Lesson 10

MULTIPLYING WHOLE NUMBERS
AND DECIMALS

Let us face facts.  The student will quickly replace written multiplication with a calculator.  (What if you don't have a calculator?  What if you don't have a pencil  Lesson 9.)  Therefore, what should the student know about written multiplication that deserves to be called educational?  The student should know that it is based on decomposing the multiplicand. And the student should know the basics of placing the decimal point.

First:

"7 times 8 is 56."  Write 6, carry 5.

"7 times 2 is 14, plus 5 is 19."  Write 9, carry 1.

"7 times 6 is 42, plus 1 is 43."  Write 43.

That is,

7 × 628 = 4396.

Why can we calculate it that way?  Because we have distributed 7
to the ones, tens, hundreds of 628:

On the left, it is the 5 tens of  7 × 8 that we add to 7 × 2 tens, making 19 tens; and 1 hundred of  7 × 20 that we add to 7 × 6 hundreds, making 43 hundreds.

*

When the multiplier has more than one digit --

-- follow the same procedure for each digit.  Now, when we multiply by 3 tens, the product is 1884 tens.  It is not necessary to write 18840.(Lesson 2.) Simply begin the 4 of 1884 in the tens column.

When we multiply by 2 hundreds, the product is 1256 hundreds, and so we write 6 in the hundreds column.

We can state the rule as follows:

Anticipating the next Question, if there were decimal points --

-- the multiplication would proceed in exactly the same way. In the answer, we would then separate as many decimal digits as there are in the two numbers together; in this case, three.

On multiplying by 7 ones, write 2 in the ones column.

On multiplying by 3 tens, write 8 in the tens column -- because the partial product 288 means  288 tens.

Rather than do the traditional method of "carrying," the student should be able to calculate each partial product mentally by distributing from left to right.  We have indicated this above.

Example 2.   Multiply 45 × 236.

It does not matter which number you choose as the multiplier, that is, to place on the bottom.  However it is more efficient to choose 45, because then there will be only two multiplications rather than three.

Example 3.   0's within the multiplier.   

On multiplying by 8 ones, write 6 in the ones column.

Any number times 0 is 0, therefore it is not necessary to write anything in the tens column.

On multiplying by 3 hundreds, write 1 in the hundreds column.

It is not necessary to write rows of 0's.  They add nothing to the product.

We are now about to see that we can do arithmetic only with whole numbers -- and then correctly place the decimal point

Example 4.   .2 × 6.03

Solution.  Ignore the decimal points.  Simply multiply

2 × 603 = 1206         

Now we must place the decimal point.  Together,  .2 and 6.03 have three decimal digits.  (Lesson 4, Problem 11.)  Therefore, starting from the right, separate three digits:

1.206

When we ignore a decimal point, we have in effect moved the point to the right:

6.03 → 603

We have multiplied by a power of 10. (Lesson 4, Question 2.) Therefore, to name the right answer,
we must divide by that power, we must separate
the same number of decimal digits. For, a number
does not change if we multiply it and then divide
by the same number. (Lesson 11.)

Example 5.   .04 × .011

Solution.   When we ignore the decimal points --

04 × 011

-- then those are leading 0's. They add nothing to the value.  We simply multiply

4 × 11 = 44.

Now, .04 and .011 together have five decimal digits.  Therefore the product will have five decimal digits -- it will look like this:

._ _ _ _ _

Here is the product:

.0 0 0 4 4

Example 6.   200 × .012

Solution.  Ignore the decimal point.  Multiply

200 × 12 = 2400

Again, to multiply whole numbers that end in 0's, first ignore the 0's, then replace them.  (Lesson 9, Question 2.)  But replace only the 0's on the end of whole numbers.  Do not replace the 0 of .012

Now separate three decimal digits (.012):

2.400 = 2.4

(Lesson 3, Question 8.)

6 × 45 = 6 × 40  +  6 × 5 = 240 + 30 = 270.

Therefore, on separating two decimal digits,

6% of $45 is $2.70.

We can justify this as follows.  Since

1% of $45 is $.45

(Lesson 4), then 6% will be 6 times 1%, that is, 6 × .45.

Alternatively, we could say, Change 6% to .06 (Lesson 4), and multiply .06 × 45. Either way, we have to divide by 100, that is, separate two decimal digits.

Example 7.   How much is 9% of $84?

 Solution.   Multiply 9 × 84, then separate two decimal digits.

9 × 84 = 720 + 36 = 756.

Therefore, 9% of $84 = $7.56.

Example 8.   How much is 3% of $247?

 Solution.   3 × 247 = 600 + 120 + 21 = 741.

Therefore, 3% of $247 = $7.41.

Example 9.   How much is 11% of $76?

 Solution.   11 × 76 = 760 + 76 = 836.

Therefore, 11% of $76 = $8.36.

These are simple problems that do not require a calculator.  For more such simple problems, see Lesson 29.  To learn how to do percent problems with a calculator, see Lesson 14.

Area of a rectangle

What is "1 square foot"?

1 square foot is a square figure in which each side is 1 foot.

We abbreviate "1 square foot" as 1 ft².

Now here is a rectangle whose base is 3 cm and whose height is 2 cm.

What do we call the small shaded square?

Since each side is 1 cm, we call it "1 square centimeter."  And we can see that the entire figure is made up 2 × 3 or 6 of them

In other words, the area of that rectangle -- the space enclosed by the boundary -- is 6 square centimeters:  6 cm².

If the rectangle were 3 by 3 -- that is, if it were a square -- then it would be made up of 9 cm².  If it were 3 by 4, the area would be 12 cm². And so on.  In every case, to calculate the area of a rectangle, simply multiply the base times the height.

When the length is measured in centimeters, the area is measured in square centimeters:  cm².  And similarly for any unit of length.

We have illustrated this with whole numbers, but it will be true for any numbers.

If the base is 12 in, and the height is 6.5 in, then to find the area, multiply

12 × 6.5

Now,

12 × 65 = 10 × 65  +  2 × 65 = 650 + 130 = 780.

Therefore on separating one decimal digit (6.5):

Area = 78 in².

The order property

Let us return to the order property of multiplication (Lesson 9):

If two numbers take turns at being multiplied by each other, then the numbers produced in each case will be equal.

(Euclid, VII. 16.)

For example, if the two numbers are 32 and 5, then if we repeatedly add 32 five times, we will get the same number as when we add 5 thirty-two times.

32 + 32 + 32 + 32 + 32 = 5 + 5 + 5 + . . . + 5 + 5.

And we could prove that without naming the product

But to take a simpler example, let the two numbers be 3 and 4.  Then

Four 3's = Three 4's.

And we do not mean simply that each one is 12.

Look at this figure:

The bottom row, let us say, is made up of 3 cm².  The entire rectangle is made up of 4 such rows.  That is, it is made up of

4 × 3 cm².

On the other hand, the first column on the left is made up of 4 cm².   And there are 3 such columns, so that from that point of view, the rectangle is made up of

3 × 4 cm².

In other words,

3 × 4 cm² = 4 × 3 cm².

If we add 4 cm² three times we will get the same result as when we add 3 cm² four times -- and to know that we did not need to know that 3 × 4 = 12

That is is the order property of multiplication.

*

You sometimes see

3 cm × 4 cm = 12 cm²,

which to be honest make no sense. The multiplier (on the left) shows the number of times to repeatedly add the multiplicand (on the right). Therefore the multiplier must always be a pure number.

Please "turn" the page and do some Problems.

or

Continue on to the Next Lesson.

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