Natural numbers arithmetic

What is a natural number

To start off, we will look at natural numbers. A natural number refers to numbers used for counting.

When counting we have 10 unique numbers to work with. In order, they are...

If you have anything that can be counted in terms of whole numbers (\(1, 2, 3, \ldots\)) that number can be considered a natural number (I have 5 apples, the jar has 450 jelly beans, there is 602214076000000000000000¹ atoms in 12 gram of carbon, etc...).

Once we have used up all the numbers, we add a new 'place' to the left, increment it by one, and then start over. Once that new place gets filled up, we repeat the process. Here is an animation that I think will make it a bit clearer.

The right most digit is called either the ones place or the units place. Moving to the left we call that the tens place, this is because every circle that is filled in this place refers to a full set of ten circles that have been filled in the ones place. Moving left again to the number in the third position, we refer to as the hundreds place, because similar to the tens, every circle in the hundreds place refers to ten circles filled in the tens place which refers to a hundred circles filled in the ones place. Below is a table of the first few places for later reference.

For the number

\[19,083,564\]

we have
- \(4\) in the 1's place (representing 4 'units' of what we are counting)
- \(6\) in the 10's place (representing 6 sets of 10 of the units we are counting)
- \(5\) in the 100's place (representing 5 sets of 100 of the units we are counting)
- \(3\) in the 1000's place (" " '' 3 sets of 1000 of the unit)
- \(8\) in the 10,000's place ('" " " 8 sets of 10,000 of the unit)
- \(0\) in the 100,000's place (" " " 0 sets of 100,000 of the unit)
- \(9\) in the 1,000,000's place(" " " 9 sets of 1,000,000 of the unit)
- \(1\) in the 10,000,000's place(" " " 1 set of 10,000,000 of the unit)

We could also say that there is a \(0\) in the 100,000,000's place, which would represent 0 sets of a hundred million of the unit. Technically we could write the number as something like

\[\ldots0,000,000,000,019,283,564\]

with any amount of \(0\)'s preceding the \(19,083,564\) While this is not wrong, because we can always add more places and zeros, we generally don't show zero's to the left of a number. However there are examples where you would show the leading zero's such as in a date \((02/05/2000)\). It's fine to have them if it helps with clarity, however in most cases it doesn't make sense to add them.

If you notice the 100,000's (hundred thousands) place it's 0, but because we have a non-zero number to the left of it (\(9\)) it still contributes to the total number. You could also think of there being 90 sets of 100,000 (which is equivalent to 0 sets of 100,000 and 9 sets of 1,000,000).

We've already talked a bit about combining sets of things, but lets formalize it and go over our first operation, addition.

Adding two natural numbers

The process of adding means combining the two different sets of things into one. An alternate way of saying this is 'finding the sum of' the two things. It is denoted with a plus sign \((+)\). If we have 4 things, and we add 3 more things, this means putting the two sets of 4 and 3 things together and seeing how many we have total, In this case we have we have 7 things total.

A good way to think about and visualize adding is with a number line.

On this number line, each space is equally distanced. on the left it starts at \(0\) and on the right the arrow pointing (\(\rightarrow\)) tells you that it keeps going. So while we can only see up to \(16\) on this one, the numbers would keep going if not for the size constraints of the paper your using/this webpage. If we mark each space as 'one unit' of what we are counting and fill it in corresponding with each set we are adding, we can get an interesting geometric intuition of what addition is.

We fill in up to \(4\) on the number line associated with the first set.

We fill in up to \(3\) on a number line associated with the second set

Finally by combining the two lengths we can see how far they reach when put side by side. As we can see it reaches to the answer we found which was \(7\)

At its core, adding natural numbers is just counting. The number 135 (meaning 1 in the hundreds place, or 100 of the unit. 3 in the 10's place, or 30 of the unit, and 5 in the 1's place, or 5 of the unit) can be thought of as \(100 + 30 + 5\). Lets look at what I mean with an example that uses smaller numbers.

Starting off we have a big set of marbles. Counting them individually we can see we have \(24\) of them

We can break this up into two sets of \(10\) and one set of \(4\)

If we add those 3 sets together \((10 + 10 + 4)\) we can get back to \(24\), notice that nothing really changed throughout each of these steps we just rearranged and thought about the groups of marbles in a slightly different way.

Here we can see that when using a number line to show \(10 + 10 + 4\) we do indeed get back to \(24\)

So why is this a useful way to think about numbers? To see, lets look at the example \(13 + 11\). Similar to how we broke up \(24\) into \(10 + 10 + 4\), we can break both the \(13\) and \(11\) into simpler numbers and add them together. Lets turn \(13\) in \(10 + 3\) and \(11\) into \(10 + 1\) visually with sets we have

and numerically we have \(10 + 3 + 10 + 1\), By adding the \(3\) and \(1\) we can get 4 leaving our equation to be \(10 + 10 + 4\) which is the same as what we had above!

Additionally with our number line visual we can see that \(10 + 10 + 4\) is the same as \(13 + 11\) which is also the same as \(24\). One thing I want you to take note of is that while red is first and blue is second, if we were to switch the order to blue then red, we would still get to the same result.

We can see that visually here. Another way to say this is 13+11 is the same as 11 + 13. This is the Commutative low of Addition

Commutative Law of Addition

\[a + b = b + a\]

The Commutative Law of Addition simply states that we can swap the order of numbers and still get the same answer

Note that we are using letters instead of numbers here. What this notation generally means is that \(a\) and \(b\) can be any number, but all instances of \(a\) are the same number and all instances of \(b\) are the same number. for example

• \(4 + 68 = 68 + 4\)

  • in this case \(a\) would be 4 and \(b\) would be 68.

• \(593+229 = 229+593\)

  • in this case \(a = 593\) and \(b=229\)

\(\text{any number } 'a' + \text{any number } 'b' = \text{the same number } 'b' + \text{the same number }'a'\)

Here is a table of adding two single digit numbers for reference as well as an interesting visual of the Commutative property. You can see that the whole table is mirrored on the diagonal because \(a + b = b + a\)

A similar thing could be said for the \(10 + 10 + 4\) number line. As long as we are stacking them side by side, its length will always reach 24.

It doesn't matter what order we add add numbers in \(10 + 10 + 4\) is the same as \(10 + 4 + 10\) is the same as \(4 + 10 + 10\). This is The Associative law of Addition

Associative Law of Addition

\[(a + b) + c = a + (b + c)\]

The Associative Law of Addition just states that we can groups of numbers in any order and still get the same number. Similar to above, \(a\), \(b\), and \(c\) are all numbers, Additionally all the parenthesis mean is that we do add those numbers first. In the above example...

\[\begin{align*} (10 &+ 10) + 4 = 10 + (10 + 4) \\ &\Downarrow \qquad\qquad\qquad\qquad\,\,\,\,\Downarrow\\ &20 + 4 \quad\,\,\,=\quad\,\,\, 10 +14 \\ &\quad\,\,\Downarrow\qquad\qquad\qquad\,\,\Downarrow\\ &\quad24\qquad\, =\,\,\,\qquad 24\end{align*}\]

Note that it we can do this with more than 3 numbers

\[1 + 2 + 3 + 4 = 10\]

---

Adding the first two numbers

• \(3 + 3 + 4 = 10\)

Adding from left to right

• \(6 + 4 = 10\)
• \(10 = 10\)

---

Now lets add the 2 + 3 first

• \(1 + 5 + 4 = 10\)

Adding the 5 + 4

• \(1 + 9 = 10\)
• \(10 = 10\)

Adding Larger numbers

For this Next problem we are going add two large numbers that we can't use our fingers for. Once we work through it with the symbols to see the process, I'll go over an algorithm to quickly add any two numbers.

We are going to add squares, because there are so many of them we will define a triangle to be the same as 10 squares, and we will define a star to be the same as 10 triangles which is the same as 100 squares.

We're going to add \(88\) squares to the \(156\) squares that we have.

Writing this with our symbols (and remembering that we can write \(155\) as \(100 + 50 + 5\) and \(88\) as \(80 + 8\)) we have this

Using the associative law (we can add things in any order) lets group all the squares, triangles, and stars together to see how many of each we have

With all of the shapes grouped we can see that we have more than 10 squares, so lets remove 10 and make them a triangle as we defined triangles as 10 squares above.

This leaves us with 3 squares left over, but now we also have more than 10 triangles, so lets turn 10 of them into a star

With that there is no more simplifying we can do and we are left with 2 stars, 4 triangles, and 3 squares which equates to 243 squares total. So \(155 + 88 = 243\)

Lets look at the above problem but with numbers

\[\begin{align*}155\\+ \,\,88\\\hline\,\end{align*}\]

Following the same steps as before lets break it into separate numbers, ill also include the 'quicker' way on the right to see the direct comparison with the shortcut.

\(\qquad\qquad\qquad\begin{align*}100 + 50 + 5\\+ \qquad\,\,\,\,\,80 + 8\\\hline\,\end{align*}\) \(\qquad\qquad\qquad\begin{align*}155\\ +\,\,\, 88\\\hline\,\end{align*}\)

starting with the far right we add 8 and 5.

\(\qquad\qquad\qquad\begin{align*}100 + 50 + \textcolor{red}{5}\\+ \qquad\,\,\,\,\,80 + \textcolor{red}{8}\\\hline\,\end{align*}\) \(\qquad\qquad\qquad\begin{align*}15\textcolor{red}{5}\\ +\,\,\, 8\textcolor{red}{8}\\\hline\,\end{align*}\)

\(8 + 5=13\), We can break that \(13\) into \(10 + 3\)

\(\qquad\qquad\qquad\begin{align*}100 + 50 + 0\\ \qquad\,\,\,\,\,80 + 0\\+\qquad\quad \textcolor{red}{10 + 3}\\\hline\,3\end{align*}\) \(\qquad\qquad\qquad\begin{align*}\textcolor{red}{1}\,\,\,\\155\\ +\,\,\, 88\\\hline\,\textcolor{red}{3}\end{align*}\)

\(0+0+3 = 3\), now we can move onto the 10's place

\(\qquad\qquad\qquad\begin{align*}100 + \textcolor{red}{50} + 0\\ \qquad\,\,\,\,\,\textcolor{red}{80} + 0\\+\qquad\quad \textcolor{red}{10} + 3\\\hline\,3\end{align*}\) \(\qquad\qquad\qquad\begin{align*}\textcolor{red}1\,\,\,\\1\textcolor{red}{5}5\\ +\,\,\, \textcolor{red}{8}8\\\hline\,3\end{align*}\)

\(50 + 80 + 10 = 50 + 90 = 140\), we can break that into \(100 + 40\). Remember that if there is nothing to the left of a number, we can think of it as a \(0\) in that place.

\(\qquad\qquad\qquad\begin{align*}100 +\,\,\, 0 + 0\\ \qquad\,\,\,\,\,0 + 0\\\qquad\quad 0 + 3\\+\,\textcolor{red}{100 + 40}\,\,\,\,\quad\\\hline\,40+3\end{align*}\) \(\qquad\qquad\qquad\begin{align*}\textcolor{red}{1}1\,\,\,\\155\\ +\,\,\, 88\\\hline\,\textcolor{red}{4}3\end{align*}\)

\(0 + 0 + 0 + 40 = 40\), now we can move onto the hundred's place

\(\qquad\qquad\qquad\begin{align*}\textcolor{red}{100} +\,\,\, 0 + 0\\ \qquad\,\,\,\,\,0 + 0\\\qquad\quad 0 + 3\\+\,\textcolor{red}{100} + 40\,\,\,\,\quad\\\hline\,200+40+3\end{align*}\) \(\qquad\qquad\qquad\begin{align*}\textcolor{red}{1}1\,\,\,\\\textcolor{red}{1}55\\ +\,\,\, 88\\\hline\,243\end{align*}\)

With that we have found our answer.

For the algorithm to solve these, it's equivalent to add the numbers in the same place, if it is over 10, we take the value in the 1's place and add 1 to the next place. We repeat until we have gone through all of the places and have found our answer.

To finish off this section i'll give a few examples of adding different natural numbers

Examples

1.

• Starting out we have \(8 + 4 = 12\) so we put a \(2\) in the unit's place, and move the \(1\) to the \(10\)'s place
• We then add the \(10\)'s place and get a \(2\) (equivalent to \(20\))
• We move on to the hundreds place and 'add it to 0' which essentially means just bringing it down. With that we have our answer of \(122\)

2.

• Starting with the \(1\)'s place we add the two numbers, because it stays under 10 we don't carry anything to the next place.
• At the \(10\)'s place we have \(9+8=17\), so we leave the unit's place (\(7\)) and carry the \(1\) to the next place
• At the \(100\)'s place we have \(7 + 2 + 1 = 10\), even though it's a \(0\) in the unit's place, we still leave it and carry the \(1\) to the next place.
• At the \(1000\)'s place we only have \(1\) or \(1 + 0 + 0 = 1\), so we bring it down and find our answer.

3.

Even though the smaller number is on top, the same process still works because of the commutative property (\(a+b = b+a\))
- starting with the 1's, we have \(4+2=6\) and then move on to the \(10's\) place
- in the \(10\)'s place we have \(8 + 9 = 17\), so we leave the \(7\), and move the \(1\) over to the next place.
- in the \(100\)'s place, we have \(1 + 0 + 3 = 4\) and with that we have our final answer

Practice

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