Whole number arithmetic

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Whole number arithmetic APPENDIX

DIVISION

How many times would you have to add 5 to itself in order to get 15?
How many times would you have to add 7 to itself in order to get 14?

This is the concept of division.

Division is the opposite of multiplication, just as subtraction was the opposite of addition. A division is basically asking you "how many groups of a specified size are required to reach a given number?" You will see division illustrated for you in the next two examples.

Example 1

15 divided by 5 which is written as 15 ¸ 5.

The problem is asking us "how many groups of 5 are required to reach a total of 15?"
Well here are our 15 circles.

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
We could draw these 15 circles in any pattern we wish. We would still have 15 circles. For example here are the same 15 circles.

Here are the 15 circles redistributed in groups of 5.

0 |  0 |  0
0 |  0 |  0
0 |  0 |  0 
0 |  0 |  0
0 |  0 |  0

As you can see there are three groups of 5 making up the 15 circles. Therefore we need three groups of five to reach a total of 15. This can be written as 15 ¸ 5 = 3 (15 divided by 5 is equal to 3).
This can be reinterpreted as saying "3 groups of 5 is equal to 15." As you are well aware the previous statement can be written as "5 x 3 = 15."

Before going on to the next example let's take another look at that last diagram.

0 |  0 |  0
0 |  0 |  0
0 |  0 |  0 
0 |  0 |  0
0 |  0 |  0

Suppose you read the groups horizontally rather than vertically. Then you would have the following picture

0   0   0
__________
0   0   0
_________
0   0   0 
__________
0   0   0
__________
0   0   0

Here you have 5 groups of 3. And they still make up the same 15 circles.
We could write this as 15 ¸ 3 = 5. Fifteen divided by three is equal to five. In other words we would need five groups of three to reach a total of 15.

This can be reinterpreted as saying "5 groups of 3 is equal to 15." As you are well aware the previous statement can be written as "3 x 5 = 15."

Furthermore I want you to observe one extremely important relationship.
15 ¸ 3 = 5 and 15 ¸ 5 = 3.
This is becasue 5 x 3 = 3 x 5 = 15.

Example 2

10 divided by 5 which is written as 10 ¸ 5.

The problem is asking us "how many groups of 5 are required to reach a total of 10?"
Well here are our 10 circles.

0 0 0 0 0 0 0 0 0 0
We could draw these 10 circles in any pattern we wish. We would still have 10 circles. For example here are the same 10 circles.

Here are the 10 circles redistributed in groups of 5.

As you can see there are two groups of 5 making up the 10 circles. Therefore we need two groups of five to reach a total of 10. This can be written as 10 ¸ 5 = 2 (10 divided by 5 is equal to 2).
This can be reinterpreted as saying "2 groups of 5 is equal to 10." As you are well aware the previous statement can be written as "5 x 2 = 10."

Before going on to the next example let's take another look at that last diagram.

Suppose you read the groups horizontally rather than vertically. Then you would have the following picture

0   0  
__________
0   0  
_________
0   0   
__________
0   0  
__________
0   0

Here you have 5 groups of 2. And they still make up the same 15 circles.
We could write this as 10 ¸ 2 = 5. Ten divided by two is equal to five. In other words we would need five groups of two to reach a total of 10.

This can be reinterpreted as saying "5 groups of 2 is equal to 10." As you are well aware the previous statement can be written as "2 x 5 = 10."

Furthermore I want you to observe one extremely important relationship.
10 ¸ 2 = 5 and 10 ¸ 5 = 2.
This is becasue 5 x 2 = 2 x 5 = 10.

There are specific names for the parts of a division problem. The number that is being divided is called the dividend. The number that we are "dividing by" is called the divisor. The solution is called the quotient.

The division problem, "10 ¸ 2 = 5"
The dividend is 10, The divisor is 2, and the quotient is 5.

The division problem, "10 ¸ 5 = 2"
The dividend is 10, The divisor is 5, and the quotient is 2.

The division problem, "15 ¸ 5 = 3"
The dividend is 15, The divisor is 5, and the quotient is 3.

The division problem, "15 ¸ 3 = 5"
The dividend is 15, The divisor is 3, and the quotient is 5.

Rather than continually draw pictures every time you wan to divide some numbers, we have a reinterpretation of the multiplication table known as the division table. As you will soon see, the division table contains the answers to 100 division problems. In each of these problems the number that we are "dividing by" (i.e. the divisor) will be either 10 or a single digit number. The number that we are dividing (i.e. the dividend) will either be one or two digits. The solution to the division problem (i.e. the quotient) will be either 10 or a single digit number.

Examine the following table
Study it!
Drill yourself on it.
Quiz yourself on it using flash cards that you either bought or made yourself.
Commit it to memory.

0¸1=0 since 0x1=0	1¸1=1 since 1x1=1	2¸1=2 since 2x1=2	3¸1=3 since 3x1=3	4¸1=4 since 4x1=4	5¸1=5 since 5x1=5	6¸1=6 since 6x1=6	7¸1=7 since 7x1=7	8¸1=8 since 8x1=8	9¸1=9 since 9x1=9	10¸1=10 since 10x1=10
0¸2=0 since 0x2=0	2¸2=1 since 1x2=2	4¸2=2 since 2x2=4	6¸2=3 since 3x2=6	8¸2=4 since 4x2=8	10¸2=5 since 5x2=10	12¸2=6 since 6x2=12	14¸2=7 since 7x2=14	16¸2=8 since 8x2=16	18¸2=9 since 9x2=18	20¸2=10 since 10x2=20
0¸3=0 since 0x3=0	3¸3=1 since 1x3=3	6¸3=2 since 2x3=6	9¸3=3 since 3x3=9	12¸3=4 since 4x3=12	15¸3=5 since 5x3=15	18¸3=6 since 6x3=18	21¸3=7 since 7x3=21	24¸3=8 since 8x3=24	27¸3=9 since 9x3=27	30¸3=10 since 10x3=30
0¸4=0 since 0x4=0	4¸4=1 since 1x4=4	8¸4=2 since 2x4=8	12¸4=3 since 3x4=12	16¸4=4 since 4x4=16	20¸4=5 since 5x4=20	24¸4=6 since 6x4=24	28¸4=7 since 7x4=28	32¸4=8 since 8x4=32	36¸4=9 since 9x4=36	40¸4=10 since 10x4=40
0¸5=0 since 0x5=0	5¸5=1 since 1x5=5	10¸5=2 since 2x5=10	15¸5=3 since 3x5=15	20¸5=4 since 4x5=20	25¸5=5 since 5x5=25	30¸5=6 since 6x5=30	35¸5=7 since 7x5=35	40¸5=8 since 8x5=40	45¸5=9 since 9x5=45	50¸5=10 since 10x5=50
0¸6=0 since 0x6=0	6¸6=1 since 1x6=6	12¸6=2 since 2x6=12	18¸6=3 since 3x6=18	24¸6=4 since 4x6=24	30¸6=5 since 5x6=30	36¸6=6 since 6x6=36	42¸6=7 since 7x6=42	48¸6=8 since 8x6=49	54¸6=9 since 9x6=54	60¸6=10 since 10x6=60
0¸7=0 since 0x7=0	7¸7=1 since 1x7=7	14¸7=2 since 2x7=14	21¸7=3 since 3x7=21	28¸7=4 since 4x7=28	35¸7=5 since 5x7=35	42¸7=6 since 6x7=42	49¸7=7 since 7x7=49	56¸7=8 since 8x7=56	63¸7=9 since 9x7=63	70¸7=10 since 10x7=70
0¸8=0 since 0x8=0	8¸8=1 since 1x8=8	16¸8=2 since 2x8=16	24¸8=3 since 3x8=24	32¸8=4 since 4x8=32	40¸8=5 since 5x8=40	48¸8=6 since 6x8=48	56¸8=7 since 7x8=56	64¸8=8 since 8x8=64	72¸8=9 since 9x8=72	80¸8=10 since 10x8=80
0¸9=0 since 0x9=0	9¸9=1 since 1x9=9	18¸9=2 since 2x9=18	27¸9=3 since 3x9=27	36¸9=4 since 4x9=36	45¸9=5 since 5x9=45	54¸9=6 since 6x9=54	63¸9=7 since 7x9=63	72¸9=8 since 8x9=72	81¸9=9 since 9x9=81	90¸9=10 since 10x9=90
0¸10=0 since 0x10=0	10¸10=1 since 1x10=10	20¸10=2 since 2x10=20	30¸10=3 since 3x10=30	40¸10=4 since 4x10=40	50¸10=5 since 5x10=50	60¸10=6 since 6x10=60	70¸10=7 since 7x10=70	80¸10=8 since 8x10=80	90¸10=9 since 9x10=90	100¸10=10 since 10x10=100

Now you try your hand at division.

Click on the "generate a problem" button to see your problem. Place your answer in the text boxt and then click the "check my answer" button. You'll be told whether or not your answer was correct. After you finish the problem, click on the "reset" button and then click the "generate a problem" button to see another problem. Keep on generating problems until you feel confortable with division.

Text and source code c2003 Martin Selditch