Remember that whole numbers are the set: {0, 1, 2, 3, 4, …}. In this section we’ll go over some of the definitions and processes of the basic operations you’ll do with whole numbers. If you’d like to see the rules for doing operations with negatives (integers), head down to the section *Positive and Negative Numbers*.

Addition is the simplest way to combine two values. It answers questions like: “If I have $25 and you have $32, how much do we have in total?” In this case, you can write the equation 25 + 32 = 57. The two parts added together (25 and 32) are called the *addends* of the equation and the answer (57) is called the *sum*. A good method to add values that contain multiple digits is to place them in columns, with place values aligned, and add values that are in the same column, starting from the *ones* digit and moving left.

The 5 and 2 in the *ones* place value add up to 7 and the 2 and 3 in the *tens* place value add up to 5. Of course a slight issue arises if two of the numbers add up to more than 9. Let’s look at this example:

The numbers 5 and 7 in the *ones* place add up to the two digit number 12. The number 12 is made up of 1 *ten* and 2 *ones*. Therefore, we split up the digits of 12 by putting 2 as the solution to the number of *ones* and placing a 1 above the 4 to show that there is one more group of *tens*.

Now you can carry through with the rest of your addition, remembering to include any numbers that were written above the original problem while you added previous columns.

To subtract means to take away an amount. Take this subtraction problem: 5 - 3 = 2. In this example, 5 is called the *minuend*, 3 is called the *subtrahend*, and 2 is the *difference*. Similar to addition problems, it’s often convenient to stack the minuend and subtrahend and then subtract the digits in each place value column, again, always starting from the rightmost column.

Sometimes, an issue occurs when the digit in the minuend is less than the digit below. Take the following problem for example:

Notice that 2 - 7 in the *ones* place doesn’t make sense when dealing with whole numbers (the answer is negative). And, while you *know* that the top number is larger than the bottom and we *should* be able to subtract, this column just doesn’t work.

We’ve developed a method to regroup (or borrow) one of the tens from the *tens* place. In this case, take one from the 5 in the *tens* place and bring it into the *ones* column to make 12 ones. It’s important to remember to change the 5 into a 4 since you took 1 away. This example has the columns separated to show what’s happening.

For a subtraction problem, you can always check your answer using addition. Simply add the *difference* and the *subtrahend* and you should get the *minuend*.

Recently, there has been trend toward teaching subtraction more as backward addition. It eliminates the need for this *borrowing* process above. Say we want to solve the following: 30 - 25 . Really, we could be asking 25 plus what number equals 30? Easy: 5.

Now try with a more difficult example: 72 - 38 = ?. In other words, 38 plus what number equals 72? Try a multi-step approach to find that number. Think:

38 + 2 = 40 (the next whole group of *tens*)

40 + 30 = 70 (the closest group of *tens* below 72)

70 + 2 = 72

The answer will be the sum of all numbers you added: 2 + 30 + 2 = 34. So, 72 - 38 = 34.

Try this example 351 - 269. (269 plus what number equals 351?)

So

Multiplication is a way to combine *multiple* copies of the same number. For example, If I have 4 nickels and I want to try to find the value, I could do the following:

Or, using multiplication:

In this case, 5 and 4 are called the *factors* and 20 is called the *product*. It’s very helpful to memorize your *times tables* using all numbers from 1 to 9.

For multi-digit multiplication, once again, you should arrange the numbers in columns. Then, take the digit in the *ones* place of the bottom factor and multiply by each digit above, right to left. Here’s the first step, .

Now, the second step and the third step .

Now let’s add one digit to the second factor to see how that changes the problem.

To do this, again, start with the digit in the *ones* place of the bottom factor (2) and multiply it by each digit above (which we’ve already done).

Now, we’ll deal with the digit in the *tens* place of the bottom factor (1) and multiply it by each digit above, right to left. As we place our answer, start a new row, but skip the ones place and place the first product in the tens place column (which is directly underneath the 1 in this example).

*Note: If it helps, you can use a blank (underline) or a zero (0) to mark the ones place during this step and any purposely skipped places when working on complicated multiplication problems. Some people prefer to do this so they don’t accidentally record a digit in the wrong place value column. This is what it would look like:*

*or*

Continue to multiply 1 by the next two digits above, 4 and 1.

Now add the two products to find the final product.

So, .

We’ve used small numbers (only 1 through 4) for convenience sake, so now let’s investigate what happens with larger numbers. Let’s try .

In this case, the first thing to do is multiply 9 by 5, which is 45. Just like with addition of larger numbers, this number won’t fit in the *ones* column, so we should remember to *add* 4 to the answer for the *tens* column *after* we finish multiplying there. With pencil and paper, you could make a small 4 at the top of the next column to remind you.

*or* you can pencil it in the product space and put it in parentheses so you know it’s not the final answer for that column.

The next step is to multiply the 9 by 3 (27). Now, we have to remember to add the 4 from the first step to 27 (31) and write that number down.

Notice, we didn’t have to put the 3 in parentheses (or pencil it in lightly) because there are no more numbers to multiply.