Multiplication is a repeated addition:

$$

\begin{align}

a \times b & = \overbrace{a + a + a + … + a}^{\mbox{b times}} \\

& = \overbrace{b + b + b + … + b}^{\mbox{a times}}

\end{align}

$$

We can visualize it using sets:

or the number line:

### Writing multiplications

The following are common forms of describing the product of a and b:

$$

\begin{align}

a \times b &\hspace{.2in}&\mbox{e.g., } 3 \times 5\\

a \cdot b& &\mbox{e.g., } a \cdot a^{-1}\\

ab& &\mbox{e.g., }3ab + c\\

a\,(b)& &\mbox{e.g., }5\,(3 + a)\\

(a)(b)& &\mbox{e.g, }(x + 1)(x + 2)\\

a \star b& &\mbox{used in computer languages}

\end{align}

$$

and, of course

$$

\begin{align}

a&\\

\underline{\times\,\,b}&\\

\end{align}

$$

### Multiplying values with units

It is useful to think of the units as multiplying the values that they modify, i.e., ‘5 meters’ means ‘5 x meters’. For example, we can multiply unit-less values:

3 x 5 = 15

or values with units to unit-less values:

5 cookies x 3 = (5 x cookies) x 3 = (5 x 3) x cookies = 15 x cookies = 15 cookies

or values with the same units, as when we are looking for an area or a volume:

3 m x 5 m = (3 x m) x (5 x m) = (3 x 5) x (m x m) = 15 x m^{2} = 15 m^{2}

or values with different units, e.g., if it takes 5 masons 10 days to build a wall then we can say that the work needed to build the wall was

(5 masons) x (10 days) = (5 x 10) x (masons x days) = 50 x mason-days = 50 mason-days

The ‘-‘ in ‘masons-days’ is not a minus sign but a dash that should be interpreted as multiplication; yes, math is a language that, like any other language, sometimes has ambiguities.

### Operation precedence

The associative and commutative laws of addition allow us to manipulate the terms of a sum so we can add them in any way we want:

$$

3 + 4 + (-5) = (-5) + 4 + 3 = 4 + (-5) + 3

$$

This is also the case for operations that only involve multiplications

$$

3 \times 4 \times (-5) = (-5) \times 4 \times 3 = 4 \times (-5) \times 3

$$

but it is not the case if there are both multiplications and additions, even when the terms are not commuted:

$$

-35 = (3 + 4) \times (-5) \neq 3 + (4 \times (-5))= -17

$$

To evaluate an expression correctly we follow this order:

- parenthesization
- exponentiation
- multiplication and division from left to right
- addition and subtraction from left to right

For example,

$$

3 + 2^2 / 2 \times 5 – 2 = (3 + (((2^2) / 2) \times 5)) – 2

$$

### Multiplying a digit by a digit

Multiplication is repeated addition so can find a product by adding one digit the number of times indicated by the other digit, i.e,

$$

\begin{align}

4 \times 5 &= \overbrace{4 + 4 + 4 + 4 + 4}^{\mbox{5 times}}\\

&= 20

\end{align}

$$

or

$$

\begin{align}

5 \times 4 &= \overbrace{5 + 5 + 5 + 5}^{\mbox{4 times}}\\

&= 20

\end{align}

$$

However, this is inefficient; multiplication is used so often that it pays to simply memorize the multiplication tables from 0 to 10. These are 110 products so it sounds difficult but it is not as hard:

- \(0 \times a = 0\)
- \(1 \times a = a\)
- \(10 \times a = a\) with a zero at the end (e.g., \(10 \times 7 = 70\))
- \(9 \times a = (10 \times a) – a\) (e.g., \(9 \times 7 = 10 \times 7 – 7 = 70 – 7 = 63\))
- \(2 \times a = a + a\) (e.g., \(2 \times 7 = 7 + 7 = 14\), easy because we know the addition facts)
- \(a \times b\) is even unless both a and b are odd (i.e., 75% of the single-digit products are even)

So now we only need to memorize the tables of 3, 4, 5, 6, 7, and 8, from 2 to 8.. that is 6 × 6 = 36 products, instead of 110, which is much better, and since multiplication is commutative, e.g., 3 × 7 = 7 × 3, we only need to memorize half of these 36 products, i.e., we only need to memorize 18 products.

Most people have little problem memorizing the tables of 3, 4, and 5, and memorizing the squares, i.e., the products of the digit by itself, like 7×7 and 8×8; we are left with only two ‘difficult’ products to memorize: 6 × 7 and 7 × 8.

6 × 7 = 42 is a prestigious number since it is divisible by 1, 2, 3, 6, 7, 14, 21, and 42, which is a lot, plus, according to Douglas Adam’s “The Hitchhiker’s Guide to the Galaxy”, 42 is “The Answer to the Ultimate Question of Life, the Universe, and Everything”, as found by the supercomputer “DeepThought”.

This leaves us with the dreaded 7 × 8 = 56 that we can rewrite as

$$

56 = 7 \times 8

$$

yes… “five, six, seven, eight”. Simple, right?

### Multiplying a digit by a multi-digit number

The U.S. standard multiplication algorithm uses ‘carries’:

$$

\begin{align}

424\,\,\,13\,\,\,&\\

785,927&\\

\underline{\times\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,5}&\\

3,929,635&\\

\end{align}

$$

Another way to multiply these numbers is by multiplying by place values; it is longer but avoids errors due to forgotten carries:

$$

\begin{align}

&\,\,\,\,\,\,785,927\\

&\underline{\times\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,5}\\

&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,35\hspace{.2in}⇐ 7 \times 5\\

&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,100\hspace{.2in}⇐ 20 \times 5\\

&\,\,\,\,\,\,\,\,\,\,\,4,500\hspace{.2in}⇐ 900 \times 5\\

&\,\,\,\,\,\,\,\,25,\hspace{.5in}⇐ 5 \times 5 \times 1,000\\

&\,\,\,\,\,400,\hspace{.5in}⇐ 80 \times 5 \times 1000\\

+\,&\underline{3,500,\hspace{.4in}}⇐ 700 \times 5 \times 1000\\

&3,929,635\\

\end{align}

$$

### Multiplying multi-digit numbers

The U.S. standard multiplication algorithm arranges the result of a series of single-digit partial products into a sum. For example, to multiply 961 by 392 we first find the single-digit products of 961 × 3 = 2,883, 961 × 9 = 8,649, and 961 × 2 = 1,922, and arrange them as a sum, positioning them under the respective digits, and then we must add:

$$

\begin{align}

\,\,\,\,\,\,\,\,\,\,\,\,961&\\

\,\,\,\,\,\,\underline{\times\,392}&\\

\,\,\,\,\,\,\,\,\,1,922&\hspace{.2in}⇐ 961 \times 2 \mbox{; first partial product}\\

86,49\,\,\,&\hspace{.2in}⇐ 961 \times 9 \mbox{; second partial product}\\

\underline{+\,288,3\,\,\,\,\,}&\hspace{.2in}⇐ 961 \times 3 \mbox{; third partial product}\\

376,712&

\end{align}

$$

The standard multiplication algorithm is not an “in place” algorithm but instead we need to find the partial products separately and then bring the results as partial products that we need to add; if there is an error, we have to look for it in a different place.

In contrast to the standard algorithm, we can multiply “in place” using place values:

$$

\begin{align}

\,\,\,\,\,\,\,\,\,\,\,\,961&\\

\,\,\,\,\,\,\underline{\times\,392}&\\

2&\hspace{.2in}⇐ 2 \times 1\\

120&\hspace{.2in}⇐ 2 \times 6 \times 10\hspace{.2in}&\mbox{, i.e., }2 \times 60\\

1,800&\hspace{.2in}⇐ 2 \times 9 \times 100&\mbox{, i.e., }2 \times 900\\

90&\hspace{.2in}⇐ 9 \times 1 \times 10&\mbox{, i.e., }90 \times 1\\

5,400&\hspace{.2in}⇐ 9 \times 6 \times 100&\mbox{, i.e., }90 \times 60\\

81,\,\,\,\,\,\,\,\,\,\,&\hspace{.2in}⇐ 9 \times 9 \times 1,000&\mbox{, i.e., }90 \times 900\\

300&\hspace{.2in}⇐ 3 \times 1 \times 100&\mbox{, i.e., }300 \times 1\\

18,\,\,\,\,\,\,\,\,\,\,&\hspace{.2in}⇐ 3 \times 6 \times 1,000&\mbox{, i.e., }300 \times 60\\

\underline{+\,270,\,\,\,\,\,\,\,\,\,}&\hspace{.2in}⇐ 3 \times 9 \times 10,000&\mbox{, i.e., }300 \times 900\\

376,712&

\end{align}

$$

If we need the partial products, we can add the terms generated by each digit and end up with the same set of sums that we get from the U.S. standard algorithm.

### Writing products of sequences

When writing the product of a sequence we usually use dots, or the symbol for a product:

$$

\begin{align}

\prod_{i = 1}^{10}i &= 1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 \times 9 \times 10\\

&= 1 \times 2 \times \dots \times 10

\end{align}

$$

The same applies for infinite products:

$$

\prod_{i = 1}^{\infty}i = 1 \times 2 \times 3 \times \dots

$$

### Factorials

The product of sequential numbers starting from 1 is used so much that we give it a particular name: factorial. The factorial of a number is written as the number followed with an exclamation mark, e.g.,

$$

\begin{align}

5! &= 1 \times 2 \times 3 \times 4 \times 5\\

&= 1 \times 2 \times \dots \times 5\\

&= 120

\end{align}

$$

which we read as “five factorial”.

Now let’s consider some factorials, expressing n! in terms of the previous factorial (n-1)!

$$

\begin{align}

1! &= 1 = 1\\

2 !&= 1! \times 2 = 2\\

3! &= 2! \times 3 = 6\\

4! &= 3! \times 4 = 24\\

\vdots\\

n! &= (n-1)! \times n

\end{align}

$$

This is a pattern that we can reverse, expressing n! in terms of the next factorial (n+1)!; this allow us to assign a value to 0!:

$$

\begin{align}

n! &= (n+1)! / (n+1)\\

\vdots\\

4! &= 5! / 5 = 120 / 5 = 24\\

3! &= 4! / 4 = 24 / 4 = 6\\

2 !&= 3! / 3 = 6 / 3 = 2\\

1! &= 2! / 2 = 2 / 2 = 1\\

0! &= 1! / 1 = 1 / 1 = 1

\end{align}

$$

So, in summary, we define the factorial of any non-negative integer as

$$

n! =

\left\{

\begin{array}{ll}

1&\mbox{if}\,\,\,n = 0\\

n \times (n-1)!&\mbox{if}\,\,\,n \neq 0

\end{array}

\right.

$$

### The laws of multiplication

These laws are analogous to the laws of addition.

The law of associativity says that we can multiply the factors of a product in any order:

**P5.** \(a \times (b \times c) = (a \times b) \times c\)

Since regrouping does not affect the result of multiplication, we usually omit the parentheses but, in reality, the parentheses are implicitly there; it just happens that it does not matter where they are.

The existence of a multiplicative identity introduces the number 1; it says that there exists one unique number, namely 1, that does not change the value of $%a$% when we multiply it by it.

**P6.** \(a \times 1 = 1 \times a = a\)

Now we can use this number 1 to bring forward the fractions with the existence of a multiplicative inverse:

**P7.** \(a \times a^{-1} = a^{-1} \times a = 1\hspace{.5in}\mbox{for a }\neq 0\)

The number a^{-1} is called the reciprocal, or the multiplicative inverse, of a.

Finally, the commutative law states that the product of two numbers does not depend on the order in which we multiply them:

**P8.** \(a \times b = b \times a\)

### The Distributive law

This law connects addition with multiplication:

**P9.** \(a \times (b + c) = (a \times b) + (a \times c)\)

### Problems

**1.** Show that 2 cookies x 3 = 3 cookies x 2

**2.** Multiply 2 meters by 3 meters by 5 meters

**3.** Add parenthesis between any two terms following the order precedence of operations

12 + 34 + 56

12 – 34 + 56

12 x 34 x 56

12 x 34 + 56

12 + 34 + 56

12 x 34 / 56

12 / 34 + 56

12 + 34 x 56 / 78

12 / 34 x 56 / 78

**4.** Evaluate

3 x 3 x 3

3 x (-3) x 3

3 x (-3) x (-3)

(-3) x (-3) x (-3)

-[(-3) x (-3) x (-3) x (-3)]

**5.** Multiply

2 x 8

7 x 4

6 x 7

8 x 7

**6.** Multiply using both the U.S. standard and decimal decomposition algorithms

34,908 x 3

**7.** Multiply using both the U.S. standard and decimal decomposition algorithms

$$

34,908 \times 356

$$

**8.** Evaluate

$$

\prod_{i=1}^{10}i^2

$$

**9.** Evaluate

6!

**“take all the time you want” problems:**

For the following problems use…

**P1.** \(a + (b + c) = (a + b) + c\)Additive associativity

**P2.** \(a + 0 = 0 + a = a\)Additive identity

**P3.** \(a + (-a) = (-a) + a = 0\)Additive inverse

**P4.** \(a + b = b + a\)Commutativity

**P5.** \(a \times (b \times c) = (a \times b) \times c\)Multiplicative associativity

**P6.** \(a \times 1 = 1 \times a = a\)Multiplicative identity

**P7.** \(a \times a^{-1} = a^{-1} \times a = 1\mbox{ for }a\neq 0\)Multiplicative inverse

**P8.** \(a \times b = b \times a\)Multiplicative commutativity

**P9.** \(a \times (b + c) = (a \times b) + (a \times c)\)Distributivity

to show that …

**10.** \(a \times 0 = 0\)

**11.** \(a\cdot(-b) = -(a \cdot b)\)

**12.** \((-a)\cdot b = -(a \cdot b)\)

**13.** \(-a = -1 \cdot a\)

**14.** \(-(-a) = a\)

**15.** \((-a)\cdot(-b) = a\cdot b\)