SHORT DIVISION

FRACTIONS x WHOLES

MULTIPLICATION

DIVISION

REDUCING

FRACTIONS ON THE ABACUS

MENU

SHORT DIVISIONTo begin the fraction lesson teach short division on the abacus. I recommend using a story or sequence of images to direct the solution process for students. I might tell the students that a family of three squirrels searched for nuts on an autumn morning and found fourteen nuts. They each took one nut in turn from their collection until there were not enough left for them each to have another. They left the extra for the winter birds. How many nuts did each squirrel get.

FRACTIONS

Begin to push where indicated on abacus A. Abacus A shows a small triangle with beads in its base equal the number of squirrels, the divisor. Abacus B and abacus C show how a number of beads equal the nuts collected by the squirrels, the dividend, are pushed from the right side to the left side. Do not push bottom roll beads.

NO!

Abacus C and abacus D show that the triangle remaining on the right side has a base number of beads equal the number of nuts each squirrel got, the whole number part of the quotient. The last column of beads, of the dividend, equals what remained for the winter birds. Students can now be shown how remainders can be expressed as fractions.

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FRACTIONS TIMES WHOLESOnce students can divide on the abacus then they can explore multiplying a whole number by a fraction. The same image sequence used above can direct the solution process.

For the example demonstrated above, two-thirds (times or of) twelve, tell the students a family of three squirrels collected twelve nuts and each took one nut in turn from the collection until there were no nuts left. How many nuts does each squirrel get? As shown on abacus A the number of squirrels is represented by the triangle on the left with three beads in its base, and the collected nuts by the rectangle of beads above that triangle. And as before, the number of beads in the base of the triangle to the right is the quotient.

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TWO-THIRD OF THREECORRESPONDS TOTWO-THIRDS OF TWELVE

To direct the solution process for two-thirds (of or times) twelve ask how many nuts would two of the squirrels get altogether. Abacus B and abacus C show how you can push two-thirds of the base beads of the triangle representing the family of squirrels to the right and separate out two-thirds of the rectangle representing the collected nuts. In this way, the answer eight is displayed.

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I invited over a friend, baked a pie,Then, decided a little piece I'd try.In an hour my friend arrived.What remained made them cryBut, just a bit they ate with a sigh.So all alone, I finished the pie. 

Once students can multiply a whole number by a fraction on the abacus, a sequence of manipulations can easily be learned to solve multiplication and division of fraction problems. It is again helpful to direct the solution process with a story or image sequence. The poem above may be the bases for such a sequence.

MULTIPLICATION OF FRACTIONS

I invited over a friend, baked a pie,Then, decided a little piece I'd try.In an hour my friend arrived.What remained made them cryBut, just a bit they ate with a sigh.So all alone, I finished the pie. 

In the fraction problem shown below, the second fraction, reading from left to right, is how much of the pie remained when the friend arrived, and the first fraction is how much of what remained the friend ate.

To begin the solution process have the students multiply the denominators, as shown on abacus A, to see into how many coins the pie is sliced. This product, twelve, is the denominator of the solution. Direct students to write it under the fraction bar of the solution fraction.

THREE-FOURTHS OF FOUR CORRESPONDS TOTHREE-FOURTHS OF TWELVE

Now to find the numerator students must figure out how many coins of the whole pie the friend ate. Have students first find how many coins remained when the friend arrived, by taking three-fourths of twelve, the whole pie. As shown over Abacus B and abacus C, the answer is nine.

TWO-THIRD0F THREECORRESPONDS TOTWO-THIRDSOF NINE

Then students can find out how much of the three-fourths or nine coins the friend ate, by taking two-thirds of the nine coins. As shown over abacus D and abacus E the answer is six. Direct students to write six above the fraction bar of the solution fraction 

I invited again my friend for pie.Have no fear this is why;For - I baked two with pride.I ate most of one, but no need to cry.Here's another for my friend to try.Eat my friend and don't be shy.

Division of fractions can be shown to be a comparison of one fraction to another. The solution process can be directed by continuing our poem.

DIVIDING FRACTION

Have the students, as before, multiply the denominators to see into how many equal coins the pie is sliced (twelve), but position the quadrilateral of beads between the triangles, as shown on abacus A. Now they are prepared to take a fraction of the pie.

BC

Have students multiply the second fraction times twelve, the number of slices. As shown over abacus B, abacus C and abacus D the answer is nine. This fraction of beads is represented to students as the slices eaten of the first pie and is the denominator of the solution fraction. Have students write it under the solution fraction bar.

DTHREE-FOURTHSOF FOURCORRESPONDS TOTHREE-FOURTHSOF TWELVE

EFG

Now, have students multiply the first fraction times twelve, the products of the denominators. As shown over abacus E, abacus F and abacus G the answer is eight. This fraction of beads is represented to the students as the slices eaten of the second pie by the friend and is the numerator of the solution fraction. Have student write it over the solution fraction bar. 

TWO-THIRDS OF THREECORRESPONDS TOTWO-THIRDSOF TWELVE

REDUCING FRACTIONS ON THE PYTHAGOREAN ABACUS To reduce a fraction such as 4/6 on the abacus separate from the array a triangle with a number of base beads equal the denominator. Above the bottom roll find the roll with a number of beads equal the numerator. Zigzagging up and down the columns from the bottom row to the top row and down again to the bottom can now identify beads representing the reduced fraction.

The numbers on the beads in the picture to the left indicate the zigzag path to take from row to row. To select identified beads as you go along the path label with small note tags. Take note that when you "zag" down to the bead numbered (5) you cannot "zig" back to the top or the numerator row. So count to the left along the bottom row a number beads equal the numerator. If the beginning bead lies a number of beads equal the numerator from the last "zag" bead, which is here the case, the reducing process is complete. The number of

selected beads in the top row, zig" beads, over the number of selected beads in the bottom row, "zag" beads, is the reduced fraction, in this case 2/3.

To the right beads are colored to represent the fraction 4/10. Below we can "zig" up to (2) and "zag" down to (3), but we can not "zig" again back up to the numerator row. So we must count across the bottom row a number beads equal the numerator

In this case the beginning bead does not lie the a number of beads equal the numerator from the last "zag" bead, so we must "zig" up again from the bead that does, in this case numbered (4). That will take us up to the "zig" bead (5) and back down to the "zag" bead (6). Since we cannot "zig" back up to the numerator row we must count again across the bottom row a number of beads equal the numerator from the last "zag" bead (6). In this case bead (7) is selected, but we can not "zig" again back up to the numerator row so we continue on to the beginning bead, which lies a number of beads equal the numerator from

the "zag" bead (7). Now, the reducing process is complete. There are two "zig" beads in the numerator row and five "zag" beads in the denominator row so the reduced fraction equals 2/5.

2

4

1

5

3

If the numerator (4) and denominator (6) or represented as a square and a hexagon the beads of the abacus may represent the points at the vertices of these polygons. Now from a beginning point or vertex, count clockwise around the polygons to select points separated by a number equal the difference between the numerator and denominator, in this case (2) . The number of selected points for each polygon will be less than the number of points of the given polygons by a common factor. The triangular array of the abacus coordinates this property of polygons.