Beyond the Integers

We've seen now with subtraction that any time we run into "issues" with our operations, as we don't know how to do a particular operation with some values, we can expand our concepts of what numbers are, and extend the amount of mathematics we're able to do.

Definition

Here we introduce the concept of fraction, and this opens up a whole new branch of numbers for us to deal with.

A fraction is a number that is not an integer/or whole number, and this encodes a part that can be express through a division of two whole numbers.

$$a = \frac{x}{y}$$

Before, when we had to deal with numbers between the integers, we just basically gave up and defined the operation to be impossible. Now instead, we introduce the set of fractional numbers. The set $\mathbb{Q}$:

$$\mathbb{Q} = \{ \frac{x}{y} | x \in \mathbb{Z}, y \in \mathbb{Z}, y \neq 0 \}$$

The above defintion says that the set $\mathbb{Q}$ is made up by elements of the form $\frac{x}{y}$, where both $x$ and $y$ are integers. The easiest and most common comparison of these numbers is a cake. Assume that the number $1$ represents a whole cake. Then the number $\frac{y}{x}$ represents taking $y$ slices of a cake divided in $x$ equal sizes.

With actual numbers, $\frac{2}{7}$ represents taking 2 slices out of a cake divided in 7 equal parts. Sadly, the comparison with actual food stops here as the numerator (the top number in the fraction) does not have to be always lower than the bottom part. Ergo we could express things such as 7 slices of a cake divided in 5, but we would be borrowing 2 slices from another cake... This could be expressed then as:

$$\frac{7}{5} = \frac{5+2}{5} = \frac{5}{5} + \frac{2}{5} = 1 + \frac{2}{5}$$

To explain step by step the above, we know that $7=5+2$, so we replace our "7" with "5+2". Then as I mentioned in the previous article, division is commutative over addition, and our fraction is nothing more than the result of $(5+2)/5$. Therefore we separate the division on each single "+" term, and we know that $\frac{x}{x}$ is $1$, (due to how the inverse operation of $1*x=x$).

Some examples of expression manipulation

It's also important to note that, while "+" is distributed, the same is not possible for $*$. Let's take a number "x", and see how we can transform it around.

$$x = x*1$$

So far, nothing impossible, we know that any number multiplied by 1 is the same exact value.

$$x*1 = x*\frac{y}{y}$$

We only replaced "1" with "y/y", but with a caveat that $y\neq 0$. Now, intuitively, we can replace the "*" symbol, with a repeated addition, as we defined it way back.

$$\overbrace{\frac{y}{y} + \ldots + \frac{y}{y}}^{x \text{ times}}$$

And now, as we saw above, we have an addition of many fractions with all the same "slice size" or also called denominator (the number at the bottom), and therefore we can regroup them.

$$\frac{\overbrace{y+\ldots+y}^{x \text{ times}}}{y}$$

We can regroup the addition back into a multiplication.

$$\frac{x*y}{y}$$

Therefore, a single division works for the entirety of the multiplicative term above. While an addition on the numerator part breaks the fraction in multiple smaller fractions, a multiplication (talking about the main operators on precedence of course) does not do the same. Due to this property, we can also write things like:

$$1 = 1*\frac{y}{y} = \frac{1*y}{y} = \frac{y*1}{y} = y * \frac{1}{y}$$

And treating every denominator as a multiplication by its inverse ($\frac{1}{\text{expression}}$) makes very clear that, if a fraction is multiplied by its denominator, only the numerator remains. As an example:

$$\frac{x+z}{y}*y=x+z$$

Therefore, when solving equations, it's important to keep in mind that we can always multiply both sides of the equation by the denominator, in order to try and get rid of the fraction.

$$4+\frac{5}{x}=8$$

Multiply by "x":

$$4*x+5=8*x$$

Shuffle around...

$$5=8*x-4*x$$

Simplify:

$$5=4*x$$

Divide both terms by 4:

$$\frac{5}{4} = \frac{4*x}{4}$$ $$\frac{5}{4} = \frac{4}{4}*x$$ $$\frac{5}{4} = x$$

A bit more explicit, but it shows every step, without any shortcuts.

Another thing to note is that, whenever a common term is multiplied on both numerator and denominator, we can "elide" it from both sides, provided that it's a term that multiplies the whole upper part and the whole lower part.

$$\frac{x*(\text{expression}_1)}{x*(\text{expression}_2)}=\frac{\text{expression}_1}{\text{expression}_2}$$

And as an example:

$$\frac{3*x+3}{6x+3} = \frac{3*(x+1)}{3*(2*x+1)} = \frac{x+1}{2*x+1}$$

In the above expression, we cannot easily simply further, as the "+" on the denominator is what is sadly stopping us. In the first step, the "multiplication" was as the last term, and that made it possible to group terms, but in this way, we cannot find any common multiplicative factor.

Also, dividing by "-1" is equal to multiplying by "-1", and let's make it explicit:

$$\frac{x}{-1} = \frac{x*1}{-1} = x*\frac{1}{-1}$$

So far applying only what we've seen, we know that $(-1)*(-1)=-(-1)=1$, so we change the 1 in the numerator, and the denominator is simply expanded as being multiplied by "1" (which does not change the value):

$$x*\frac{1}{-1}=x*\frac{(-1)*(-1)}{(-1)*1}$$

We now see a common "-1" to group up between the numerator and denominator:

$$x*\frac{(-1)*(-1)}{(-1)*1} = x*\frac{-1}{1}$$

We know that $y/1=y$, that is, dividing by 1 does not change the value, so our "-1" stays:

$$x*(-1) = -x$$

Therefore, dividing by "-1" is equal to negating the numerator. Since every negative number can be also expressed as "-1*x", we can repeat this whole process, and simply negate the numerator:

$$\frac{y}{-x} = \frac{-y}{x}$$

A small preview

While we've seen what fractions are, we'll first introduce very informally decimal numbers. Usually seen in prices, numerals are written in the form "<digits>.<digits>" or depending on the country "<digits>,<digits>".

As an example, $4.53\$$ (or €) is a commonly seen form to indicate a price. The ".53" part is called in "cents", because it's expressed in the form

$$\frac{53}{100}$$

. Therefore the price above would be a short form for:

$$4+\frac{53}{100}$$

What if we need to indicate more precise fractionary parts? In the notation we add a digit at the end, like $4.536$, which now means:

$$4+\frac{536}{1000}$$

Let's try to take it apart, expressing each digit as a sum:

$$4+\frac{536}{1000} = 4+\frac{500+30+6}{1000}$$

We know the sum can be taken apart:

$$4+\frac{500}{1000}+\frac{30}{1000}+\frac{6}{1000}$$

Now we can take apart these numbers by knowing that $500=5*100$, and that $1000=10*100$, which gives us the same "100" to group up and remove:

$$4+\frac{5*100}{10*100}+\frac{3*10}{10*100}+\frac{6}{1000}$$

We remove the common terms from each fraction (which we can because each numerator/denominator is a single multiplication):

$$4+\frac{5}{10}+\frac{3}{100}+\frac{6}{1000}$$

And that's a very easy way to express a decimal number (decimal because it's all fractions of ten... and powers of ten, basically the inverse of 10,100,1000,....). If we added another digit, it would be simply of the form $\frac{x}{10000}$ and so forth. We can also intuit that the "4" here is not special, and could be just expressed as $\frac{4}{1} = 4$.

An extra property

When we get to adding fractions together (or subtracting), we do so only on their numerators, provided the denominator is the same.

$$\frac{a}{b} + \frac{c}{d} \neq \frac{a+c}{b+d}$$

How do we simplify the above in a single fraction? We just reuse the knowledge that :

• Any number "x" has property $x=x*1$.

• And 1 can be expressed as such $1=\frac{y}{y}$.

Therefore:

$$\frac{a}{b}*\frac{y}{y}=\frac{a*y}{b*y}$$

So our conundrum above can be solved by making sure the denominator becomes the same:

$$\frac{a}{b} + \frac{c}{d} = \frac{a}{b}*\frac{d}{d} + \frac{c}{d}*\frac{b}{b}$$

With our properties above, we can see that when we have two fractions being multiplied, the numerators are multiplied together, and so are the denominators.

This works also for a whole number:

$$\frac{a}{b}*y=\frac{a}{b}*\frac{y}{1}=\frac{a*y}{b*1}=\frac{a*y}{b}$$

$$\frac{a}{b}*\frac{d}{d} + \frac{c}{d}*\frac{b}{b} = \frac{a*d}{b*d} + \frac{c*b}{d*b}$$

A bit of commutativity on the multiplication on the denominator:

$$\frac{a*d}{b*d} + \frac{c*b}{d*b} = \frac{a*d}{b*d} + \frac{c*b}{b*d}$$

We are now finally sure the denominators are the exact same so we sum together the numerators:

$$\frac{a*d}{b*d} + \frac{c*b}{b*d} = \frac{a*d + c*b}{b*d}$$

And that's the true result of that addition. Since the quantities $a,b,c,d$ are all abstract, we could replace them with any expression, and reapply this final result.