply them.

An example problem for this is: $\frac{4}{3} - \frac{7}{8}$

First, let's find out what the LCM is, for this we write down some multiples of both of the denominators:
$3 \implies 6 \implies 9 \implies 12 \implies 15 \implies 18 \implies 24 \implies 27 \implies 30$
$8 \implies 16 \implies 24 \implies 32 \implies 40 \implies 48 \implies 56 \implies 64 \implies 72 \implies 80$

What number is the lowest common factor of $3$ and $8$? It's $24$. After a while, you will be able to see this instantaneously.

Now, all we have to know is: by what number do we have to multiply $3$ to get $24$? And the same for $8$. The answers are respectively $8$ and $3$.

The final solution becomes:
$\frac{4}{3} - \frac{7}{8} = \frac{4 \cdot 8}{3 \cdot 8} - \frac{7 \cdot 3}{8 \cdot 3} = \frac{32}{24} - \frac{21}{21} = \frac{32 - 21}{24} = \frac{11}{24}$

I hope that you know understand, and that I didn't make it too complicated.