SHORT DIVISION

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FRACTIONS ON THE ABACUS
SHORT DIVISION FRACTIONS x WHOLES MULTIPLICATION D
IVISION REDUCING
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FRACTIONS
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.
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NO!
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.
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PUSH HERE
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.
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TWO-THIRD OF THREECORRESPONDS TOTWO-THIRDS OF
TWELVE
PUSH HERE
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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PUSH HERE
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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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. 
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.
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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. 
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.
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THREE-FOURTHS OF FOUR CORRESPONDS
TOTHREE-FOURTHS OF TWELVE
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.
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TWO-THIRD0F THREECORRESPONDS TOTWO-THIRDSOF
NINE
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.
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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 
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DIVIDING FRACTION
I invited again my friend for pie.Have no fear
this is whyFor - 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.
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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.
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THREE-FOURTHSOF FOURCORRESPONDS
TOTHREE-FOURTHSOF TWELVE
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.
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TWO-THIRDS OF THREECORRESPONDS TOTWO-THIRDSOF
TWELVE
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. 
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4
1
3
5
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.