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 no matter which two numbers you pick.

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 hold true no matter which number replaces the letter.

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 stand for two particular numbers. Substitute any pair you like into $a + b = b + a$ , and it comes out true; that is what makes the line a claim about every pair, not a report about one. That shift, from computing with fixed numbers to reasoning about all numbers together, is one of the two moves that take you from arithmetic into algebra. Naming an unknown you mean to find, as with $x + 3 = 7$ , is the other.

The laws every number obeys

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 idea carries over to numbers that are not counts of stones, including negative ones, using the number line, though negatives need one more idea: a rightward move and a leftward move partly undo each other. When $a$ and $b$ point the same way, both positive or both negative, there is nothing to undo: moving $a$ then $b$ , or $b$ then $a$ , just lays two same-direction rods end to end, and exactly as with the stone piles the order does not change how far the combined rod reaches. When either one is $0$ , that move covers no ground at all, so it trivially does not matter where it sits in the order.

When one is positive and the other is negative, call the positive one’s size $p$ and the negative one’s size $n$ , so the negative number is $-n$ . The two moves overlap over whichever is shorter, a stretch of length $\min(p, n)$ , and that overlap always cancels, however the moves are ordered.

If $p \ge n$ : moving right $p$ then left $n$ retraces the last $n$ units of that rightward trip, canceling them, and leaves $p - n$ to the right of the start. Moving left $n$ first then right $p$ : the first $n$ units of that rightward trip retrace the same ground back to the start, and the remaining $p - n$ continues past it, landing at the identical $p - n$ to the right.

If $n > p$ : moving right $p$ then left $n$ : the first $p$ units of that leftward trip retrace the rightward trip back to the start, and the remaining $n - p$ continues past it, landing at $n - p$ to the left. Moving left $n$ first then right $p$ retraces the last $p$ units of that leftward trip, leaving the identical $n - p$ to the left.

Either way, swapping the order only changes which trip gets retraced first; the two trips still cancel over their shared length and leave the same leftover in the same direction, so $a + b$ and $b + a$ mark the same point. (Check it: with $p = 3$ and $n = 5$ , so $n > p$ , right $3$ then left $5$ lands at $3 - 5 = -2$ , and left $5$ then right $3$ lands at $-5 + 3 = -2$ , the same point, $n - p = 2$ to the left.)

Nothing in either argument depended on which numbers $a$ and $b$ actually were: the counting argument works the same way whatever the two piles hold, and the number-line argument works the same way whatever the two moves are. So the equality holds for every choice of $a$ and $b$ , not just the ones checked above. 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 for no values, for one value, for several values, or for every value. 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 can also land in between, true for some values and false for others, though none of the three examples above happen to do that.)

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 that holds no matter which number you substitute. 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 nothing more needs checking.

The other direction 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. Two agreements were never a guarantee that it holds everywhere, which is exactly why checking a value or two can fool you, and why the one disagreement at $n = 3$ is what actually settles the question.

Check your understanding

A student claims that $n^2 = 4n$ for every number $n$ . Which value of $n$ shows the claim is false?

n = 0

n = 4

n = 2

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 turning those words into symbols is a skill worth having early. Name the unknown with a letter, then convert each phrase into an operation: “more than,” “the sum of,” or “increased by” adds; “times” or “the product of” multiplies; “divided by” or “per” divides. “Less than” also subtracts, but it reverses the order you read: ” $5$ less than $n$ ” means $n - 5$ , not $5 - n$ , because $n$ is the amount you start from and $5$ is what comes off it. Two kinds of results come out of this, and they are not the same.

A word phrase turns into an expression: it names a value, and makes no claim about it. A word statement that says two amounts are equal, using a word like is or equals for the equals sign, turns into an equation: a claim you can test. An expression is something you evaluate; an equation is a claim that can turn out true for no values, one value, several values, or every value, as the three equations near the start of this lesson began to show.

Worked example 4 Translate two statements into algebra

Name the unknown number $n$ in each, then read carefully for a claim that two amounts are equal.

“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 true for exactly one value, $x = 4$ , and false for every other. 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 at some values and fail at others ( $n + n = n \cdot n$ agrees at $n = 0$ and $n = 2$ but fails at $n = 3$ ). 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.
Reading a “less than” phrase in word order. ” $5$ less than $n$ ” means $n - 5$ , not $5 - n$ ; find the amount you start from before subtracting.
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 every number, no matter which one, 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 statement that claims two amounts are equal becomes an equation (has an equals sign, and may be solved, or may hold for every value as an identity).

From Arithmetic to Algebra

Learning goals

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

The laws every number obeys

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)

Core practice