# Arithmetic/Multiplication and Division

## Multiplication

Multiplication is denoted by an asterisk (*), ${\displaystyle \times }$ , or ${\displaystyle \cdot }$  sign. However, the “×” sign is normally not used in algebra, and is instead limited to very basic elementary math, as it can easily be confused with an “x” variable. The generic multiplication operator will take any two numbers, called factors, as operands. The result is called the product of the two numbers. If the multiplicants are not both written as numbers, the multiplication sign can be left out. Thus, the following example expressions are equivalent:

${\displaystyle 3*\!(a*b)=3\times \!(a\times b)=3\!\cdot \!(a\!\cdot \!b)=3\!\cdot \!(ab)=3(a\!\cdot \!b)=3(ab)=3ab}$

Multiplication is a form of repeated addition. For example ${\displaystyle 3\times 5}$  means

${\displaystyle 3+3+3+3+3\quad \operatorname {or} \quad 5+5+5}$

Multiplication is also commutative. This means that the multiplication of two numbers (factors) will give the same product regardless of the order in which the numbers are multiplied together. The following expressions are also equivalent:

${\displaystyle 3\!\cdot \!(ab)=a\!\cdot \!(3b)=b\!\cdot \!(3a)}$

Numbers with exponents that are whole numbers larger than 1 indicate the number of factors to be multiplied, thus that number is multiplied by itself as many times as the exponent shows. Numbers with an exponent of 1 have only one factor, and therefore are equal to the number. Any number with an exponent of 0 has no factors at all, and the result is 1. Examples:

${\displaystyle 5^{3}=5\times 5\times 5\qquad \qquad 5^{1}=5\qquad \qquad 5^{0}=1}$

### Long Multiplication

Long Multiplication is the multiplication of numbers more than 12, but usually only the facts from 1 through 9 are used. Before you attempt long multiplication, please make sure you know the facts 1 through 9. The others are optional, but makes long division a bit easier. The steps for the vertical multiplication method are:

1. Write the numbers down.

  52
19
------


2. Multiply 9 times 2. If there is a tens place for the answer, regroup. Multiply 9 times 5, add the regroup, and write the numbers down. You should have:

  1
52
19
------
468


3. Multiply 1 times 2, and 1 times 5, regrouping if needed, but this time shift the answer one space to the left. If you want to, you can put a zero under the 8 instead. You should now have:

  52
19
------
468
520


 52
19
-----
468
520
-----
988


If you are multiplying decimals, then multiply without the decimal point. Count the number of decimal places in both numbers, and add the number of decimal places. In the answer, count that number of spaces to the left. Put the decimal point there.

5. In a summary:

Write the problem down, vertically. Multiply the last digit of the second number to the last digit of the first number. Regroup if there is a tens place in the answer. Multiply it by the second to last number, and ADD the regroup. Repeat the process for the second to last digit of the second number, but put a 0 at the end of the line under the number you got before, if a third line put 2 0s, and repeat until the problem is done.

### Fast Multiplication

Fast Multiplication is the method in which you can reasonably multiply numbers greater than 10 and reasonably less than 1000 by simply multiplying by the method of "10's." This is done by recognizing how many digits there are in the numbers. Here are some steps which are useful for multiplying numbers really fast

1. See if you can recognize any zeros on the end and simply "add them" to your answer.

  45,300 x 5
The easy way to do this is "taking away" the 2 zeros for now and reserving them for later.

the number is now 453 x 5, which is much too mind boggling to do.
Now here comes the interesting part of the method
of "10's"


2. Break down the number into its "10's" parts

  What this means is basically breakup the number by its place value.

  453 = 4 (hundreds place) + 5 (tens place) + 3 (ones place)

  Knowing that, this become 400, 50 and 3.


3. Multiply and apply the "10's part"

  okay now simply multiply:

  So, here is a step where we essentially take out the "0's" out for a bit and put it back
in when were done.

  so, its now 5 x 400. in order to make it easier, "take out" the zeroes for now and
multiply 4 x 5 = 20. Now
heres the magic. Since you magically took away the 2 zeroes, you will now suddenly make
the 2 zeroes reappear!
20 + "00" = 2000! AMAZING! (the quotes means they're magic zeroes, and simply not the value
of zero!)

  50 is done the same away. Take away the "0" and multiply 5 x 5 = 25. Now add it back, 25 +
"0" = 250

  Simply 3 x 5 = 15


  2000
250
15
-----
2265!


5. Now take the two zeroes you reserved in the beginning (from the original 45,300), and tack them onto the end to get your answer: 226,500.

Step 2: Multiplying numbers that are not zero friendly

  2102 x 52


Using the step from before recognize that:

2102 = 2 (thousand) + 1(hundred) + 0 (tens) + 2 (ones)
52 = 5 (tens) + 2 (ones)

Now, to make it easier on yourself, circle the number 2 of "52" and put it in your magic
hat. (2)

Now the problem becomes 2102 x 50. Look familiar? First of all, take out the magic "0" and
put it in the hat, too.
Since we recognized that 50 is basically 5 with an added magical "0" to it, we now see the
problem as

2102 x 5!

Now break down the bigger, uglier number and start multiplying:
2000 x 5 (take away the magic zeroes) = 2 x 5 = 10 + "000"(now put them back!) = 10,000 (notice it has 4 zeroes)
100 x 5 (take away the magic zeroes) = 1 x 5 = 5 + "00"  (now put them back!) =   500 (2 zeroes)
2 x 5  (sadly, no magical zeroes)  = 2 x 5 = 10                             =    10 (1 zero)

 Remember, after every step, be sure to put your friendly magical "0" back in:
10,000 + "0" =100,000
500 + "0" =  5,000
10 + "0" =    100
(notice how the number of zeroes on the left side equal the number of zeroes on the right
side)

Now add them all together:

 100000
5000
100
-----
105100....... That's not all yet folks! Do you remember the 2 in your magic hat? Lets get it to work:

  2 x 2102 =
2000 x 2 = 2 x 2 = 4 = 4000
100 x 2 = 1 x 2 = 2 =  200
2 x 2             =    4
total:  4204

  So the answer should be
105,100
4,204
-------
109,304! Wow!


### Outside Resources

To practice this concept, I recommend Developmental Mathematics, volume 8, along with cards to memorize multiplication facts. This is my favorite for developing multiplication skills, but if you prefer something else, try Miquon Math. This is a complete program for grades 1-3 and is an excellent program for developing advanced math skills. Another option is Progress in Mathematics, a standards-based math program for Sadlier-Oxford. Of course, advance multiplication skills until the student/s is doing 5-digit multiplication with ease.

## Division

Division uses the ÷ sign. It may also be signified by the slash(/), :, or the fraction bar. The generic division operator will take any two numbers as operands. The number before the ÷ sign is called the dividend and the number after the ÷ sign is called the divisor. The result is called the quotient of the two numbers.

Division is not a commutative operation. Switching the dividend and the divisor will likely give a different quotient (but sometimes not). The division with a divisor of 0 is not defined. There is no answer for it.

Example:

• ${\displaystyle 4/2=2}$  and ${\displaystyle 2/4=0.5}$

### Long Division

In arithmetic, long division is an algorithm for division of two real numbers. It requires only the means to write the numbers down, and is simple to perform even for large dividends because the algorithm separates a complex division problem into smaller problems. However, the procedure requires various numbers to be divided by the divisor: this is simple with single-digit divisors, but becomes harder with larger ones.

A more generalized version of this method is used for dividing polynomials (sometimes using a shorthand version called synthetic division).

In long division notation, 500 ÷ 4 = 125 is denoted as follows:

${\displaystyle {\begin{matrix}\quad 125\\4{\overline {)500}}\\\end{matrix}}}$

The method involves several steps:

1. Write the dividend and divisor in this form:

${\displaystyle 4{\overline {)500}}}$

In this example, 500 is the dividend and 4 is the divisor.

2. Consider the leftmost digit of the dividend (5). Find the largest multiple of the divisor that is less than the leftmost digit: in other words, mentally perform "5 divided by 4". If this digit is too small, consider the first two digits.

In this case, the largest multiple of 4 that is less than 5 is 4. Write this number under the leftmost digit of the dividend. Divide the multiple by the divisor (in this case, 4 divided by 4) and write the result (in this case, 1) above the line over the leftmost digit of the dividend.

${\displaystyle {\begin{matrix}1\\4{\overline {)500}}\\4\end{matrix}}}$

3. Subtract the digit under the dividend from the digit used in the dividend (in this case, subtract 4 from 5). Write the result (remainder) (in this case, 1) underneath and in the same column, then drop the second digit of the dividend (in this case, the first zero) to the right of it. This gives you a new number to divide by the divisor.

${\displaystyle {\begin{matrix}1\\4{\overline {)500}}\\{\underline {4}}\\\;\,10\end{matrix}}}$

4. Now apply the same steps 2 and 3 to this new number, and write the results in the corresponding columns (in this case, the unit column is aligned with the second digit of the original dividend): multiple and remainder underneath the new number, and the answer above the line.

${\displaystyle {\begin{matrix}\,\,12\\4{\overline {)500}}\\{\underline {4}}\\\;\,10\\\quad {\underline {8}}\\\quad \;\,20\end{matrix}}}$

5. Repeat step 4 until there are no digits remaining in the dividend. The number written above the bar is the quotient, and the last remainder calculated is the remainder for the entire problem.

${\displaystyle {\begin{matrix}\quad 125\\4{\overline {)500}}\\{\underline {4}}\\\;\,10\\\quad {\underline {8}}\\\quad \;\,20\\\quad \;\,{\underline {20}}\\\qquad 0\end{matrix}}}$