Algebra 1 · 7.NS.A.1A · 7.NS.A.1C

Identity and inverse elements

Is there a number that changes nothing when you add it or multiply by it, and can the effect of every number be undone?

You need

Opposites and zero pairsMath 6

Where you are 3 in Properties of real numbers

Number9

Irrational and real numbers5

Properties of real numbers3

Closure Commutative, associative and distributive properties

Identity and inverse elements

Why this exists

Every equation you solve ends the same way: the letter stands alone on one side. Getting there means removing whatever has been done to the letter. In n + 15 = 6 something has been added to n; in \displaystyle \frac{2}{5}n = 6 it has been multiplied by something. Which number removes each, and what exactly is left when it has?

The answers are two pairs of ideas. An identity is a number that does nothing: adding 0 leaves a number as it was, and so does multiplying by 1. An inverse is a number that undoes another: adding -15 undoes adding 15, and multiplying by \displaystyle \frac{5}{2} undoes multiplying by \displaystyle \frac{2}{5}. With these named, every step in solving an equation has a reason, and you can see which steps are always allowed and which one never is.

Definition

Identities. For every real number a,

a + 0 = a \qquad\text{and}\qquad 0 + a = a.

Adding zero changes nothing, so 0 is called the additive identity. In the same way, for every real number a,

a \cdot 1 = a \qquad\text{and}\qquad 1 \cdot a = a,

and 1 is called the multiplicative identity.

Inverses. An inverse is defined by aiming at an identity. The additive inverse of a is the number that, added to a, gives the additive identity 0. It is the opposite of a, written -a:

a + (-a) = 0.

The multiplicative inverse of a nonzero number a is the number that, multiplied by a, gives the multiplicative identity 1. It is the reciprocal of a, written \displaystyle \frac{1}{a}:

a \cdot \frac{1}{a} = 1 \qquad (a \neq 0).

The reciprocal of a fraction \displaystyle \frac{p}{q}, with p and q both nonzero, is the fraction turned over, \displaystyle \frac{q}{p}, because \displaystyle \frac{p}{q} \cdot \frac{q}{p} = \frac{pq}{qp} = 1. A negative number has a negative reciprocal, since the product has to be the positive number 1: the reciprocal of -4 is \displaystyle -\frac{1}{4}. A decimal or a mixed number is easiest to turn into a fraction first.

“Inverse” always means “the number that brings you back to the identity”, so the identity has to be known first. And the condition a \neq 0 is not decoration: 0 has no reciprocal at all, as the results below show.

The word identity has a second meaning in algebra. An equation that is true for every value of its variable, such as 3(y - 4) = 3y - 12, is also called an identity, and equations with no solution or infinitely many are about that kind. That meaning names an equation; the meaning here names a number.

Intuition

Think of adding as a jump along the number line. Adding 3 is a jump of three steps to the right, adding -3 is a jump of three steps to the left, and adding 0 is a jump of no length at all, which is why it changes nothing. Figure 1 makes one jump and then its inverse: the second jump lands exactly where the first began. Doing something and then its inverse is the same as doing the identity.

Figure 1Adding -3 undoes adding 3. The two jumps together land where they began, which is what adding 0 does.

Multiplying by a positive number stretches or shrinks distances from 0. Multiplying by 3 makes every number three times as far from 0; multiplying by \displaystyle \frac{1}{3} shrinks it back. Multiplying by 1 is a stretch by a factor of one, so nothing moves. Figure 2 sends 2 to 6 and back.

Figure 2Multiplying by \displaystyle \frac{1}{3} undoes multiplying by 3. The two steps together are what multiplying by 1 does.

The same picture shows why 0 is different. Multiplying by 0 squashes every number onto 0: 5, -2 and 1000 all land on the same point, and nothing can tell them apart again. No multiplication carries 0 back to 1.

Results

Identity properties. For every real number a: a + 0 = 0 + a = a and a \cdot 1 = 1 \cdot a = a.

Inverse properties. For every real number a: a + (-a) = (-a) + a = 0. For every real number a \neq 0: \displaystyle a \cdot \frac{1}{a} = \frac{1}{a} \cdot a = 1.

Zero has no reciprocal. No real number b makes 0 \cdot b = 1, because 0 \cdot b = 0 for every b. Every other real number has a reciprocal.

Only one of each. Every number has exactly one opposite, and every nonzero number exactly one reciprocal. So -(-a) = a: the number a adds to -a to give 0, and -a has no other opposite. In the same way the reciprocal of \displaystyle \frac{1}{a} is a.

Subtraction and division go through inverses.

a - b = a + (-b) \qquad\qquad a \div b = a \cdot \frac{1}{b} \quad (b \neq 0).

For integers the first is subtracting by adding the opposite, and for fractions the second is dividing by multiplying by the reciprocal. They are one idea seen twice: subtracting undoes adding, and dividing undoes multiplying.

Why it is true

For counting numbers the identity properties are what adding and multiplying mean: putting a collection together with an empty one leaves it as it was, and one group of a things is a things. For all real numbers, the identity and inverse properties are taken as starting rules, called axioms, like the commutative, associative and distributive properties. Everything else here is proved from those rules.

Why 0 has no reciprocal. First, 0 \cdot b = 0 for every b. Since 0 + 0 = 0, the distributive property gives

0 \cdot b = (0 + 0) \cdot b = 0 \cdot b + 0 \cdot b.

Add the opposite of 0 \cdot b to both sides. On the left, a number plus its opposite is 0. On the right, regroup (associative property) so that one copy of 0 \cdot b meets its opposite and becomes 0, and adding 0 leaves the other copy unchanged. So 0 = 0 \cdot b. A reciprocal of 0 would be a number b with 0 \cdot b = 1, and that would need 0 = 1. There is no such b.

Only one opposite. Suppose a + b = 0 and a + c = 0. Then

\begin{aligned} b &= b + 0 && \text{identity} \\ &= b + (a + c) && \text{since } a + c = 0 \\ &= (b + a) + c && \text{associative} \\ &= 0 + c && \text{since } b + a = a + b = 0 \\ &= c && \text{identity} \end{aligned}

The step b + a = a + b is the commutative property. The same argument with \cdot in place of + and 1 in place of 0 shows that a nonzero number has only one reciprocal.

Why subtracting is adding the opposite. a - b is the number that, added to b, gives a. The number a + (-b) does exactly that:

\big(a + (-b)\big) + b = a + \big((-b) + b\big) = a + 0 = a,

by the associative, inverse and identity properties in turn. Division works the same way, with multiplying, the reciprocal and 1, which is why it needs b \neq 0.

Worked examples

Example 1

Finding both inverses

Find the additive inverse and the multiplicative inverse of each number: \displaystyle -\frac{11}{7}, 0.35, \displaystyle 2\frac{3}{5} and 0.

Solution

The additive inverse is the opposite, the same distance from 0 on the other side.

The opposite of \displaystyle -\frac{11}{7} is \displaystyle \frac{11}{7}.
The opposite of 0.35 is -0.35.
The opposite of \displaystyle 2\frac{3}{5} is \displaystyle -2\frac{3}{5}.
The opposite of 0 is 0, since 0 + 0 = 0.

The multiplicative inverse is the reciprocal, the number whose product with the original is 1.

Turn \displaystyle -\frac{11}{7} over and keep the sign: \displaystyle -\frac{7}{11}. Check: \displaystyle \left(-\frac{11}{7}\right)\left(-\frac{7}{11}\right) = \frac{77}{77} = 1.
Write 0.35 as a fraction first: \displaystyle 0.35 = \frac{35}{100} = \frac{7}{20}, so its reciprocal is \displaystyle \frac{20}{7}. Check: \displaystyle \frac{7}{20} \cdot \frac{20}{7} = 1.
The reciprocal has to undo the whole number, not just its fraction part, so write \displaystyle 2\frac{3}{5} = \frac{13}{5} first. The reciprocal is \displaystyle \frac{5}{13}. Check: \displaystyle \frac{13}{5} \cdot \frac{5}{13} = 1.
0 has no reciprocal: 0 times any number is 0, never 1.

Example 2

Simplifying with every property named

Simplify each expression and name the property behind each step: (a) 14p + 9 + (-14p); (b) \displaystyle \frac{3}{8}\left(\frac{8}{3}q\right).

Solution

(a) The two terms that cancel are not next to each other, so bring them together first:

\begin{aligned} 14p + 9 + (-14p) &= \big(14p + (-14p)\big) + 9 \\ &= 0 + 9 \\ &= 9 \end{aligned}

The first line reorders and regroups the terms, which is the commutative and associative properties of addition. The second line is the inverse property of addition, and the third the identity property of addition. The p terms cancel whatever p is.

(b) Regroup so that \displaystyle \frac{3}{8} meets its reciprocal:

\frac{3}{8}\left(\frac{8}{3}q\right) = \left(\frac{3}{8} \cdot \frac{8}{3}\right)q = 1 \cdot q = q.

The steps are the associative property of multiplication, the inverse property of multiplication and the identity property of multiplication.

Example 3

Subtracting and dividing by way of inverses

Rewrite each subtraction as an addition and each division as a multiplication, then evaluate: (a) \displaystyle \frac{7}{12} - \frac{5}{6}; (b) \displaystyle -\frac{8}{15} \div \frac{4}{9}; (c) 2.4 \div (-0.15).

Solution

(a) Subtracting \displaystyle \frac{5}{6} is adding its opposite, and \displaystyle \frac{5}{6} = \frac{10}{12}:

\frac{7}{12} - \frac{5}{6} = \frac{7}{12} + \left(-\frac{10}{12}\right) = -\frac{3}{12} = -\frac{1}{4}.

(b) Dividing by \displaystyle \frac{4}{9} is multiplying by its reciprocal, \displaystyle \frac{9}{4}:

-\frac{8}{15} \div \frac{4}{9} = -\frac{8}{15} \cdot \frac{9}{4} = -\frac{72}{60} = -\frac{6}{5}.

(c) As a fraction, \displaystyle -0.15 = -\frac{15}{100} = -\frac{3}{20}, so its reciprocal is \displaystyle -\frac{20}{3}:

2.4 \div (-0.15) = 2.4 \cdot \left(-\frac{20}{3}\right) = -\frac{48}{3} = -16.

Check (c) by multiplying back: -16 \cdot (-0.15) = 2.4.

Subtraction and division are neither commutative nor associative, but addition and multiplication are both. Once a subtraction is written as an addition, its terms can be reordered and regrouped freely, which is what the previous example did.

Example 4

Solving an equation, one property per step

Solve \displaystyle \frac{4}{7}n - 9 = 11, giving the reason for every step.

Solution

The letter has been multiplied by \displaystyle \frac{4}{7} and then had 9 taken away. Undo the last thing first, the subtraction, by adding the opposite; then undo the multiplication with the reciprocal.

Statements	Reasons
\displaystyle \frac{4}{7}n - 9 = 11	given
\displaystyle \frac{4}{7}n + (-9) = 11	subtracting is adding the opposite
\displaystyle \frac{4}{7}n + (-9) + 9 = 11 + 9	adding the same number to both sides keeps them equal
\displaystyle \frac{4}{7}n + 0 = 20	inverse property of addition
\displaystyle \frac{4}{7}n = 20	identity property of addition
\displaystyle \frac{7}{4}\left(\frac{4}{7}n\right) = \frac{7}{4} \cdot 20	multiplying both sides by the same number keeps them equal
\displaystyle \left(\frac{7}{4} \cdot \frac{4}{7}\right)n = 35	associative property of multiplication
1 \cdot n = 35	inverse property of multiplication
n = 35	identity property of multiplication

Check: \displaystyle \frac{4}{7} \cdot 35 - 9 = 20 - 9 = 11.

In practice the lines with +\,0 and 1 \cdot n are done in your head. Written out once, they show the pattern: an inverse brings a number to the identity, and the identity property makes the identity disappear. A reason beside every line is justifying each step.

Example 5

A decimal coefficient with the letter on the right

Solve 3.2 = -1.6t - 4.8.

Solution

The letter sits on the right, which changes nothing: an equation reads the same in either direction.

Undo the subtraction first. -1.6t - 4.8 is -1.6t + (-4.8), so add its opposite, 4.8, to both sides. On the right, -4.8 and 4.8 make 0 (inverse property), and adding 0 changes nothing (identity property):

3.2 + 4.8 = -1.6t \qquad\text{so}\qquad 8 = -1.6t.

Now t is multiplied by \displaystyle -1.6 = -\frac{8}{5}, whose reciprocal is \displaystyle -\frac{5}{8}. Multiply both sides by it:

-\frac{5}{8} \cdot 8 = \left(-\frac{5}{8}\right)\left(-\frac{8}{5}\right)t = 1 \cdot t,

so t = -5.

Check: -1.6 \cdot (-5) - 4.8 = 8 - 4.8 = 3.2. Dividing both sides by -1.6 is the same step under another name.

Example 6

The one step that cannot be undone

The equation 3x + 1 = 13 has the single solution x = 4. What happens to the solutions when both sides are multiplied by \displaystyle -\frac{2}{5}, and what happens when both sides are multiplied by 0?

Solution

Multiplying both sides by \displaystyle -\frac{2}{5} gives

-\frac{2}{5}(3x + 1) = -\frac{26}{5}.

It looks different, but it has exactly the same solutions. Any x that makes the old sides equal makes the new sides equal, and the step can be reversed: multiplying both sides by \displaystyle -\frac{5}{2}, the reciprocal of \displaystyle -\frac{2}{5}, gives 1 \cdot (3x + 1) = 1 \cdot 13, the original equation. So x = 4 is still the only solution.

Multiplying both sides by 0 gives 0 \cdot (3x + 1) = 0 \cdot 13, which is 0 = 0. That is true for every x: x = 4 and x = -100 both satisfy it. The solutions jumped from one number to every real number. The step cannot be reversed, because reversing it would mean multiplying by the reciprocal of 0, and there is none.

This is the reason behind every “do the same to both sides” step. Adding any number to both sides is safe, because every number has an opposite. Multiplying both sides by a number is safe exactly when the number is not 0, because that is exactly when a reciprocal exists.

Boundaries

The reciprocal needs a nonzero number. Every real number has an opposite, but only nonzero numbers have reciprocals. The same applies to letters: multiplying or dividing both sides of an equation by an expression such as x or x - 2 is reversible only where that expression is not 0, and doing it without checking can lose a solution.

An inverse depends on which numbers are allowed. Among the integers, 5 has no reciprocal, because \displaystyle \frac{1}{5} is not an integer; among the rational numbers it does. Whether an answer stays inside a set of numbers is the question of closure.

Subtraction and division have no identity of their own. It is true that a - 0 = a, but 0 - a = -a, which is not a unless a = 0. Likewise a \div 1 = a but 1 \div a is usually not a. An identity has to work on both sides.

Where this goes next

Every method for solving equations runs on these properties: one-step equations use a single inverse, two-step equations use two, and multi-step equations tidy each side first and then do the same.

Asking what a^{-1} should mean so that the rules of exponents keep working gives \displaystyle a^{-1} = \frac{1}{a}, the reciprocal again, and zero and negative exponents works this out. The identity function f(x) = x is the function that changes nothing, giving back every input as it was. And inverse functions undo each other the way opposites and reciprocals do.

Exercises

Finding inverses

Find the additive inverse and the multiplicative inverse of each number.

12
\displaystyle -\frac{7}{9}
1.25
\displaystyle 2\frac{2}{7}
Which number is its own additive inverse? Which numbers are their own multiplicative inverse?

Naming and using the properties

Name the property each statement shows: (a) 0 + (-2.7) = -2.7; (b) \displaystyle \frac{11}{4} \cdot \frac{4}{11} = 1; (c) -19 \cdot 1 = -19; (d) 3k + (-3k) = 0.
Fill each blank: (a) \displaystyle \frac{5}{14} \cdot \square = 1; (b) -4.5 + \square = 0; (c) \square \cdot y = y for every y; (d) 9w + \square = 9w for every w.

Simplifying

Simplify, naming the property behind each step.

26r + 31 + (-26r)
\displaystyle \left(-\frac{5}{12}\right) z \left(-\frac{12}{5}\right)
0.125(8h)
\displaystyle \frac{1}{9}(9n) + (-n)

Subtracting and dividing through inverses

Rewrite each division as multiplying by a reciprocal, or each subtraction as adding an opposite, then evaluate.

\displaystyle -\frac{14}{15} \div \frac{7}{10}
5.4 \div (-0.09)
\displaystyle -\frac{3}{8} - \left(-\frac{11}{12}\right)

Solving equations

Solve, naming the inverse you use at each step.

\displaystyle \frac{2}{9}x + 7 = -1
-1.5 = 1.2q + 2.7
Solve \displaystyle -\frac{3}{4}y = 12. Then find what multiplying both sides by \displaystyle \frac{4}{3} gives instead, and explain why that step does not leave y alone.

Reasoning

Explain why 0 has no reciprocal. Then explain why 0 \div 7 has a value but 7 \div 0 does not.
Among the whole numbers 0, 1, 2, 3, \ldots, which numbers have an additive inverse that is also a whole number? Among the integers, which numbers have a multiplicative inverse that is also an integer? Explain.
Dividing both sides of x^2 = 11x by x leaves x = 11. Is that every solution of x^2 = 11x? Explain using reciprocals.

Answers

Additive inverse -12; multiplicative inverse \displaystyle \frac{1}{12}, since \displaystyle 12 \cdot \frac{1}{12} = 1.
Additive inverse \displaystyle \frac{7}{9}; multiplicative inverse \displaystyle -\frac{9}{7}, since \displaystyle \left(-\frac{7}{9}\right)\left(-\frac{9}{7}\right) = 1.
Additive inverse -1.25. As a fraction \displaystyle 1.25 = \frac{5}{4}, so the multiplicative inverse is \displaystyle \frac{4}{5} = 0.8.
Additive inverse \displaystyle -2\frac{2}{7}. As a fraction \displaystyle 2\frac{2}{7} = \frac{16}{7}, so the multiplicative inverse is \displaystyle \frac{7}{16}.
0 is its own opposite: a + a = 0 means 2a = 0, so a = 0. The numbers 1 and -1 are their own reciprocals: a \cdot a = 1 means a^2 = 1, so a = 1 or a = -1.
(a) Identity property of addition. (b) Inverse property of multiplication. (c) Identity property of multiplication. (d) Inverse property of addition.
(a) \displaystyle \frac{14}{5}, the reciprocal. (b) 4.5, the opposite. (c) 1, the multiplicative identity. (d) 0, the additive identity.
31. Reorder and regroup (commutative and associative properties of addition) to get \big(26r + (-26r)\big) + 31, then 0 + 31 (inverse property), then 31 (identity property).
z. Reorder and regroup (commutative and associative properties of multiplication) to get \displaystyle \left(-\frac{5}{12}\right)\left(-\frac{12}{5}\right)z = 1 \cdot z (inverse property), which is z (identity property).
h. Regroup to (0.125 \cdot 8)h (associative property). Since \displaystyle 0.125 = \frac{1}{8}, this is 1 \cdot h (inverse property), which is h (identity property).
0. First \displaystyle \frac{1}{9}(9n) = \left(\frac{1}{9} \cdot 9\right)n = 1 \cdot n = n (associative, inverse and identity properties of multiplication). Then n + (-n) = 0 (inverse property of addition).
\displaystyle -\frac{4}{3}. The reciprocal of \displaystyle \frac{7}{10} is \displaystyle \frac{10}{7}, and \displaystyle -\frac{14}{15} \cdot \frac{10}{7} = -\frac{140}{105} = -\frac{4}{3}.
-60. As a fraction \displaystyle -0.09 = -\frac{9}{100}, with reciprocal \displaystyle -\frac{100}{9}, and \displaystyle 5.4 \cdot \left(-\frac{100}{9}\right) = -\frac{540}{9} = -60. Check: -60 \cdot (-0.09) = 5.4.
\displaystyle \frac{13}{24}. Subtracting \displaystyle -\frac{11}{12} is adding its opposite, \displaystyle \frac{11}{12}: \displaystyle -\frac{3}{8} + \frac{11}{12} = -\frac{9}{24} + \frac{22}{24} = \frac{13}{24}.
x = -36. Add -7, the opposite of 7, to both sides: \displaystyle \frac{2}{9}x = -8. Multiply both sides by \displaystyle \frac{9}{2}, the reciprocal of \displaystyle \frac{2}{9}: \displaystyle x = -8 \cdot \frac{9}{2} = -36. Check: \displaystyle \frac{2}{9}(-36) + 7 = -8 + 7 = -1.
q = -3.5. Add -2.7 to both sides: -4.2 = 1.2q. As a fraction \displaystyle 1.2 = \frac{6}{5}, so multiply both sides by \displaystyle \frac{5}{6}: \displaystyle q = -4.2 \cdot \frac{5}{6} = -3.5. Check: 1.2(-3.5) + 2.7 = -4.2 + 2.7 = -1.5.
y = -16. The reciprocal of \displaystyle -\frac{3}{4} is \displaystyle -\frac{4}{3}, and \displaystyle y = 12 \cdot \left(-\frac{4}{3}\right) = -16. Check: \displaystyle -\frac{3}{4}(-16) = 12. Multiplying by \displaystyle \frac{4}{3} instead gives -y = 16, because \displaystyle \frac{4}{3} \cdot \left(-\frac{3}{4}\right) = -1, not 1: it is not the reciprocal, so it does not bring the coefficient to the identity.
A reciprocal of 0 would be a number b with 0 \cdot b = 1, but 0 \cdot b = 0 for every b, so there is none. Then \displaystyle 0 \div 7 = 0 \cdot \frac{1}{7} = 0, which is fine because 7 has a reciprocal. But 7 \div 0 would be 7 times the reciprocal of 0, which does not exist. Put another way, 7 \div 0 = c would mean c \cdot 0 = 7, and no number times 0 is 7.
Among the whole numbers only 0 has an additive inverse in the set, since the opposite of any other whole number is negative. Among the integers only 1 and -1 have an integer reciprocal: 1 \cdot 1 = 1 and (-1)(-1) = 1, while for any other nonzero integer a the reciprocal \displaystyle \frac{1}{a} lies strictly between -1 and 1 and is not 0, so it is not an integer.
No. The solutions are x = 0 and x = 11. Dividing by x is multiplying by \displaystyle \frac{1}{x}, which exists only when x \neq 0, so that step assumes x \neq 0 and loses that case. Directly, 0^2 = 0 = 11 \cdot 0, so x = 0 is a solution too.