In this module, we’ll provide a review of the basic mathematical operations (addition, subtraction, multiplication & division) with whole numbers.

Module 2 Learning Objectives

At the completion of this module (and its respective sections), you will be able to:

  1. Read and write whole numbers in the base-ten system (CLO 1)

  2. Perform computations involving Addition, Subtraction, Multiplication, and Division of Whole Numbers (CLO 1, 4)

  3. Apply the rule of rounding to estimate sums and differences (CLO 5)

Whole Numbers

Writing Whole Numbers

Place Value System

The Base-10 System (our normal number system) is a place-value system that depends on 3 main objectives.

  1. The tens-digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9

  2. The placement of each digit

  3. The value of each place where the digit is located.

Example 1

The number 954 can be broken down into the following pieces:

  • 9 is located in the 100’s place

  • 5 is located in the 10’s place

  • 1 is located in the 1’s place

In addition, 954 can be written using expanded notation:

\[ 9(100) + 5(10) + 4(1) = 900 + 50 + 1 = 954 \]

Example 2

The number 6507 can be broken up into the following pieces:

  • 6 is located in the 1000’s place

  • 5 is located in the 100’s place

  • 0 is located in the 10’s place

  • 7 is located in the 1’s place

In addition, 6507 can be written using the expanded notation

\[ 6\times 1000 + 5\times 100 + 0 \times 10 + 7\times 1 = 6000+500+0+7 = 6507 \]

Definition - Whole Numbers

The Whole Number System (sometimes called the positive Integers) are the counting numbers. Remember, these are numbers you can count on your fingers and toes. Mathematically, the numbers are:, 𝕎 = 0 , 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , …

 

Alignment of Whole Numbers

Figure 1: Placement of Numbers
Figure 1: Placement of Numbers
   

The figure above outlines the placement, values, and names of the numbers. There are a few things to remember when you’re reading and writing whole numbers:

  1. Digits are read in groups of three
  2. Separation -
    1. In the U.S. we use commas to separate the groups of three digits.
    2. In other parts of the word, a period is used to separate the groups of three digits.
  3. The word and does not appear when we write. It only shows up when we say it.
  4. Hyphens are used to write English words for the two-digit numbers from 21 to 99, except for those that end in zero.

Example 3

  • \(653\) is standard notation

  • \(600 + 50 + 3\) is the expanded notation

  • six hundred fifty-three is the English word equivalent.

Example 4

  • \(75,900\) is the standard notation

  • \(70,000 + 5,000 + 900 + 0 + 0\) is the expanded notation

  • Seventy-five thousand, nine hundred is the English word equivalent.

Example 5

Write one hundred eighty thousand, five hundred forty-three in standard notation.

To do this, lets look at each group of three.

  • One hundred eighty thousand is \(180,000\)

  • Five hundred forty-three is \(543\)

Then we write it in expanded notation: \(180,000 + 543\)

Combining it, then we have \(180,543\)

Adding and Subtracting Whole Numbers

To start this section off, we have to cover a few fundamental (and abstract) definitions. Bear with me on this.

Definition - Commutative Property of Addition

For any two whole numbers a and b, a + b = b + a In other words, if you're dealing with addition, the order on how you add the numbers does not matter. You'll still end up with the same answer.

 

Definition - Associative Property of Addition

For any whole number a, b and c, a + b + c =   ( a + b ) + c = a + ( b + c ) In other words, if you're dealing with addition, the order on how you add the numbers does not matter. You'll still end up with the same answer.

 

Definition - Additive Identity Property of Addition

For any whole number, a, a + 0 = 0 + a = a This means, the additive identity is zero, because if you add zero to a number, it's still the same number.

 

Example 1

\((15+20)+3 = 15 + (20 + 3)\) is an example of the Associative Property

Example 2

\(8+4 = 4+8 = 12\) is an example of the Commutative Property

Example 3

\(15 + 0 = 15\) is an example of the Identity Property of Addition.

Addition of Whole Numbers

To add numbers, yes, we’ll 99.99% of the time use a calculator. However, if there’s that 0.01% chance you may not have a calculator, then you may need to do it by hand. Or perhaps, check to make sure the calculator is correct (yes that does happen with programming errors).

The most common method to add by hand is vertically. If you remember a few key pieces of information:

  • When you add numbers in one column and if that sum is greater than 9, you keep the ones digit in there and carry the tens digit over to the next column.

  • You work from right to left when you add these numbers.

Example 4

Let’s add the following numbers, \(328\) and \(604\)

\[\begin{equation*}\begin{array}{c} \phantom{\times99}\color{green}1\\ \phantom{\times99}328\\ \underline{+\phantom{99}604}\\ \phantom{\times99}932\\ \end{array}\end{equation*}\]

In this example (starting right to left), we add the \(8\) and the \(4\) to get \(12\). Keep in mind, we keep the \(2\) in the ones column, but we “carry” over the \(1\) into the next column, which we have as green on the top. Then we add \(1\), \(2\) and \(0\) to get \(3\). Since \(3\) isn’t greater than \(9\), we can move over to the next column, which then, we’ll add \(3\) and \(6\) to get \(9\). Therefore our final answer is \(932\).

Example 5

Let’s take it up a notch and add the following numbers: \(328\), \(605\), \(517\), and \(192\)

\[\begin{equation*}\begin{array}{c} \phantom{\times9}\color{green}{1 2}\\ \phantom{\times99}328\\ \phantom{\times99}605\\ \phantom{\times99}517\\ \underline{+\phantom{99}192}\\ \phantom{\times9}1642\\ \end{array}\end{equation*}\]

Still, we’ll arrange them vertically, just like we did in Example 4. In the one’s column (far right), we add \(8\), \(5\), \(7\), and \(2\) to get \(22\). The rightmost 2 goes in the one’s column, the left most 2 goes up to the top on the column moving left (denoted in Green). Then we add 2, 2, 0, 1, 9 to get 14. The 4 goes at the bottom, the 1 goes up to the top on the next column (denoted in Green). Then we add 1, 3, 6, 5, 1 to get 16. Since we don’t have any more columns, we keep the 16 at the bottom. Our final answer would be \(1,642\).

Example 6

Dr. McMurrough bought a brand new TV for the Chemistry class. She spent $859 for the TV, $697 for the sound system and $1,285 for the podium that it’s housed in. How much did she spend all total?

\[\begin{equation*}\begin{array}{c} \phantom{\times9}\color{green}{1 2 2}\\ \phantom{\times99}859\\ \phantom{\times99}697\\ \underline{+\phantom{9}1285}\\ \phantom{\times9}2841\\ \end{array}\end{equation*}\]

Subtraction of Whole Numbers

Unfortunately, subtraction isn’t as easy as addition. The associative and commutative properties of addition can’t work with subtraction. Let’s take a look:

\(5-2\) isn’t the same thing as \(2-5\). The only property that does work would be the identity property; \(15-0\) is still 15.

To subtract, we need to take the act of “borrowing” from the next column, still working from left to right.

Example 7

Subtract the following numbers: \(496 - 342\)

\[\begin{equation*}\begin{array}{c} %\phantom{\times9}\color{green}{1 2 2}\\ \phantom{\times9}496\\ \underline{-\phantom{9}342}\\ \phantom{\times9}154\\ \end{array}\end{equation*}\]

Here, we’ll work from right to left, just as we did in Addition. 6 “takeway” 2 is 4, then 9 “takeway” 4 is 5, then 4 “takeway” 3 is 1. Therefore, our final answer will be \(154\). I used “takeway” because I’ve seen a lot of my students use money or a number line to visualize subtraction.

Example 8

Subtract the following numbers: \(65 - 28\)

\[\begin{equation*}\begin{array}{c} %\phantom{\times9}\color{green}{1 2 2}\\ \phantom{\times9}65\\ \underline{-\phantom{9}28}\\ \phantom{\times9}37\\ \end{array}\end{equation*}\]

From right to left, we can’t take away 8 from 5, so we have to “borrow” a “ten” from the next column. In other words:

\[\begin{equation*}\begin{array}{c} \phantom{\times999}\color{green}{15}\\ \phantom{\times9}5 \\ \underline{-\phantom{9}28}\\ \phantom{\times9}37\\ \end{array}\end{equation*}\]

When we “borrow” a ten from the next column, the \(6\) becomes a 5, then the \(5\) has a \(10\) added to it, which now becomes 15. So, now, \(15 - 8\) is 7. Then we move to the next column, \(5-2\) is 3. Therefore, our final answer would be \(37\).

See, it’s a bit more complex when we have to the borrowing technique. Let’s look at another one:

Example 9

Subtract \(258\) from \(536\). In other words, it’ll be \(536 - 258\). I said it like that on purpose because sometimes, you’ll hear people say it like that so you have to make sure to get the numbers in the correct order.

\[\begin{equation*}\begin{array}{c} %\phantom{\times9}\color{green}{1 2 2}\\ \phantom{\times99}536\\ \underline{-\phantom{99}258}\\ \phantom{\times99}278\\ \end{array}\end{equation*}\]

From right to left, we can’t take away 8 from 6, so we’ll have to “borrow” from the next column. But wait, we see that we’ll encounter the same problem in the second column, so we’ll have to “borrow” from the third column. Now, after we track our borrowing, we’ll have:

\[\begin{equation*}\begin{array}{c} \phantom{\times99}\color{green}{12}\\ \phantom{\times99999}\color{green}{16}\\ \phantom{\times}4\\ \underline{-\phantom{99}258}\\ \phantom{\times99}278\\ \end{array}\end{equation*}\]

To explain this, the \(5\) becomes a 4, then the 3 becomes a \(13\). However, we need to “borrow” from that column for the farthest right column, so the \(13\) becomes a 12. The farthest right column now becomes a \(16\). Now, we can subtract our numbers. \(16-8\) is 8, \(12 - 5\) becomes 7, and \(4-2\) becomes 2. Therefore, our final answer is \(278\).

   

Remark

For subtraction, we can only subtract two numbers at one time. If we are subtracting more than 2 numbers, you can only do it two at a time. It's because the properties of association and commutative don't apply with subtraction.

Estimating

When it comes to estimations, it’s important to know the placement of your numbers; the thousands, hundreds, etc. (Refer back to the image in the Alignment of Whole Numbers)

 

Two main rules you want to keep in mind for Estimation. Look at the digit to the right of what's in your question.

  • If the digit to the right is less than 5, then you round down.
  • If the digit to the right is 5 or greater, then you round up.

 

Example 1

Round 46 to the nearest ten.

Since the question asks for the nearest ten, we look at the number in the ones place. The number is \(6\), which is larger than 5. That means the number in our tens spot will jump up to a 5. Therefore, rounding 46 to the nearest ten will be \(50\).

Example 2

Round 438 to the nearest hundred.

The question asks for the nearest hundred, that means we ask ourselves, will we move the 4 up to a 5, or keep it the same?

We look at the number next to it, which is a 3. Since the three is less than 5, we will keep the 4 the same. Therefore, 438 rounded to the nearest hundred is \(400\).

Example 3

Round 8500 to the nearest thousand.

Since 8 is in the thousand spot, we ask “do we keep it as 8, or do we round up?”. The digit to the right of \(8\) is 5, so, by rule, we’ll round up. In this case, \(8500\) will be rounded to \(9,000\).

 

Remark

When adding or subtracting using estimation, you want to round the left-most digit of your numbers.

 

Example 4

Estimate the sum (addition) of the following problem, then find the actual sum. We want to estimate the sum of \(468\), \(936\), and \(688\)

First thing’s first, let’s round each of the numbers to the nearest hundred. This will make the mental calculations much easier. Therefore, \(468\) is rounded up to \(500\), \(936\) is rounded down to \(900\) and \(688\) is rounded up to 700.

\[\begin{equation*}\begin{array}{c} %\phantom{\times99}\color{green}{12}\\ %\phantom{\times99}\color{green}{16}\\ \phantom{\times99}500\\ \phantom{\times99}900\\ \underline{+\phantom{99}700}\\ \phantom{\times9}2100\\ \end{array}\end{equation*}\]

Therefore, the estimated sum is \(2,100\). To find the actual sum, we can add the numbers like we have previously.

\[\begin{equation*}\begin{array}{c} %\phantom{\times99}\color{green}{12}\\ %\phantom{\times99}\color{green}{16}\\ \phantom{\times99}468\\ \phantom{\times99}936\\ \underline{+\phantom{99}688}\\ \phantom{\times9}2092\\ \end{array}\end{equation*}\]

The actual sum is \(2,092\) which isn’t far off from our estimated sum. Sometimes, it’s pretty convenient to estimate.

Example 5

Estimate the difference (subtraction) for the following numbers then find the actual difference, \(759 - 288\).

First thing’s first, estimate your numbers to the nearest hundred, which will be \(800\) and \(300\).

\[\begin{equation*}\begin{array}{c} %\phantom{\times99}\color{green}{12}\\ %\phantom{\times99}\color{green}{16}\\ \phantom{\times99}800\\ \underline{-\phantom{99}300}\\ \phantom{\times99}500\\ \end{array}\end{equation*}\]

Now to find the actual difference, we subtract our numbers, like we have in previous lessons.

\[\begin{equation*}\begin{array}{c} %\phantom{\times99}\color{green}{12}\\ %\phantom{\times99}\color{green}{16}\\ \phantom{\times99}759\\ \underline{-\phantom{99}288}\\ \phantom{\times99}471\\ \end{array}\end{equation*}\]

471 is pretty far from 500, so even though it’s easier to estimate, it becomes less accurate at times. However, it all depends on the context and applications you’re working in.

Multiplication and Division

Multiplication of Whole Numbers

Multiplication and Division are designed to make addition much easier to do when you’re dealing with larger numbers, and repetitive operations. For example, lets take a look at a relatively simple one.

\[ 3 + 3 + 3+3+3+3+3 = 21 \]

Instead of writing out all of those threes, we can say 3 is added to itself 7 times. Or, as we know it, 3 times 7, which equals 21.

\[ 3\times 7 = 21 \]

Symbols for Multiplication

  • \(\times\) The X-Sign: Most familiar. Ex: \(5\times 8\)

  • \(\cdot\) The Center Dot: Used by most math folks. Ex: \(7\cdot 4\)

  • \(()\) Parenthesis: that’s is also used quite a bit. Ex. \(9(6)\)

   

Definition - Commutative Property of Multiplication

For any two whole numbers a and b, a ⋅ b = b ⋅ a In other words, if you're dealing with multiplication, the order on how you multipliy the numbers does not matter. You'll still end up with the same answer.

 

Example 1

In other words, we can say that \(3\cdot 5\) is the same thing as \(5\cdot 3\). Well, let’s see.

\(3+3+3+3+3 = 15\) and \(5+5+5 = 15\) so the commutative property holds true for multiplication.

   

Definition - Associative Property of Multiplication

For any whole numbers a, b and c, a ⋅ b ⋅ c =   ( a ⋅ b ) ⋅ c = a ⋅ ( b ⋅ c ) In other words, if you're dealing with multiplication, the grouping of the numbers does not matter. You'll still end up with the same answer.

   

Definition - Zero Property of Multiplication

For any two whole numbers a and b, a ⋅ b = b ⋅ a In other words, if you're dealing with multiplication, the order on how you multipliy the numbers does not matter. You'll still end up with the same answer.

 

Definition - Identity Property of Multiplication

If we multiply any whole number, call it a, by 1, we get a, a ⋅ 1 = a In other words, if we mulitply anything by 1, we get that same number.

 

Example 1

Multiply the following numbers: \(38\) and \(9\)

\[\begin{equation*}\begin{array}{c} %\phantom{\times99}\color{green}{12}\\ \phantom{\times9}\color{green}{4}\\ \phantom{\times99}38\\ \underline{\times \phantom{999}6}\\ \phantom{\times9}228\\ \end{array}\end{equation*}\]

To explain this, we still work from right to left. We perform the indicated multiplications. 8 times 6 is 48. We keep the 8 and bring the 4 up to the top. Then we multiply 3 and 6 to get 18, then add the 4 to get 22. We join the 22 and the 8 together to get 228 as our final answer.

It gets a bit more complex when we multiply numbers with more than 1 digit. Let’s check this out!

Example 2

Multiply the following numbers \(43\) and \(38\)

\[\begin{equation*}\begin{array}{c} %\phantom{\times99}\color{green}{12}\\ \phantom{\times99}\color{green}{2}\\ \phantom{\times999}43\\ \underline{\times \phantom{999}38}\\ \phantom{\times}\color{green}{1}\\ \phantom{\times99}344\\ \underline{+ \phantom{9}1290}\\ \phantom{\times9}1634\\ \end{array}\end{equation*}\]

Let’s walk through this. Still working from right to left. Multiply the \(3\) and the \(8\) to get \(24\). We bring the 4 down, the \(2\) goes to the top (in green). Then we multiply \(4\) and \(8\) to get 32, then we add the 2 to get 34. The 34 now gets attached to the 4 we brought down and that’s where the 344 comes from. But wait, there’s more!

Now, because we have two digits to multiply, we place a zero in the one’s place under the \(4\), then we go back to our original numbers and take 43 and multiply it by the 3. \(3\) times \(3\) becomes 9, and 4 times 3 becomes a \(12\), which is where the \(1290\) comes from.

Lastly, we’ve done all the multiplication we need. Now we can add \(344\) and \(1290\) together. \(4+0\) is 4, \(4+9\) is 13 so we bring the 3 to the bottom, and the one goes to the top (in green), then we add \(1+3+2\) which is 6, then \(0+1\) is 1. Put it together, we get \(1634\) as our final answer.

I know this looks extremely complicated, but this is actually how we learned it in elementary school! We’ve just be so accustomed to calculators that we forgot how it’s broken down.

Example 3

Multiply the following numbers \(27\) and \(280\).

\[\begin{equation*}\begin{array}{c} \phantom{\times9}\color{red}{1}\\ \phantom{\times9}\color{green}{5}\\ \phantom{\times999}280\\ \underline{\times \phantom{9999}27}\\ \phantom {\times} \color{green}{1 }\\ \phantom{\times99}1960\\ \underline{+ \phantom{99}5600}\\ \phantom{\times99}7560\\ \end{array}\end{equation*}\]

The explanation, we work from right to left again. \(0\) times \(7\) is \(0\), \(8\) times \(7\) is \(56\) so we bring the 6 down, and the 5 goes up. Then we take \(2\) times \(7\) to get \(14\) then we add 5 to get \(19\), which is where the \(1960\) comes from.

Since we have to multiply another digit, we place a zero in the ones place under the 1960, then we repeat the process and taken \(280\) and multiply it by 2, to get \(560\) which, putting it together, we get \(5600\).

Now that we’re done with the multiplication, we can add \(1960\) and \(5600\) to get \(7,560\) as our final answer.

Division of Whole Numbers

Division basically unravels Multiplication. However, the same rules for multiplication do not apply to division.

Symbols for Division

  • \(\div\) this is the most common symbol that we’re taught. It’s used, mainly on calculators.

  • \(\frac{a}{b}\) a fraction is typically seen more often. While most people aren’t fond of fractions, they’re still division.

  • \(\require{enclose} \begin{array}{rll} \\ \text{b} \enclose{longdiv}{\text{a}}\end{array}\) this is a classic division symbol that we used when we had to do division by hand, which is what we’ll see shortly.

 

Long Division by Hand

There are a few terms to recall when we’re dividing by hand. I’m pulling back from Elementary School on this, so bear with me.

\[\require{enclose} \begin{array}{rll} \text{Quotient} \\ \text{Divisor} \enclose{longdiv}{\text{Dividend}}\end{array} \]

  • The Dividend is what is inside the house. Usually, this is the bigger number.

  • The Divisor is how many of these go into the dividend. This goes to the “party”.

  • The Quotient is how many times the Divisor can go into the Dividend evenly.

  • The Remainder is what we have left, when the divisor doesn’t go into the quotient evenly.

 

Example 1

What is \(185\div 7\) ? We can ask this question a few different ways.

  • What is 185 divided by 7?

  • How many 7’s go into 185?

\[ \require{enclose} \begin{array}{rll} 026 \\ 7 \enclose{longdiv}{185}\\ \underline{-14\phantom{0}} \\ 4\phantom{}5 \\ \underline{-\phantom{0}42\phantom{}} \\ \phantom{0}3 \end{array}  \]

To explain this, we take our divisor and see how many times it goes into our dividend, left to right. \(7\) doesn’t go into 1, so have a zero in place of that on the Quotient. Then we move left, how many times does \(7\) go into \(18\), well this is 2 times. Then, the \(2\) goes in our Quotient, \(7\times 2\) is 14, and we place that underneath the 18. The Division Algorithm (which is mathematically what this is called), says we always subtract, so \(18-14 = 4\). Then, we bring down the next number, which is \(5\) and we repeat our process.

How many times does \(7\) go into \(45\)? The answer is at least \(6\), so the \(6\) goes in our quotient, and \(7\times 6 = 42\) which now we place that under the 45, then subtract to get a remainder of 3.

To explain our answer, there’s a few ways to do this.

  • 7 goes into 185, 26 times with a remainder of 3.

  • \(26 \frac{3}{7}\) is the final answer, which is usually what we’d be left if we did this soley by hand.

However, since I know we’re always going to have a calculator, \(3\div 7 =0.4285\) so we’ll have \(26 + 0.4285 = 26.4285\) as our final answer.

When you do this division in your calculator, \(185 \div 7 \approx 26.428571\) so that’s spot on!

   

Answers in Long Division

To write your answers in Long Division, Quotient + Remainder Divisor

 

Checking Division

To check your division, there's a formula to use: Divisor × Quotient + Remainder = Dividend

 

Therefore, for us to check our answer, we’ll have:

\[ \begin{eqnarray*} 7 \times 26 + 3 &\stackrel{?}{=}& 185\\ 182 + 3 &\stackrel{?}{=}&185\\ 185 &=& 185\\ \end{eqnarray*} \]

Example 2

Divide the following numbers: \(2076 \div 8\)

\[ \require{enclose} \begin{array}{rll} 259 \\ 8 \enclose{longdiv}{2076}\\ \underline{-16\phantom{00}} \\ 47\phantom{0} \\ \underline{-\phantom{0}40\phantom{0}} \\ \phantom{0}76 \\ \underline{-\phantom{0}72\phantom{}}\\ \phantom{0}4 \end{array}  \]

Therefore, our answer is \(259 \frac{4}{8}\) or \(259\frac{1}{2}\) or how your calculator would display it, \(259.5\)

Example 3

Divide the following numbers: \(2451 \div 12\)

\[ \require{enclose} \begin{array}{rll} 204 \\ 12 \enclose{longdiv}{2451}\\ \underline{-24\phantom{00}} \\ 051\phantom{} \\ \underline{-\phantom{0}48\phantom{}} \\ \phantom{0}3 \end{array}  \]

One thing to note in this process is that when the \(24\)’s subtract out, then we have 0 and then drop down the 5. Since \(12\) doesn’t go into 5, we place a zero in our quotient to indicate that. Then we drop down the \(1\) to get 51. Therefore, \(12\) goes into \(51\) at most 4 times, to get \(48\). Therefore, our remainder is 3.

Our final answer is \(204 \frac{3}{12}\) or \(204 \frac{1}{4}\) or \(204.25\)

Example 4

Dr. McMurrough purchased a classroom set of cadavers, containing 18 cadavers. What is the unit price of one cadaver if the final bill was $545?

In other words, find the price of each cadaver. The term Unit is used quite frequently in business contexts.

\[ \require{enclose} \begin{array}{rll} 030 \\ 18 \enclose{longdiv}{545}\\ \underline{-54\phantom{0}} \\ 05\phantom{} \end{array}  \]

Since 18 can’t go into \(5\), that’s where we stop. Therefore, we can say that the cadavers cost a little over $30. If you want to be exact,

\[ \$ 30 \frac{5}{18} = \$30 + \$0.2777 = $30.27 \text{ or } $30.28 \]

Dividing by Zero

One fundamental rule in Mathematics, You Can Not Divide By Zero. In this case, your Divisor can not be zero.