tiny soroban gif LOCATING THE DECIMAL: MULTIPLICATION & DIVISION


There are plenty of instruments out there that don't have unit rods marked on the beam. The following techniques are one solution to solving the question, "Where should the decimal point go?"


MULTIPLICATION:
Here's a method to consider for locating the decimal point in problems of multiplication.

Rule: Simplify both numbers to read x.xxx... Take the net of the decimal point movement in both the multiplicand and in the multiplier. (This yields a number.) In the final analysis, undo by that number by moving in the opposite direction. This is probably best explained in the following examples:


Example 1: 31.68 x 0.00082

Put both numbers in the form x.xxx .....

Move the decimal in the multiplicand one place to the left (call it L1).
Move the decimal in the multiplier four places to the right (call it R4).

This gives : 3.168 x 8.2

Take the net of these : L1 plus R4 = R3.

In the final analysis remember to *undo* R3 by moving the decimal point L3.

Now, look again at the equation : The approximation is 3 x 8 = 24

Undo R3 by moving the decimal point L3. This leaves 0.024

The actual answer to the problem is : 31.68 x 0.00082 = 0.0259776


Example 2: 5326.879 x 0.00000079

Put both numbers in the form x.xxx .....

Think : L3 for the multiplicand and R7 for the multiplier

5.326879 x 7.9

Take the net of these : L3 plus R7 = R4

Remember to "undo" R4 at the end by moving the decimal point L4.

The approx answer is 5 x 8 = 40

Undo R4 by moving the decimal point L4. This leaves 0.0040

The actual answer to to the problem is : 5326.879 x 0.00000079 = 0.004208234


DIVISION:

Here's a method to consider for locating the decimal point in problems of division. The method is similar to that done for problems of multiplication in that we simplify both numbers to read x.xxx.... However, it's a little more complicated but once gotten used to it's very easy.

Rule: Simplify both numbers to read x.xxx... Take the net of the decimal point movement in the dividend and the *opposite* of the decimal point movement in the divisor. (This yields a number.) In the final analysis, undo by that number by moving in the opposite direction. Probably best explained by example. :)


Example 1: 0.68 390

Put both numbers in the form x.xxx .....

Move the decimal in the dividend one place to the right (call it R1)
Move the decimal in the divisor two places to the left (call it L2)

The result is : 6.8 3.9

Take the net of the decimal movement in the dividend and the *opposite* of the movement in the divisor. That is;

R1 plus R2 = R3

6.8 3.9 is approximately 7 4 = 1.75

Undo R3 by moving the decimal L3 : this equals 0.00175

The actual solution : 0.68 390, is 0.00174358


Example 2: 0.073 0.0054

Put both numbers in the form x.xxx .....

Think : R2 for the dividend and R3 for the divisor

The result is : 7.3 5.4.

Take the net of the decimal movement in the dividend and the *opposite* of the movement in the divisor. That is;

R2 plus L3 = L1

7.3 5.4 is approximately 7 5 = 1.4

Undo L1 by moving the decimal R1 : this equals 14.00

Actual solution : 0.073 .0054 = 13.518...


Example 3: 897 0.00061

Put both numbers in the form x.xxx .....

Think : L2 for the dividend and R4 for the divisor

8.97 6.1

Take the net of the decimal movement in the dividend and the *opposite* of the movement in divisor. That is;

L2 plus L4 = L6

8.97 6.1 is approximately 9 6 = 1.5

Undo L6 by moving the decimal R6 : this equals 1500000.00

Actual solution : 897 0.00061 = 1470491.803

 

 

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Gary Flom   Atlanta Georgia   USA
Email
gsflom[at]bellsouth[dot]net