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If the remainder is negative you must put the minus sign in front of the remainder in these exercises. If you were dividing a number and showing the quotient without the R for remainder and expressing your answer as a mixed number like -2 3/4, the fraction part of the number is assumed to have the same sign as the whole number part and you wouldn't put the minus sign in front of 3/4. When you put the R in and express -2 3/4 you would write it as -2 R. -3/4.

We found that multiplication was multiple adding. For example 3X4=, means to add 4, 3 times like in the addition problem 4+4+4=12. Also recall that in multiplication, one number is the multiplier and tells how many times to add the other number which is called the multiplicand. In the above case 3 is the multiplier and 4 is the multiplicand.

Division is multiplication in reverse. We can look at division as multiple subtraction. Say we have the division problem 8/2=, which means 8 divided by 2. We can find the answer by finding out how many times we can take 2 away from 8. Lets run through the problem and see how it works out. 8-2=6, so we can take 2 from 8 at least once. 6-2=4, So we can take 2 from 8 at least twice. 4-2=2, So we can take 2 from 8 at least 3 times. Finally 2-2=0, So we can take 2 from 8, 4 times. This means that 8 divided by 2=4 or 8/2=4. To check and make sure that 8/2=4 we multiply 2X4 to see if it does equal 8. 2X4= means to add 4, twice like in the addition problem 4+4=8. We are counting how many times we can perform the subtraction operation of taking 2 from 8. This is the reverse of multiplication. Open the scratchpad if it isn't open yet by clicking on the scratch- pad icon at the top of the screen. Drag 8 number 1 cards to an empty area of the screen. Now drag 2 cards away and count this as 1 operation. Take another 2 cards away and count this as 2 operations. Take another 2 cards away and count this as 3 operations. Take the last 2 cards away and count this as 4 operations. You can clear the screen of cards by clicking the Clear Scratchpad icon or you can drag cards you are done with to the Recycle Bin. Another way we look at division problems is to say how many times can we put 2 into 8 for the problem 8/2=. Say we have 2 units that are full or shaded and a box made up of 8 empty units the same size as the units in our 2 shaded units.

If we take the 2 full units and put them into the 8 empty units as many times as we can, we will end up with something like the illustration above. On the right side each of the 2 full units that is placed in the 8 empty units is numbered and you can see that 4 of the 2 full units fits into the 8 empty units. You can do this in the Scratchpad by dragging number 2 cards into a group on the screen until the sum of the cards reaches the value of 8 and then counting how many 2 cards were needed. These are good ways to look at how division works and to understand what division means but if you used the multiple subtraction method on a large number it would take you a long time to find the quotient or answer to the division problem and you might loose track of how many times you have performed your subtractions. There is a method called long division that can help speed the process up, works great on large numbers and is accurate too. we'll start with 6 divided by 3 first to show the basic process and then do a larger number. The steps in the illustrations are numbered and go along with the numbered steps in the text. It will be helpful to use a scrap of paper at first and draw out the division problems.

1.) We make our long division symbol, put the number we are dividing(6) in the division box. We also call this number the dividend. Next we put the number(3) we are dividing our dividend by, just outside the long division symbol on its left. We call the number we are dividing the dividend by, the divisor. Next we ask ourselves, how many times can 3 go into 6 without going over 6. 3x1=3, 3x2=6, So, 3 will go into 6, 2 times.

2.) We put a 2 right over the 6. Now we multiply our divisor(3) times our partial quotient or answer(2) and get 6.

3.) We place this 6 directly under the 6 we are working on dividing and then subtract it from the six we are working on dividing. In this case 2 is our final answer. What happens if we divide 6 by 4. We get an answer with a remainder or decimal value on the end of it depending on how you want to express your answer. We use Remainders in this exercise. Lets do 6/4=?

1.) Set up the division problem with the number you're dividing, inside the long division symbol. This number is called the dividend. Put your divisor or number you are dividing the dividend by just outside and to the left of the long division symbol. Find the number of times that 4 will go into 6 without going over 6. 1X4=4, 2X4=8, So 4 will go into 6, 1 time without going over 6. Make a 1 directly over the 6.

2.) Multiply the divisor(4) times the partial answer or quotient(1) and place the product(4) directly under the 6, put a minus sign in front of it and draw a line under it.

3.) Subtract the 4 from 6. this is the remainder. we write it R.2/4 to show that 2 still needs to be divided by 4.

4.)In school they may have you just write 1 R.2 and not show the 4 in the denominator of the remainder. Either way 1 R.2 or 1 2/4 are correct but 1 and 2/4 shows more infor- mation about the number and that is why i show my remainders that way. Since 4 doesn't go into 6 evenly we end up with this left over part. Fraction means part of a number. The remainder is shown in what we call a fraction. The top number in the fraction is called the numerator The bottom number is called the denominator. 2/4 Means 2 divided by 4. The numerator corresponds or is like the dividend in long division. The denominator corresponds to the divisor or is like the divisor in long division. In this exercise you don't do any further reducing of the remainder but you could reduce this answer a little more if you wanted to. To reduce fractions, find numbers that you can divide both the numerator and denominator by without having any remainders. Divide the numerator and denominator by the same number. repeat the process until you can't find any more numbers that you can divide both the numerator and denominator by without having any remainders. When you reduce in this way the value of the fraction stays the same. Below is a graphical illustration of 6/4=1 2/4 to help you visualize it.

If we had a box divided into 4 equal parts and shaded them in. Made a box of 6 equal parts the same size as the ones in the box containing 4 shaded parts. Put the 4 shaded parts into the 6 un-shaded parts and you will have something like the illustration above. Looking at the illustration above you can see that if we tore the 4 equal shaded parts in half, we would have 2 shaded parts that would fit into the last empty parts in the 6 equal parts box. This also illustrates that 2/4=1/2. To check and see if 6 divided by 4 is 1 R. 2, multiply 4X1 and add the R.2. 4 Times 1 is 4. 4 + The remainder 2 is 6, so that is correct. To check and see if 6 divided by 4 is also equal to 1 2/4 we multiply 4 X 1 2/4. To do this we make the 4 into 4/1. Any number divided by one is equal to that number, so 4/1=4. We need to make the 1 2/4 into an improper fraction to multiply it by 4/1. We multiply the denominator 4 times the whole number part 1 and add this to the numerator. 4X1+2=6 so we get 6/4. Now we can multiply 4/1 X 6/4. 4x6=24 and 1x4=4 so we end up with 24/4. When we multiply fractions we go across the top and multiply the numerators then across the bottom and multiply the denominators and then reduce the answer. To reduce the improper fraction we divide the numerator by the denominator. to make it a proper fraction first. In this case 24 divided by 4 = 6 and 4x1 2/4 = 6 so this answer is also correct. One example of a real life situation where using long division and remainders is measuring with a tape measure having feet and inch graduations. Each foot has 12 inches in it. You measure a distance of 123 inches and want to know how many feet that is. 123/12 or 123 divided by 12 gives you the number of feet in 123 inches since each foot has 12 inches in it. 10 Feet R. 3/12 or 10 Feet 3 inches is the quotient to this problem.

This is where things get a little interesting. Remember the multiple subtraction method for solving division problems. If we used it to solve 7/3 rather than 7/-3, we could say 7-3=4, so 3 goes into 7 at least 1 time. 4-3=1, So 3 goes into 7, at least 2 times. 1-3=-2, so we will show 1 as a remainder. Our answer will be 7/3=2 1/3. When we try the multiple subtraction method on 7/-3=, we say 7-(-3)=10. I have had lots of people say no that can't be right at this point, so we need to look at it closer for those of us that it just doesn't make sense to at first, before we go any farther. We can use algebra to show that --3 is really +3. Algebra's main rule say that if we perform the same mathematical operations on both sides of the equation or equals sign, both sides will remain equal. In the equation 7--3=10 we can take 7 away from both sides like this 7-7--3=10-7. On the left side, the 2 sevens cancel out and on the right side we have 10-7=3, giving us a new equation, --3=3. This also shows that when solving equations that have 2 minus signs in front of a number, cancel the double minus signs and change to plus signs or you will get erroneous answers. We can follow our normal long division process but make our quotient a negative number. In step 1 we ask ourselves, how many times will -3 go into 7? -3X-1=3, -3X-2=6, -3X-3=9, So -3 goes into 7 at least -2 times. -2 Is the quotient. -3X-2=6, Take 6 away from 7 and the remainder is 1/-3. The final answer is -2 R. 1/-3. Like multiplication when you divide a positive number by a negative number you get a negative quotient. Dividing a negative by a positive also results in a negative. Dividing a negative by an negative results in a positive. Dividing a positive by a positive will result in a positive. We do the same thing for dividing larger numbers. Lets run through a division problem involving a larger number to show how it works. 62/2=.

1.) We ask ourselves, how many times will 2 go into 6? Since 2X3=6, 2 will go into 6, 3 times.

2.)Put a 3 right above the 6 that you are working on dividing to show this. Next multiply the divisor(2) by the part of the quotient(3) you are working on and place the product(6) directly under the 6 you are working on dividing, put a minus sign in front of it and subtract it from the 6 we are working on at the moment.

3.) Bring down the next digit in the dividend(2) from our dividend. Place the 2 in front of the results of the last subtraction.

4.) Ask yourself, how many times will 2(the divisor) go into the 2 we just brought down? Since 2X1=2, 2 will go into 2, 1 time. Put a 1 directly over the 2 that we just brought down.

5.)Multiply the divisor(2) times the 1 we just placed in the quotient and put the product under the 2 that is the part of the dividend we are working on dividing at the moment. Place a minus sign in front of it and draw a line under it.

6.)Subtract the 2 you just put the minus sign in front of from the 2 above it. Since the difference is 0, we are done and the answer or quotient of 62/2= is 31.

7.) We begin to multiply the quotient by the divisor to make sure that our quotient is correct. If 2X32=62 our answer is correct. 2X1=2 So, we place a 2 under our multiplication problem.

8.) 2X3=6, So, we place a 6 to the left of the last number we placed under our multiplication problem. There are no more digits to multiply so our answer is 62 and our quotient to the problem 62/2 is correct. We could also subtract 31 from 62, 2 times using the subtraction method to show that 62/31=2, reversing the order of division to check for correctness. Ok, lets look at 300/4=, a division problem where 4 doesn't go into 3. How do we deal with this?

1.) Ask yourself, how many times will 4 go into 3 without going over 3? 4 Won't go into 3. Put a 0 over the 3 you are working on dividing, because we divided 3 by 4 0 times.

2.) Multiply the divisor(4) times the partial quotient(0), 4X0=0. Put the product of 4X0 under the 3 you are working on. Draw a minus sign in front of that 0 and put a line under the 0.

3.) Subtract the 0 from the 3. 3-0=3, So put a 3 under your subtraction problem.

4.) Bring down the next digit from your dividend(0) and place it in front of the 3, the difference of 3-0.

5.) Now you have the number 30 to work on dividing. How many times will 4 go into 30 without going over 30? 1X4=4, 2X4=8, 3X4=12, 4X4=16, 5X4=20 6X4=24, 7X4=28, 8X4=32, So 4 will go into 30 7 times without going over 30. Put a 7 directly above the 0 in the 30 you are working on dividing at the moment, in your answer or quotient area.

6.) Now multiply your divisor(4) times your partial quotient(07) and put the product under the 30 you are working on dividing. Note: 7 and 07 mean exactly the same thing, just ignore the 0 to the left of the 7 and multiply 4X7. 4X7=28, So put 28 under the 30 you are working on. Draw a line under it. Put a minus sign in front of the 28.

7.) Subtract the 28 from 30. First look at the 0 in 30 and it is to small to take 8 away from it. Barrow 10 from the 3 to its left. Make a 10 over the 8, cross off the 3 and make the 3 into a 2. 10-8=2, So put a 2 under the 8. Move left in the subtraction problem and 2-2=0. You can make a 0 under the 2 or just leave it blank. 30-28=2 or 30-28=02 Depending on how you prefer to write it. Bring down the next digit in 300 and put it in front of the 2 from the last operation. You'll be looking at dividing 20 by 4 next.

8.) How many times will 4 go into 20 without going over 20? 1X4=4, 2X4=8, 3X4=12, 4X4=16, 5X4=20, So 4 will go into 20, 5 times. Make a 5 directly above the 0 in the 20 you are working on.

9.) Multiply your divisor(4) times the 5 in the quotient and put the product(20) under the 20 you are working on. Draw a minus sign in front of it and draw a line under it.

10.) Subtract 20 from 20. you can put zeros down or just leave it blank. There are no more digits in your dividend to divide and nothing left over in your last subtraction, So 300 divided by 4 is 75.

To solve we first ask ourselves how many times will 3 go into 4 and write the answer directly over the 4 in the division box. (It will only go into 4 once because 3x1=3 and 3x2=6, 2 times 3 won’t fit into 4) Next we multiply 3 times our partial answer(1), which equals 3. Write 3 directly under the 4 in the division box. Put a line under the 3 and a minus sign in front of the 3. Take that 3 away from the 4 and write the answer (1) directly under the 3 that is under the 4. This is the part left over that 3 won’t go intoevenly and is the remainder. We show this by writing R then 1/3. Your final answer is 1 1/3 that is how many time 3 will go into 4. Also written as 1 R 1/3.

- Simple Addition-single digit addition with all positive numbers. Free in Shareware version
- Simple Multiplication-Single digit multiplication with all positive numbers
- Simple Subtraction-Single digit subtraction with all positive numbers Free in Shareware version
- Subtraction With Negative Numbers
- Addition and Negative Numbers Free in Shareware version
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- Subtracting with Decimals Free in Shareware version
- Multiplying With Decimals
- You are here. Simple Division and Remainders Free in Shareware version
- Adding Fractions and Reducing
- Multiplying Fractions and Reducing Free in Shareware version
- Division Decimals and Rounding Off

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Copyright 2008 Robert Lee Thomas

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