Short Division

SHORT DIVISION

To begin the fraction lesson teach short division on the Pythabacus. For an example eight (divided by) three is demonstrated to the right. 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 eight 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.

To solve 8 / 3 first reset beads to the left. Then, push 3 bottom roll beads to the right to generate a triangle representing the divisor. Now, from the columns of beads above the triangle against the left post push beads from each column, in turn, over against the divisor triangle until 8 beads have been pushed to represent the dividend.

The whole number part of the quotient is represented by the number of bottom row beads you can push from the left to the right and generate a regular triangle. The whole number part is 2.

The remainder is the number of beads in the incomplete column of the dividend. In this case, the remainder is shown to be 2.

Therefore, 8 / 3 = 2 with a remainder of 2, the number of nuts each squirrels got, and what remained for the winter birds.

Short division can be represented graphically by students as a number of squirrel holes they fill in turn one nut at a time until each hole has an equal number of nuts. The extra nuts are placed outside the squirrel holes for the winter birds.

For short division with a single digit divisor, if there are not enough beads above the divisor triangle to represent the dividend and the tens value of the dividend is less than the divisor the following method is used for short division. For a problem such as 45 / 6 reset beads left and push the tens complement of 6 (4) against the right post. Then starting with the bottom bead closest to the left post count to the right a number of bottom roll beads equal the tens value of the dividend (4) and push the remaining beads (2) midway to the right post.



Push these two beads a little back to the left leaving two columns of beads in the middle of the Pythabacus. The four beads remaining against the left post represent the tens value of the dividend 40 and the eight beads in the columns the ones value of the dividend. You may note there is an error, the dividend was 45 not 48 and 48 is greater then 45, so the quotient, given by the sum of the bottom beads against the left and right post (8), is correct for 48 / 6 not for 45 / 6.
To get the correct dividend push one more bead from the left post to the right to add another column to the array in the middle of the Pythabacus. Push the bottom roll beads a little back to the left as before to isolate the columns. Now there are three beads against the left post representing the tens value of the dividend 30 and twelve beads in the mid-columns therefore the dividend represented is 42. But we want 45 so join three more beads from the closest column to the left to the mid-columns.

The sum of the bottom roll beads against the left and right post now give the correct whole number part of the quotient (7) and the extra beads pushed over to complete the dividend gives the remainder part of the quotient (3).