Addition and subtraction with natural numbers
Adding 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...
\[\begin{aligned} (10 &+ 10) + 4 = 10 + (10 + 4) \\ &\Downarrow \qquad\qquad\qquad\qquad\,\,\,\,\Downarrow\\ &20 + 4 \quad\,\,\,=\quad\,\,\, 10 +14 \\ &\quad\,\,\Downarrow\qquad\qquad\qquad\,\,\Downarrow\\ &\quad24\qquad\, =\,\,\,\qquad 24\end{aligned}\]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
- \(1 + 9 = 10\)
- \(10 = 10\)
Before moving onto adding some larger numbers, there is one more property of addition I want to go over. This is something called the Additive Identity
Additive Identity
Remembering that \(a\) is a number, and it can by any number. We have the property
\[a + 0 = a\]some examples with numbers
- \(0+0=0\)
- \(1+0 = 1\)
- \(42 + 0 = 42\)
- \(1337 + 0 = 1337\)
Any number we put in the ''\(a\)'' space plus \(0\) won't change the number.
A way to visualize this would be thinking about 2 marble bags, one has 4 marbles in it, the other is empty. We are going to pour all the marbles from the second bag into the first and then count how many we have total. When we pour it, nothing comes out because its empty. And 'intuitively' the number of marbles in the bag hasn't changed (we still have \(4\))
Now we have everything we need to add numbers of any size.
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
\[\begin{aligned}155\\+ \,\,88\\\hline\,\end{aligned}\]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.
\(\qquad\qquad\qquad\begin{aligned}100 + 50 + 5\\+ \qquad\,\,\,\,\,80 + 8\\\hline\,\end{aligned}\) \(\qquad\qquad\qquad\begin{aligned}155\\ +\,\,\, 88\\\hline\,\end{aligned}\)
starting with the far right we add 8 and 5.
\(\qquad\qquad\qquad\begin{aligned}100 + 50 + \textcolor{red}{5}\\+ \qquad\,\,\,\,\,80 + \textcolor{red}{8}\\\hline\,\end{aligned}\) \(\qquad\qquad\qquad\begin{aligned}15\textcolor{red}{5}\\ +\,\,\, 8\textcolor{red}{8}\\\hline\,\end{aligned}\)
\(8 + 5=13\), We can break that \(13\) into \(10 + 3\)
\(\qquad\qquad\qquad\begin{aligned}100 + 50 + 0\\ \qquad\,\,\,\,\,80 + 0\\+\qquad\quad \textcolor{red}{10 + 3}\\\hline\,3\end{aligned}\) \(\qquad\qquad\qquad\begin{aligned}\textcolor{red}{1}\,\,\,\\155\\ +\,\,\, 88\\\hline\,\textcolor{red}{3}\end{aligned}\)
\(0+0+3 = 3\), now we can move onto the 10's place
\(\qquad\qquad\qquad\begin{aligned}100 + \textcolor{red}{50} + 0\\ \qquad\,\,\,\,\,\textcolor{red}{80} + 0\\+\qquad\quad \textcolor{red}{10} + 3\\\hline\,3\end{aligned}\) \(\qquad\qquad\qquad\begin{aligned}\textcolor{red}1\,\,\,\\1\textcolor{red}{5}5\\ +\,\,\, \textcolor{red}{8}8\\\hline\,3\end{aligned}\)
\(50 + 80 + 10 = 50 + 90 = 140\), 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.
\(\qquad\qquad\qquad\begin{aligned}100 +\,\,\, 0 + 0\\ \qquad\,\,\,\,\,0 + 0\\\qquad\quad 0 + 3\\+\,\textcolor{red}{100 + 40}\,\,\,\,\quad\\\hline\,40+3\end{aligned}\) \(\qquad\qquad\qquad\begin{aligned}\textcolor{red}{1}1\,\,\,\\155\\ +\,\,\, 88\\\hline\,\textcolor{red}{4}3\end{aligned}\)
\(0 + 0 + 0 + 40 = 40\), now we can move onto the hundred's place
\(\qquad\qquad\qquad\begin{aligned}\textcolor{red}{100} +\,\,\, 0 + 0\\ \qquad\,\,\,\,\,0 + 0\\\qquad\quad 0 + 3\\+\,\textcolor{red}{100} + 40\,\,\,\,\quad\\\hline\,200+40+3\end{aligned}\) \(\qquad\qquad\qquad\begin{aligned}\textcolor{red}{1}1\,\,\,\\\textcolor{red}{1}55\\ +\,\,\, 88\\\hline\,243\end{aligned}\)
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
Addition Practice
Subtracting Natural Numbers
The process of subtraction could be seen as (and is actually how subtraction is defined) the opposite or the 'inverse' of addition. Rather than putting new things into a set, we are taking some of the things that are already in a set and pulling them out. The way we say this is finding the difference of and the notation we use is a minus sign \((-)\) Subtraction has the property Looking at the first example we had \(4+3=7\) as seen here
This could represent having \(4\) marbles in a bag, then putting \(3\) new ones into that bag. In total we have \(7\) marbles now in the bag. Now with that bag of \(7\) marbles, lets take \(3\) of them out. We can visualize this in a similar way
If we take \(3\) out, we get back to our starting \(4\). So what we have essentially done here is the equation \(4 + 3 - 3 = 4\). With this, we can see that what addition does, subtraction 'un'-does.
Definition of Subtraction
\[a-a = 0\]This states that if we subtract two numbers that are the same we get to \(0\) which is the additive identity.
Another way to think about subtraction is to think about the subtracted number as 'canceling out' the other numbers. For a visual of what I mean when you put a subtraction sign in front of the circles, it turns them into 'anti-circles' which will 'pop' the normal circles when they touch
Using the \(7-3\) problem
and breaking it down into two sets (with one of the sets elements being turned into an 'anti' versions of themselves.
You can see that we have 4 left over.
For a final way to think about subtraction (and my personal favorite), we can use the number line analogy again. Except this time instead of 'adding' on to the total length of the line, we take away the overlap, and what is left is our final answer!
Subtraction Practice
- show differences between addition and subtraction numberline method
- subtraction step by step like I did for addition