Rule of signs for multiplication

Accordingly, the rule of signs can be summarized in four standard cases:

• \( (+) \cdot (+) = + \)
• \( (-) \cdot (-) = + \)
• \( (+) \cdot (-) = - \)
• \( (-) \cdot (+) = - \)

This rule is typically introduced and memorized during the early years of middle school.

Much less attention is usually paid to the underlying reasons why the product of two negative numbers is positive, or why multiplying a negative number by a positive one results in a negative value.

The theoretical justification will be provided in the proof at the end of this page. For the moment, the focus is on a series of examples that illustrate how the rule operates in concrete situations.

Practical examples

The following examples show how the rule of signs applies in standard multiplication problems.

The product of two positive numbers is positive

The product of two positive numbers, such as \( 3 \cdot 4 = 12 \), is positive because both factors have the same sign.

$$ 3 \cdot 4 = 12 $$

Explanation. The result can be justified formally by considering absolute values. $$ 3 \cdot 4 = +(|3| \cdot |4|)= +(3 \cdot 4) = 12 $$

The product of two negative numbers is positive

When two negative numbers are multiplied, for example \( (-5) \cdot (-2) = 10 \), the product is positive because the two factors have the same sign.

Both numbers are negative, so their signs agree.

$$ (-5) \cdot (-2) = 10 $$

Explanation. $$ (-5) \cdot (-2) = +(|-5| \cdot |-2|)= + (5 \cdot 2) = 10 $$

The product of a positive number and a negative number is negative

If one factor is positive and the other is negative, as in \( 6 \cdot (-3) = -18 \), the product is negative because the signs of the factors differ.

The number 6 is positive, whereas the number -3 is negative.

$$ 6 \cdot (-3) =  -18 $$

Explanation. $$ 6 \cdot (-3) = -(|6| \cdot |-3|)= - (6 \cdot 3) = -18 $$

The product of a negative number and a positive number is negative

In the same way, multiplying a negative number by a positive one, such as \( (-7) \cdot 8 = -56 \), yields a negative result because the factors have opposite signs.

$$ -7 \cdot 8 = -56 $$

Explanation. $$ -7 \cdot 8 = -(|-7| \cdot |8|)= - (7 \cdot 8) = -56 $$

These examples show that, in algebra, the sign of a product is determined solely by whether the signs of the factors coincide or differ.

Proof

How can the rule of signs be justified? To provide a rigorous explanation, the four possible cases are analyzed separately.

A] Product of two positive numbers \( (+) \cdot (+) = + \)

In this case, the factors of the multiplication are two positive real numbers, \( a>0 \) and \( b>0 \).

The product \( a \cdot b \) can be interpreted as repeated addition of the number \( a \), taken \( b \) times.

$$ a \cdot b = \underbrace{a + a + \dots + a}_{b \ times} $$

Since a positive quantity \( a \) is added a positive number \( b \) of times, the resulting sum is necessarily positive.

This conclusion follows directly from elementary arithmetic:

$$ (+a) \cdot (+b) = c $$

where \( c \) is a positive number. Hence, \( (+) \cdot (+) = + \).

Therefore, the product of two positive quantities corresponds to an operation that increases the magnitude while preserving a positive sign.

For example, if \( a=3 \) and \( b=4 \) then $$ 3 \cdot 4 = 3 + 3 + 3 + 3 $$ $$ 3 \cdot 4 = 12 $$

B] Product of a negative number and a positive number \( (-) \cdot (+) = - \)

Here, the first factor is a negative number \( a<0 \), while the second factor is a positive number \( b>0 \).

As in the previous case, the product \( a \cdot b \) can be viewed as repeated addition of \( a \), taken \( b \) times.

$$ a \cdot b = \underbrace{a + a + \dots + a}_{b \ times} $$

Since a negative quantity \( a \) is added a positive number \( b \) of times, the final result is negative.

For example, if \( a=-2 \) and \( b=5 \) then $$ (-2) \cdot 5 = (-2) + (-2) + (-2) + (-2) + (-2) $$ $$ (-2) \cdot 5 = -10 $$

Alternative explanation

Assume initially that both numbers \( a>0 \) and \( b>0 \) are positive.

By the zero property of multiplication, any real number multiplied by zero yields zero.

$$ a \cdot 0 = 0 $$

Since the sum of a number \( b \) and its additive inverse \( -b \) is zero, zero may be written as \( b + (-b) \).

$$ a \cdot [b + (-b)] = 0 $$

Applying the distributive property of multiplication over addition gives

$$ a \cdot b + a \cdot (-b) = 0 $$

Subtracting \( a \cdot b \) from both sides and simplifying leads to

$$ a \cdot b + a \cdot (-b) \color{red}{- a \cdot b} = 0 \color{red}{- a \cdot b} $$

$$ \require{cancel} \cancel{ a \cdot b } + a \cdot (-b) \cancel{- a \cdot b} = - a \cdot b $$

$$  a \cdot (-b) = - a \cdot b  $$

This identity shows that the product of two numbers with opposite signs, \( a \) and \( -b \), is the additive inverse of the product \( a \cdot b \).

Since \( a \) and \( b \) are both positive by assumption, the product \( a \cdot (-b) \) must be negative.

This establishes that the product of a positive number and a negative number is negative.

C] Product of a positive number and a negative number \( (+) \cdot (-) = - \)

In this case, the first factor is a positive number \( a>0 \), while the second factor is a negative number \( b<0 \).

By the commutative property of multiplication,

$$ a \cdot b = b \cdot a $$

Therefore, as in the previous case, the product \( b \cdot a \) may be interpreted as repeated addition of \( b \), taken \( a \) times.

$$ a \cdot b = \underbrace{b + b + \dots + b}_{a \ times} $$

Since a negative quantity \( b \) is added a positive number \( a \) of times, the result is again negative.

For example, if \( a=5 \) and \( b=-3 \) then $$ 5 \cdot (-3) =  (-3) + (-3) + (-3) + (-3) + (-3) $$ $$ 5 \cdot (-3) = -15 $$

This result may also be obtained using the same alternative explanation given above, which is not repeated here.

D] Product of two negative numbers \( (-) \cdot (-) = + \)

In the final case, both factors are negative, that is \( a < 0 \) and \( b < 0 \).

This situation is less intuitive and therefore requires a different line of reasoning.

Recall that, by the zero property of multiplication, any real number multiplied by zero equals zero, regardless of its sign.

$$ a \cdot 0 = 0 $$

This holds because zero is the absorbing element of multiplication. Moreover, the sum of a number \( b \) and its additive inverse \( -b \) is zero.

$$ b + (-b) = 0 $$

Hence, zero can be replaced by the sum \( b + (-b) \) within the product.

$$ a \cdot [b + (-b)] = 0 $$

Applying the distributive property of multiplication over addition yields

$$ a \cdot b + a \cdot (-b) = 0 $$

Under the initial assumption that both \( a < 0 \) and \( b < 0 \), the quantity \( -b \) is positive.

Consequently, \( a \cdot (-b) \) is the product of two numbers with opposite signs and is therefore negative, by the cases already established.

$$ a \cdot b + \underbrace{a \cdot (-b)}_{<0} = 0 $$

Since the sum is zero and the second term is negative, the first term \( a \cdot b \) must be positive in order for the equality to hold.

$$ \underbrace{a \cdot b}_{>0} + \underbrace{a \cdot (-b)}_{<0} = 0 $$

Because both factors \( a \) and \( b \) are negative by assumption, this shows that the only possible result of their product is a positive number.

For example, consider the number \( -2 \). By the zero property of multiplication, the product of -2 and 0 is zero. $$ (-2) \cdot 0 = 0 $$ Zero can be expressed as the sum of the number \( 3 \) and its additive inverse \( -3 \). $$ (-2) \cdot [3 + (-3)] = 0 $$ Applying the distributive property of multiplication over addition gives $$ (-2) \cdot 3 + (-2) \cdot (-3) = 0 $$ The first term, \( (-2) \cdot 3 \), is the product of two numbers with opposite signs and is therefore negative, since \( -2 \cdot 3 = -6 \). $$ -6 + (-2) \cdot (-3) = 0 $$ In order for the sum to be zero, the product \( (-2) \cdot (-3) \) must necessarily be positive.

And so on.