From Arithmetic to Algebra

Learning goals

Read algebra as arithmetic's move from particular numbers to general claims
State the familiar laws as claims about every number
Recognize an identity as an equation true for all values
Test equivalence by substitution, and say what testing cannot prove
Disprove a general claim with a single counterexample

Arithmetic asks “what is”; algebra asks “what is always”

In arithmetic you work with numbers you can see, and every question has a numerical answer. The whole subject is about computing: combine these particular numbers and report the result. Algebra keeps all of that arithmetic but adds a new move. Instead of asking what happens to $3$ and $5$ , it asks what happens to any two numbers at once.

The tool that makes this possible is the variable: a letter that stands for a number whose name we do not know yet. Sometimes the name is unknown because we are trying to find it, as when we solve $x + 3 = 7$ for the one value that fits. Other times the name is unknown on purpose, because we want a single statement to speak about every number at the same time.

Watch how the second use replaces an endless list. You could check that

2 + 3 = 3 + 2, \qquad 6 + 1 = 1 + 6, \qquad 40 + 9 = 9 + 40,

and keep going forever without ever finishing. Or you can write the whole pattern once:

a + b = b + a.

Here $a$ and $b$ do not hide two particular numbers. They stand for all numbers simultaneously, so the one line makes a claim about every pair at once. That single shift, from computing with fixed numbers to reasoning about all numbers together, is the entire step from arithmetic to algebra.

The laws you already know, now for every number

Throughout arithmetic you leaned on a small set of rules, usually without stating them. In algebra we write those rules with letters, which turns each one into a statement about every number. These are the laws that all of algebra is built on.

Law	With addition	With multiplication
Commutative	$a + b = b + a$	$ab = ba$
Associative	$(a + b) + c = a + (b + c)$	$(ab)c = a(bc)$
Identity	$a + 0 = a$	$a \cdot 1 = a$
Inverse	$a + (-a) = 0$	$a \cdot \dfrac{1}{a} = 1$ (for $a \neq 0$ )

One more law links addition and multiplication, the distributive law:

a(b + c) = ab + ac.

Read each row as a promise that holds no matter what numbers the letters stand for. The commutative law promises that order never changes a sum or a product. The associative law promises that regrouping never changes them either. The identity law names the two special numbers that leave things alone, $0$ for addition and $1$ for multiplication. The inverse law names the partner that undoes a number, $-a$ for adding and $\tfrac{1}{a}$ for multiplying. These are not new facts to memorize; they are the arithmetic you already trust, written so that a letter can stand in for any number.

It is worth pausing to see that these laws are earned, not simply declared. Here is why one of them, the commutative law of addition, must hold for every pair of numbers.

Why $a + b = b + a$ for every pair of numbers#

Think first about counting. Adding two whole numbers means combining two collections and counting the total. Put a pile of $a$ stones together with a pile of $b$ stones. Whether you scoop the first pile into the second or the second into the first, you end up looking at the very same heap of stones. Counting that one heap gives one answer. The total cannot depend on which pile you named first, so $a + b$ and $b + a$ are the same count.

The same conclusion holds for numbers that are not counts of stones, using the number line. Read $a + b$ as “start at $a$ and move $b$ units,” and $b + a$ as “start at $b$ and move $a$ units.” Both journeys are built from the same two moves, a shift of $a$ and a shift of $b$ . Two shifts combine into a single shift equal to their total, and the order in which you take them does not change where you land.

Because neither the counting argument nor the number-line argument mentions any particular value, the equality is forced for every choice of $a$ and $b$ . That is exactly what the single line $a + b = b + a$ records.

Identities: equations that are always true

An equation is two expressions joined by an equals sign, and once a variable appears, an equation can be true in three very different ways. Compare these three:

x + 3 = 7, \qquad a + 0 = a, \qquad x + 1 = x.

The first, $x + 3 = 7$ , is true for exactly one value, $x = 4$ , and false for every other. Finding that value is what the coming lessons call solving the equation. The second, $a + 0 = a$ , is true for every value of $a$ without exception. The third, $x + 1 = x$ , is true for no value at all, since no number equals one more than itself.

An equation that is true for every value of its variables is called an identity. All the laws in the table above are identities, and so is the distributive law: each holds for every number you could substitute. An identity makes no demand on the variable; it simply reports a fact that never fails. A plain equation like $x + 3 = 7$ is a different animal, a question that singles out particular values, and telling the two apart is the first habit of algebra.

Check your understanding

Which of these equations is an identity (true for every value of the variable)?

x + 5 = 12

a \cdot 1 = a

2n = 8

x - 4 = 0

Equivalent expressions

Two expressions are equivalent when they produce the same value for every input. Equivalent expressions are exactly the two sides of an identity. If $A$ and $B$ agree no matter what the variables are, then the equation $A = B$ is an identity, and the other way around.

The distributive law is a factory for equivalent expressions. Take $2(x + 3)$ and $2x + 6$ . They look different, one a product and one a sum, yet they always agree, and the reason is worth seeing in full.

Why $2(x + 3)$ and $2x + 6$ are equivalent#

The expression $2(x + 3)$ means two copies of the quantity $x + 3$ added together:

2(x + 3) = (x + 3) + (x + 3).

Now regroup, which the associative and commutative laws permit, collecting the two $x$ ‘s together and the two $3$ ‘s together:

(x + 3) + (x + 3) = (x + x) + (3 + 3) = 2x + 6.

Every step used a law that holds for every number, and no step depended on what $x$ actually is. So $2(x + 3)$ and $2x + 6$ deliver the same value for every $x$ , which is precisely what it means for them to be equivalent.

The same picture appears in a rectangle, where the distributive law becomes a statement about area.

A rectangle of height a and width b + c has area a(b + c). A vertical cut at width b splits that area into ab and ac, so a(b + c) and ab + ac measure the same region. That is why they are equivalent for every a, b, and c.

Not every pair that looks close is equivalent. The expressions $2(x + 3)$ and $2x + 3$ differ only by where the parentheses reach, but they are not the same. Finding one value where the two expressions part ways is all it takes to prove they are not equivalent.

Worked example 1 Are these expressions equivalent?

Start with $4(n + 1)$ and $4n + 4$ . The distributive law multiplies the $4$ across both parts of the sum:

4(n + 1) = 4n + 4.

Because the distributive law holds for every number, this equality holds for every $n$ , so the two expressions are equivalent.

Now compare $4(n + 1)$ with $4n + 1$ . Here the $4$ has reached only the $n$ , not the $1$ . Test a single value, say $n = 2$ :

4(2 + 1) = 12, \qquad 4(2) + 1 = 9.

The two disagree at $n = 2$ , so $4(n + 1)$ and $4n + 1$ are not equivalent. One value where they differ is enough to settle it.

Substitution and the power of a single counterexample

To substitute is to replace a variable with a chosen number and then do the arithmetic. Writing the number in parentheses where the letter used to be keeps the operations and signs intact, so $5x$ at $x = 7$ becomes $5(7) = 35$ rather than the meaningless ” $57$ .” Substitution is how you test whether a claimed identity is actually true.

The logic of that testing is not symmetric, and the asymmetry is one of the most useful ideas in all of mathematics. To disprove a claim that an equation holds for every number, you need only one value where the two sides disagree. To prove that it holds for every number, no amount of testing is ever enough.

Why one counterexample settles a claim but examples never prove it#

Suppose someone claims an equation is an identity, true for every number. The claim is a statement about all numbers at once. A single value where the left side and the right side come out different directly contradicts the word “every,” so that one counterexample ends the matter. The claim is false, and no further checking is needed.

The other direction fails, and it fails for a simple reason. There are infinitely many numbers, so any list of values you actually test, however long, leaves infinitely many untested. Agreement on your list could be a coincidence that breaks at the very next number. Testing can therefore raise your confidence, but it can never rule out a hidden failure.

To be certain an equation holds for all numbers, you must derive it from the laws. A derivation gives that certainty because those laws are the statements already known to hold for all numbers. This is why algebra proves identities by manipulation, as in the earlier proof that $2(x + 3) = 2x + 6$ , rather than by piling up examples.

The next example shows exactly how a couple of lucky agreements can mislead you.

Worked example 2 Is $n + n = n \cdot n$ an identity?

A claim that two expressions are always equal is a claim about every number, so try a few values and stay alert for one that breaks it.

At $n = 2$ the two sides agree:

2 + 2 = 4, \qquad 2 \cdot 2 = 4.

At $n = 0$ they agree again, since $0 + 0 = 0$ and $0 \cdot 0 = 0$ . It is tempting to declare the equation an identity, but two successes prove nothing. Try $n = 3$ :

3 + 3 = 6, \qquad 3 \cdot 3 = 9.

Now the sides differ, so $n + n = n \cdot n$ is false in general. In fact it is true only at $n = 0$ and $n = 2$ and false at every other number, which is precisely why checking a value or two can fool you.

Check your understanding

A student claims that $x + x = x \cdot x$ for every number $x$ . Which value of $x$ shows the claim is false?

x = 0

x = 2

x = 3

D the claim is true

The order of operations does not change

Bringing in letters changes nothing about the order of operations. Parentheses still come first, then multiplication and division, then addition and subtraction, exactly as in arithmetic. So $2 + 3x$ means $2 + (3x)$ , with the multiplication done before the addition, and it is a different expression from $(2 + 3)x$ . When you substitute a number, the order of operations decides the result.

Worked example 3 Evaluate $7 - 2x$ when $x = 3$

Substitute $3$ for $x$ , in parentheses, then follow the order of operations:

7 - 2x = 7 - 2(3).

Multiplication comes before subtraction, so multiply first:

7 - 2(3) = 7 - 6 = 1.

The value is $1$ . It is not $(7 - 2)(3) = 15$ ; the subtraction does not happen first just because it is written first. The same order of operations you used in arithmetic still governs once a letter is present.

Check your understanding

Evaluate $3 + 4x$ when $x = 2$ .

11

14

8

20

Turning words into algebra

Most problems arrive as words, and the first real skill of algebra is turning those words into symbols. Name the unknown with a letter, then convert each phrase into an operation: “more than” adds, “less than” subtracts, “times” or “product” multiplies, “divided by” or “per” divides. Two kinds of results come out of this, and they are not the same.

A phrase with no verb becomes an expression, which names a value but claims nothing. A full statement with a verb like is or equals becomes an equation, because that verb plays the role of the equals sign. The equals sign is what separates a thing you evaluate from a sentence you can later solve.

Worked example 4 Translate two statements into algebra

Name the unknown number $n$ in each, then read carefully for a verb.

“Twice a number, decreased by five” is a phrase. Twice the number is $2n$ , and decreased by five subtracts $5$ :

2n - 5.

It names a value and asserts nothing, so it is an expression.

“The sum of a number and eight is twenty” is a full statement, and the word is acts as the equals sign. The sum of the number and eight is $n + 8$ , set equal to $20$ :

n + 8 = 20.

That equals sign makes it an equation, a sentence you could later solve to find $n$ .

Common mistakes

Confusing an identity with an equation to solve. $a + 0 = a$ is an identity, true for every $a$ , while $x + 3 = 7$ is a conditional equation, a puzzle with the single answer $x = 4$ . Decide which kind you are looking at before you do anything else.
Thinking a few checks prove an identity. Passing at two or three values proves nothing, because a claim can hold sometimes and fail elsewhere ( $n + n = n \cdot n$ holds at $n = 0$ and $n = 2$ only). One counterexample, on the other hand, disproves a claim for good.
Distributing to only part of a sum. $2(x + 3)$ is $2x + 6$ , not $2x + 3$ ; the factor must reach every term inside the parentheses.
Assuming every operation is commutative. Addition and multiplication can be reordered, but subtraction and division cannot: $5 - 3 \neq 3 - 5$ and $8 \div 2 \neq 2 \div 8$ .
Dropping the order of operations once letters appear. $2 + 3x$ means $2 + (3x)$ , not $(2 + 3)x$ . Multiplication still comes before addition.
Reading $5x$ as two digits. A number next to a variable means multiply, so at $x = 2$ the value of $5x$ is $5(2) = 10$ , not $52$ .

Key takeaways

Arithmetic computes with particular numbers; algebra reasons about all numbers at once by letting a variable stand for a number whose name we do not know yet.
The commutative, associative, identity, inverse, and distributive laws, written with letters, are identities: equations true for every value of their variables.
Two expressions are equivalent when they give the same value for every input. Equivalent expressions are the two sides of an identity, guaranteed by the laws (for example $2(x + 3) = 2x + 6$ ).
Substitution replaces a variable with a number. One counterexample disproves a claimed identity, but no finite list of examples can prove one; that requires deriving it from the laws.
The order of operations is unchanged by variables: $2 + 3x$ means $2 + (3x)$ .
A word phrase becomes an expression (no equals sign, you evaluate it); a full statement becomes an equation (has an equals sign, you solve it).

From Arithmetic to Algebra

Learning goals

Arithmetic asks “what is”; algebra asks “what is always”

The laws you already know, now for every number

Identities: equations that are always true

Equivalent expressions

Substitution and the power of a single counterexample

The order of operations does not change

Turning words into algebra

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Free Response Questions (FRQ)