Numbers are ideas we use to model quantities and establish comparisons between them.

For example, I can count the books in my office and tell you that there are $300$ books. Acknowledging this information, you can count the books in your home and tell who has more books, you or me. In this sense, numbers are being used as intermediates, since comparing the books directly would be unfeasible.

1 Natural Numbers

The idea of counting gives rise to the so called natural numbers, which are denoted in the mathematical literature as $\mathbb{N}$. Historically speaking, the first formal description of the natural numbers was given by the Italian mathematician Giuseppe Peano, and that formal description is somewhat like this:

These ideas are at the very basis of our numerical system, and they allow us to produce proofs about statements. Consult the article about proofs involving naturals if you are curious about how these principles are used to give rise to the theory of the natural numbers. For now, we shall be satisfied with the knowledge of the practical aspects.

And just before we explore more about the naturals, it is worth talking about the symbols we use to write numbers down, known as the digits.

2 Base 10 notation

Our numerical systems rely on the same ten digits to write any number down (except those that we write in symbolic form, like $\pi$). We count the first numbers using a different symbol for each one:

$$\begin{align*} \text{zero} &= 0\\ \text{one} &= 1\\ \text{two} &= 2\\ \text{three} &= 3\\ \text{four} &= 4\\ \text{five} &= 5\\ \text{six} &= 6\\ \text{seven} &= 7\\ \text{eight} &= 8\\ \text{nine} &= 9 \end{align*}$$

However, from the next number onward, we carry a one to the place value to the left in order to indicate that we have already used each digit once: $10, 11, 12, \dots, 19$. We would carry another $1$, but there is a $1$ in there already, so we add them, meaning that the leftmost digit is a $2$ now: $20, 21, 22, \dots, 29$. These spaces into which we write the digits down are called place values, and they receive their names according to what power of $10$ they represent: the ones, the tens, the hundreds, and so on.

This way, we can write down any natural that we can reach from $0$.

3 Natural numbers and their operations

The set of the natural numbers, denoted $\mathbb N$, is that containing the numbers that represent whole quantities, be that the empty amount, which is $0$, or some amount we reach by counting from $0$.

3.1 Addition

That way, we say that to add two naturals $n+m$ is to obtain the successor of $n$ $m$ times. For example, to do $5+4$ is to do five and one and one and one and one. The definition of multiplication is very similar.

Of course, it becomes very boring to add like this if the numbers are somewhat big. For example, check this passage from Alice Through the Looking Glass:

A little context about the passage: Alice was supposed to prove her abilities to the Red Queen and to the White Queen, and so she is asked that question about addition. Now, do you really think that Alice does not know addition for not being able to keep track of the many “and one and one…” that the White Queen asks? I do not think so. But it does remind us of the simplicity of what it is to add: to do a number and one and one and one and so on.

However, we should not rely on the literal definition in order to carry out the calculation. Instead, we must rely on the place values that we use in order to write the very addends down. You must know the method already, but I will explain: when adding $23 + 98$ for example, you must stack them up, tens place on top of tens place, ones on top of ones:

$$\begin{align*} &23\\ +&98\\ \hline \end{align*}$$

Then, you must look at the rightmost numbers from the addends and add them: $3 + 8 = 11$. The number you will obtain is at most a two-digit number. You will write the ones place of the sum (of $11$, in this case, so $1$) below the line, aligned with respect to the place value:

$$\begin{align*} 2&3\\ +9&8\\ \hline &1 \end{align*}$$

The tens place of the sum (which was another $1$) you will carry, which means to write it on top of the upper addend of the next place value:

$$\begin{align*} \overset{1}{2}&3\\ +9&8\\ \hline &1 \end{align*}$$

Now you will repeat the process for the next place value, but adding the carry at the end: $2 + 9 = 11$, and $11 + 1 = 12$. We write the $2$ below, but there is nowhere to carry the $1$. In that case, you can simply write the carry to the left, in the next place value:

$$\begin{align*} \overset{1}{2}&3\\ +9&8\\ \hline 12&1 \end{align*}$$

The operation of addition has a few fundamental properties. For any natural numbers $a, b, c$

$$\begin{align*} (a+b)+c=a+(b+c) &\hspace {1cm}\text{Associative Property}\\ a+0=a &\hspace {1cm}\text{Identity Property}\\ a+b=b+a&\hspace {1cm}\text{Comutative Property}\\ \end{align*}$$

Above, and anywhere we use letters as if they were numbers, they are called variables. You should think of variables as placeholders for unknown or generic numbers. In the properties listed above, for example, each of $a, b$ and $c$ can be replaced by any natural number, hence the use of letters.

An example that clarifies why we need to be aware of these properties: add $17 + 14 + 3 + 6$. First, it does not matter if I do $17 + 14$ first or $14 + 3$ or $3 + 6$. Also, since everything we have is addition, the order is irrelevant.

Therefore, we can reorganize these numbers in a more convenient way that allows us to easily add them mentally:

$$\begin{align*} 17 + 14 + 3 + 6 &=\\ 17 + 3 + 14 + 6 &=\\ (17+3) + (14 + 6) &=\\ 20 + 20 &=\\ 40 \end{align*}$$

Of course, I had to literally write the numbers above for you to see, so I did not do it mentally, but I had to show to you how to organize your thoughts in order to do it mentally.

Also, take notice that when we write a lot of numbers with a lot of operations, we may mean that the calculations should be carried out in a specific order. So far, we have only talked about one operation, and this operation is associative and commutative, which basically means that parentheses (and any other grouping symbol, like brakets) are irrelevant; also, you may operate with numbers that are anywhere in the expression. More will be said about the order of operations later on.

There is a pattern to perceive in these exercises. If it did not become clear with the last one, then the ones below should be helpful. You may use a calculator if you want.

The previous exercises hinted at the twin operation to addition: subtraction.

3.2 Subtraction

If $m, n$ are two naturals such that $m$ is greater than or equal to $n$, to subtract $n$ from $m$, notation $m-n$, is to obtain a natural $p$ such that $p + n = m$.

For example, $24 - 9 = 15$, because $15 + 9 = 24$. However, $4 - 10$ was not defined, since our definition says that the first number should be greater than or equal to the second.

The previous exercise revealed to us that, differently from addition, subtraction is not commutative. So, it makes sense to give each element involved in a subtraction its proper name: the first operand is the minuend, and the second is the subtrahend. For example, in the subtraction $5-2$, $5$ is the minuend, and $2$ is the subtrahend. These names are useful if you want to quickly refer to one of them, but you may lose on clarity, since most people forget which is which.

Also, the number we obtain after subtracting is called the difference of the minuend and the subtrahend, in this order. So, when we say “the difference of 8 and 3”, we mean $8-3$, which is $5$. In some contexts, we assume to do in the order that would make sense. For example, I am $27$ years old. So, if I ask you to calculate the differente of our ages, you would do the greater number minus the smaller number.

Now, on to a practical aspect of subtraction of larger numbers: stacking them. If you wish to subtract two given numbers, you will write them in the same way that we wrote addition. For example, you can write $312 - 189$ as

$$\begin{array}{c} & 3 & 1 & 2\\ - & 1 & 8 & 9\\ \hline \end{array}$$

However, clearly we cannot subtract the rightmost digits: $2 - 9$ is not a valid subtraction of naturals. So, again, we take advantage of the place values: we will take one from the tens place, which is worth ten times as much in the ones place. Therefore, we will make the $1$ in the tens a 0, and the $2$ in the ones we will make a $12$:

$$\begin{array}{c} & 3 & \overset{0}{\not 1} & \overset{12}{\not2}\\ - & 1 & 8 & 9\\ \hline \end{array}$$

Being able to subtract now, write the difference of the rightmost numbers:

$$\begin{array}{c} & 3 & \overset{0}{\not 1} & \overset{12}{\not2}\\ - & 1 & 8 & 9\\ \hline & & & 3\\ \end{array}$$

Again we will borrow $1$ from the left place value, since $0 - 8$ is undefined. Take notice that it is not necessary to put a slash through the carry, since you can attach the digit to the $0$, yielding a clearer notation. Compare the two writings below:

$$\begin{array}{c} & \overset{2}{\not 3} & \overset{\overset{10}{\not 0}}{\not 1} & \overset{12}{\not2}\\ - & 1 & 8 & 9\\ \hline & & & 3\\ \end{array}$$

versus

$$\begin{array}{c} & \overset{2}{\not 3} & \overset{10}{\not 1} & \overset{12}{\not2}\\ - & 1 & 8 & 9\\ \hline & & & 3\\ \end{array}$$

Then, carry out the rest of the calculations just like before:

$$\begin{array}{c} & \overset{2}{\not 3} & \overset{10}{\not 1} & \overset{12}{\not2}\\ - & 1 & 8 & 9\\ \hline & 1& 2& 3\\ \end{array}$$

A trickier case is when we have some zeros in the minuend. For example, if we do $20012 - 12398$, we can write

$$\begin{array}{c} & 2 & 0 & 0 & 1 & 2\\ - & 1 & 2 & 3 & 9 & 8\\ \hline \end{array}$$

The first step is the same as before:

$$\begin{array}{c} & 2 & 0 & 0 & \overset{0}{\not 1} & \overset{12}{\not 2}\\ - & 1 & 2 & 3 & 9 & 8\\ \hline & & & & & 4 \end{array}$$

Now, we cannot do $0 - 9$, but we cannot borrow from the $0$ in the hundreds either!

But there is a way: we borrow a $1$ from the $2$ in the ten-thousands place and make the $0$ in the thousands become a $10$:

$$\begin{array}{c} & \overset{1}{\not 2} & \overset{10}{\not 0} & 0 & \overset{0}{\not 1} & \overset{12}{\not 2}\\ - & 1 & 2 & 3 & 9 & 8\\ \hline & & & & & 4 \end{array}$$

And now we can borrow from that $10$, making it a $9$, so that we can make the $0$ in the hundreds become a $10$:

$$\begin{array}{c} & \overset{1}{\not 2} & \overset{\overset{9}{\not 10}}{\not 0} & \overset{10}{\not 0} & \overset{0}{\not 1} & \overset{12}{\not 2}\\ - & 1 & 2 & 3 & 9 & 8\\ \hline & & & & & 4 \end{array}$$

Finally, we borrow $1$ from that $10$ in the hundreds, similar to what we did above:

$$\begin{array}{c} & \overset{1}{\not 2} & \overset{\overset{9}{\not 10}}{\not 0} & \overset{\overset{9}{\not 10}}{\not 0} & \overset{10}{\not 1} & \overset{12}{\not 2}\\ - & 1 & 2 & 3 & 9 & 8\\ \hline & & & & & 4 \end{array}$$

And finally, $10 - 1 = 1$:

$$\begin{array}{c} & \overset{1}{\not 2} & \overset{\overset{9}{\not 10}}{\not 0} & \overset{\overset{9}{\not 10}}{\not 0} & \overset{10}{\not 1} & \overset{12}{\not 2}\\ - & 1 & 2 & 3 & 9 & 8\\ \hline & & & & 1& 4 \end{array}$$

However, it is clearly implied by the method that if we are to borrow $1$ from a place value that is far to the left and there are only zeroes in between, then those zeros become nines and we put that $1$ that we need, yielding a clearer notation:

$$\begin{array}{c} & \overset{1}{\not 2} & \overset{9}{\not 0} & \overset{9}{\not 0} & \overset{10}{\not 1} & \overset{12}{\not 2}\\ - & 1 & 2 & 3 & 9 & 8\\ \hline & & & & 1& 4 \end{array}$$

And we can finish just like before:

$$\begin{array}{c} & \overset{1}{\not 2} & \overset{9}{\not 0} & \overset{9}{\not 0} & \overset{10}{\not 1} & \overset{12}{\not 2}\\ - & 1 & 2 & 3 & 9 & 8\\ \hline & 0& 7& 6& 1& 4 \end{array}$$

Therefore, $20012 - 12398 = 7614$.

Here goes a video exemplifying the technique:

Now, we shall see a mathematical result about subtraction that allows us to make some of them become easier.

For example, $13000 - 1232$ will require quite a bit of carrying. However, it is easy to take 1 from each number, and we get $12999 - 1231$, which is much simpler to carry out.

Before we talk about multiplication and division, I have a theoretical question and a few comments to make about subtraction:

Since subtraction is not associative, we must agree on the order that we simplify it when we have an expression with more than one subtraction operation and no grouping symbols, like $10 - 5 - 2$. We define that we must solve from left to right. So, writing $10 - 5 - 2$ is defined to be the same as $(10 - 5) - 2$. The convention is to do the same think with addition, but since addition is associative, this is not so relevant in that case.

3.3 Multiplication

We have already defined what multiplication is: simply put, it is repeated addition. This operation is denoted in a few different ways. The notation we learn first is the cross notation, $\times$. However, there are two other ways that we will prefer to use in a higher level: the dot $\cdot$ and attaching expressions in grouping symbols.

The dot is simple to use. For example, $4 \cdot 5$ means $4$ times $5$, which gives 20.

The other notation, the one with grouping symbols, basically requires you to write down two expressions attached to one another with at least one of them wrapped in grouping symbols. Below are all ways to represent the product of $4$ and $5$:

$$4 \times 5$$ $$4 \cdot 5$$ $$4(5)$$ $$(4)5$$ $$(4)(5)$$

The operands of multiplication are called factors, and the result of multiplying the factors is their product. For example, the product of $4$ and $7$ is $28$. Two factors of $27$ are $3$ and $9$. When two factors of a number multiply to that number, we say that they form a factor pair of such number.

And since we are going to start to write expressions with multiple operations going on, we define that multiplication has priority over addition and subtraction.

The properties of multiplication are

$$\begin{align*} (a \cdot b)\cdot c &= a \cdot (b \cdot c) &\hspace{1cm}\text{Associative Property}\\ a \cdot 1 &= a &\hspace{1cm}\text{Identity Property}\\ a \cdot b &= b \cdot a &\hspace{1cm}\text{Commutative Property}\\ \end{align*}$$

We also define that a number times $0$ is $0$.

Also, there is a very useful property that relates multiplication and addition, the distributive property:

$$\begin{align*} a \cdot (b + c) = a \cdot b + a \cdot c &\hspace{1cm}\text{Distributive Property}\\ \end{align*}$$

This property will allow us to multiply bigger numbers easily. In order to see why, we need to learn how to multiply by a power of 10.

Powers, exponents and powers of 10

We need to define a new operation in order to be able to write simply about what follows.

You can also think about this definition recursively: $b^0 = 1$ and, for any number $n$ larger than $0$, $b ^ n = b \cdot b^{n-1}$.

If $b^n = x$, then we can say generically that $x$ is a power of $b$. For example, $8$ is a power of $2$ because $2^3 = 8$.

When we talk about base $10$ notation — and it generalizes to other bases as well — we are talking about expressing the number using ten different symbols (the digits) as our base and then combining different powers of $10$ in order to represent the number.

For example, the number $567$ represents $5 \cdot 10^2 + 6 \cdot 10^1 + 7 \cdot 10^0$, which is the same as $5(100) + 6(10) + 7(1)$.

Now the key point of this section: when we write a digit followed by zeros, that can be expressed as the product of the digit by a power of ten.

For example, $5000 = 5(10^3)$. $1,000,000 = 1(10^6).$ And so on.

How to carry out multiplication using the distributive property

So, given two $2$-digit numbers, for example, here goes a way to multiply them using the distributive property. Bear in mind that what follows is an explanation for the method of stacking the numbers, which maybe you already know. It does not mean that we will carry out multiplication in this way. There is a video below the explanation in case you want to skip to how it wraps up, but the reason that the stacking method works is presented below via an example.

$$\begin{align*} 23 \cdot 67 &= (23) (6 \cdot 10 + 7) \\ &= 23 \cdot 6 \cdot 10 + 23 \cdot 7 \end{align*}$$

In the last passage, we mulplied the outer factor, $23$ by each one of the addends, that is $6 \cdot 10$ and $7$.

$23 \cdot 6$ can be developed using the same method:

$$\begin{align*} 23 \cdot 6 &= (2\cdot 10 + 3) \cdot 6 \\ &= 6 \cdot 2 \cdot 10 + 6 \cdot 3 \\ &= 12 \cdot 10 + 18 \\ &= 120 + 18 \\ &= 138 \end{align*}$$

In the first step, I used distributivity and commuativity at the same time. More explicitly, $(2 \cdot 10 + 3) \cdot 6 = 6 \cdot (2 \cdot 10 + 3) = 6 \cdot 2 \cdot 10 + 6 \cdot 3$.

Similarly,

$$\begin{align*} 23 \cdot 7 &= (2\cdot 10 + 3) \cdot 7 \\ &= 7 \cdot 2 \cdot 10 + 7 \cdot 3 \\ &= 14 \cdot 10 + 21 \\ &= 140 + 21 \\ &= 161 \end{align*}$$

Back to $23 \cdot 6 \cdot 10 + 23 \cdot 7$ now:

$$\begin{align*} 23 \cdot 6 \cdot 10 + 23 \cdot 7 &= 138 \cdot 10 + 23 \cdot 7 \\ &= 1380 + 161 \\ &= 1541 \end{align*}$$

Watch this video to see how it all wraps up neatly:

3.4 Division

Division is (sort of an) inverse operation to multiplication. I say “sort of” because most natural numbers do not have multiplicative inverses. For example, what natural number can you multiply by $3$ in order to obtain $5$? None.

However, it is still possible to count as many times as possible a number into another. Using the example above, we could say that $3$ can be counted at most $1$ time into $5$.

This means that to divide $D$ by $d$ is to find the greatest natural number $q$ such that $dq \leq D$. The symbol $\leq$ should be read as is less than or equal to. Back to the previous example, we can say that $5 \div 3 = 1$ because $1$ is the greatest number $d$ such that $d \cdot 3 \leq 5$. Notice that $1 \cdot 3 = 3$ which is less than 5, but $2 \cdot 3 = 6$ which is greater.

The remainder is the difference $D - dq$. So $5 \div 3 = 1$ remainder $2$, because $5 - 3\cdot 1 = 2$. This is commonly written as $5 \div 3 = 1 R 2$.

Requiring the divisor to be different than $0$ makes division have meaning. Else, allowing for something like $4 \div 0$ to be asked would be the same as asking how many times at most can you take 0 from 4? That question does not have an answer, because there is no such greatest number of times.

Also, it is very important to keep in mind that quotient does not mean answer. For example, assume that today is Wednesday. What day of the week will it be in 7 days?

If you count from today, tomorrow is day $1$ in the counting, so Thursday is $1$, and Friday is $2$ and so on. Therefore, day $7$ occurs on Wednesday.

In all of these problems, the remainder was the relevant part of the division, because it told us how many days from Wednesday the day we were looking for was.

Here goes one more challenge:

3.5 Long division

Just like we did before for the other operations, we can write division down in a special way relying on the place values of the numbers involved in order to make it simple.

The logic relies on finding how many times the divisor goes into the lefttmost digit of the dividend first, obtaining then a remainder. That remainder pertains in that place value, so it is worth $10$ times as much in the place value to the right of it.

For example, suppose that we want to find the quotient and the remainder in $843 \div 3$. Starting from the hundreds place, $3$ goes twice into $8$, leaving a remainder of $2$. So $2$ is the digit we write in the hundreds place of the quotient, and the remainder is $2$ hundreds.

Now, when we go to the next place value, the remainder we had is worth $10$ times as much, so we will have the $4$ that was already in the tens place plus $20$, so $24$. $3$ goes $8$ times into $24$, so an $8$ is the digit we write in the tens place of the quotient. Remainder $0$.

On to the ones now: $3$ goes $1$ time into $3$, so $1$ is the ones place of the quotient.

Hence, $843 \div 3 = 281 R 0$. Watch the following video to see the same example written in long division form.

When the divisor has more digits, I like to write the first multiples of the divisor to the side as I write the division down. Check this video to see what I mean: