09-16-2009Algorithms for Adding and Subtracting Whole NumbersAn algorithm is a step by step procedure for solving a problem. In this section we will discuss severalalgorithms for adding and subtracting whole numbers. Here, we will learn about why you sometimes“carry” in addition and “borrow” in subtraction.Example 1: Consider the problem128+4573One way to think about this problem is as follows: 28 is 2 tens and 8 ones, 45 is 4 tens and 5 ones.To add 45 and 28, add the “ones” first to get 13, which is 1 ten and 3 ones. Then add up all the tens.We have 2 tens from the 28, 4 tens from the 45, and 1 ten from the ones column. So we have a total of 7tens and 3 ones and so our answer is 73. This is best seen using manipulatives like blocks or strips/mats.The process where we “carried” the one is called an exchange. The idea is that we exchanged 10ones for 1 ten.Example 2: In the strip model, make exchanges to represent 15 squares, 11 strips and 2 mats with thesmallest number of pieces.The Addition Algorithm:• With units, squares, and mats add 135 and 217.1. Represent both numbers using mats for hundreds, strips for tens, and squares for ones.2. Combine all the manipulatives and make exchanges so that you use the least number of pieces.3. Count all the pieces. The answer is 352.• Use the place value card approach to add 135 and 2171. Mark off three cards with squares for ones, tens, hundreds, and so on.2. On one card record the number of ones, tens, and hundreds for the first addend. Do the samefor the second addend.3. On the third card mark off the total number of ones, tens, hundreds, and so on from the firsttwo cards. Make exchanges so that there are no more than 9 ones, tens, hundreds, and so on.4. Read the answer of the third card.• With place value diagrams, instructional diagrams, and the final algorithm add 135 and217. Record the following1. Place value diagram100s 10s 1s1 3 52 1 73 4 (12)3 5 22. Instructional algorithm135+21712403003523. Final algorithm1135+217352• As a note, teaching students the word “carrying” is somewhat misleading. What is really happeningwhen we do addition is a sequence of exchanges. You might consider teaching your students to callthis procedure exchanging, trading, or regrouping instead.Example 3:Consider the problem23/2−284We can think about this problem as taking 2 tens and 8 ones away from 3 tens and 2 ones. Since wecan’t take 8 ones away from the 2 ones in 32, we take one of the tens and “break it” into ones. So wehave rewritten 32 as 2 tens and 12 ones. Now we take 8 ones away from the twelve to get 4 ones andtwo tens from two tens to get 0 tens.The Subtraction Algorithm:• Subtract 154 from 322 using mats, strips, and units.1. Represent both the minuend and subtrahend using mats for hundreds, strips for tens, andsquares for ones.2. Start with the units. In this case we can’t subtract 2 from 4, so we must exchange one stripfrom 322 for 10 units. Then we get 12 ones from which we take 4 to get 8 ones in total.3. We traded in one of the strips in 322, so there is only one strip left for this number. We can’tsubtract 5 tens from 1 tens so we trade one of the mats for ten strips. Then there are 11 stripsleft. From these we subtract 5 strips to get 6 strips total.4. We traded in one of the mats in 322, so there are now only 2 mats left. From the two matswe take away one and get 1.5. Now count the total number of mats, strips, and units. We get 168.• Subtract 154 from 322 using place value cards.1. Mark off 3 hundreds, 3 10 tens, and 2 ones on a card.2. We must exchange 1 ten for 10 one markers. Then we can remove 4 ones markers from the12 ones markers we have.3. Exchange 1 hundred marker for one 10 marker. Then we can remove 5 ten markers from the11 hundreds markers.4. Remove 1 hundred marker from the remaining hundreds.5. Read the answer of the card.• With place value diagrams, instructional diagrams, and the final algorithm subtract 154from 322. Record the following1. Place value diagram100s 10s 1s3 2 2− 1 5 43 1 12− 1 5 42 11 12− 1 5 41 6 82. Instructional algorithm3 2 2− 1 5 43 1 (12)− 1 5 42 (11) (12)− 1 5 41 6 83. Final algorithm2113/2/12−154168Example 4: Compute (123)five+ (104)fiveand (123)five− (104)fivein base 5.The addition and subtraction algorithm work for numbers in other bases. Place value diagrams:25s 5s 1s1 2 31 0 42 2 72 3 225s 5s 1s1 2 3− 1 0 41 1 8− 1 0 40 1 4So (123)five+ (104)five= (232)fiveand (123)five− (104)five=