Defining New Operations

Intuition

Over the last three lessons you built one big idea in stages: a function is a rule that turns inputs into an output. That function deserves its own name like $f$ , and once it has a name you can evaluate it, chain it, and reverse it. This closing lesson turns that idea loose on a puzzle that looks harder than it is. Contest problems and textbooks love to spring a brand-new symbol on you, something like $3 \star 5$ or $a \diamond b$ , defined by a rule such as $a \star b = 2a + b - 1$ . The symbol is invented, so it feels like a new kind of arithmetic nobody taught you. It is not. A defined operation is just a function of two inputs wearing an unfamiliar costume, and the only genuinely new thing is the symbol. Read the definition, substitute the inputs, and simplify with the order of operations you already trust. A rule is a rule, whatever symbol names it.

Learning goals

Evaluate a new symbol by substituting into its defining rule
Drop a whole expression into a slot as one block
Test whether the operation is commutative, since most are not
Work nested operations from the innermost outward
Look for an identity by solving both sides of the condition
Turn an equation with the symbol into an ordinary one

A made-up symbol is only a rule

A binary operation takes two inputs and produces one output, exactly the way $+$ takes $3$ and $4$ and hands back $7$ . The operations you grew up with, $+$ , $-$ , $\times$ , and $\div$ , are the familiar four. But nothing stops someone from inventing another one and naming it with a fresh symbol. When a problem writes

a \star b = 2a + b - 1,

it is doing just that. The star is a made-up name, and the right-hand side is the rule that says what to do with the two inputs $a$ and $b$ . Think of $a$ and $b$ as two empty slots. To compute a particular value like $3 \star 5$ , drop the first input into every $a$ and the second input into every $b$ , then simplify:

3 \star 5 = 2(3) + 5 - 1 = 6 + 5 - 1 = 10.

That is the whole method, and it never changes: substitute the inputs into the defining rule, then evaluate with the ordinary order of operations. The order of operations matters as much here as anywhere. In $2a + b - 1$ the multiplication $2a$ happens before the additions, so $2(3)$ becomes $6$ before anything is added. Skip that and you would get a different, wrong answer. The star is new; the arithmetic inside it is not.

A defined operation is a two-input machine. Feed the first input into every a and the second into every b, run the rule, and read off the single output. Computing 3 star 5 just means substituting a = 3 and b = 5 into 2a + b - 1.

Worked example 1 Evaluating a custom operation, including negatives

Let $a \star b = 2a + b - 1$ . Compute $4 \star 7$ , then $(-2) \star 5$ , then $6 \star (-3)$ .

For $4 \star 7$ , substitute $a = 4$ and $b = 7$ , and do the multiplication first:

4 \star 7 = 2(4) + 7 - 1 = 8 + 7 - 1 = 14.

For $(-2) \star 5$ , the first input is negative, so $2a$ becomes $2(-2) = -4$ . Keep track of the sign:

(-2) \star 5 = 2(-2) + 5 - 1 = -4 + 5 - 1 = 0.

For $6 \star (-3)$ , the second input is negative, and it lands in the $+b$ slot, so you add a negative:

6 \star (-3) = 2(6) + (-3) - 1 = 12 - 3 - 1 = 8.

Each answer came from the same two moves: put the inputs in their slots, then simplify. The negatives needed care only because signs always need care, not because the star does anything unusual.

Check your understanding

A new operation is defined by $a \triangle b = 3a - b$ . What is $4 \triangle 5$ ?

7

17

2

-7

When an input is an expression

Nothing says the inputs have to be numbers. A slot can hold a whole expression, and the rule still works the same way: whatever sits in the slot gets substituted in one piece. This is the exact substitution you used to build composite functions last lesson, where the entire rule for $g(x)$ dropped into $f$ . Watch the order of operations especially closely when an input is squared or wrapped in parentheses.

Worked example 2 Substituting an expression into a slot

Let $a \diamond b = a^2 - b$ . Compute $5 \diamond 2$ and $(-4) \diamond 3$ , then simplify $(k + 1) \diamond 4$ .

For $5 \diamond 2$ , square the first input before subtracting the second:

5 \diamond 2 = 5^2 - 2 = 25 - 2 = 23.

For $(-4) \diamond 3$ , the first input is negative, and squaring a negative gives a positive:

(-4) \diamond 3 = (-4)^2 - 3 = 16 - 3 = 13.

Now let the first input be the expression $k + 1$ . It goes into the $a^2$ slot as a single block, so the whole thing gets squared:

(k + 1) \diamond 4 = (k + 1)^2 - 4 = (k^2 + 2k + 1) - 4 = k^2 + 2k - 3.

The square applies to all of $k + 1$ , not just the $k$ , which is why the middle term $2k$ appears. Treat the slot as a container: the entire input drops in, then you simplify.

The order of the two inputs usually matters

Ordinary addition does not care about order, since $3 + 5$ and $5 + 3$ are both $8$ . It is tempting to assume every operation is that polite, but most invented ones are not. Test the star on the two orders of $3$ and $5$ :

3 \star 5 = 2(3) + 5 - 1 = 10, \qquad 5 \star 3 = 2(5) + 3 - 1 = 12.

Same two numbers, but $10$ and $12$ are different, so $3 \star 5 \ne 5 \star 3$ . The reason is built into the rule. The definition $2a + b - 1$ doubles the first input but not the second, so the two slots are treated differently. Swap the inputs and the one that gets doubled changes, which changes the answer. An operation where the order of the inputs can change the result is called not commutative.

You met exactly this warning back when you studied composition earlier in this chapter. Composition was not commutative either, because the outer function acted last and decided the kind of answer. A custom operation carries the same caution for a related reason: whenever the rule handles its two slots differently, reversing the inputs reshuffles the work.

The same rule with its two inputs swapped. Because 2a + b - 1 doubles the first input only, sending 3 and 5 through in one order gives 10, and in the other order gives 12. The two results differ, so the order of the inputs matters.

Because most operations are not commutative, commutativity is a property to check, never to assume. Some invented operations really are commutative, and the only way to be sure is to test the two orders or to argue from the rule itself. Here is one that passes.

Worked example 3 One operation that is not commutative, and one that is

Compare the two orders for two different rules.

First take the star, $a \star b = 2a + b - 1$ . Its two orders on the inputs $3$ and $5$ gave $10$ and $12$ , so it is not commutative. One counterexample is enough to settle that.

Now take a different rule, $a \oplus b = a + b + ab$ , and test it on $2$ and $7$ :

2 \oplus 7 = 2 + 7 + (2)(7) = 2 + 7 + 14 = 23,

7 \oplus 2 = 7 + 2 + (7)(2) = 7 + 2 + 14 = 23.

This time the two orders agree. A single matching pair is only a hint, not a proof. But the match is no accident, and the next result shows why $\oplus$ is commutative for every choice of inputs.

The operation $a \oplus b = a + b + ab$ is commutative#

To show $\oplus$ is commutative, we must show $a \oplus b = b \oplus a$ for every pair of numbers $a$ and $b$ , not just for one lucky example. Start from the definition applied in each order:

a \oplus b = a + b + ab, \qquad b \oplus a = b + a + ba.

Now compare the two right-hand sides using facts you already know about ordinary numbers. Addition of numbers is commutative, so $b + a$ is the same as $a + b$ . Multiplication of numbers is commutative too, so $ba$ is the same as $ab$ . Replacing each piece of $b \oplus a$ with its equal turns it into

b \oplus a = a + b + ab,

which is exactly the expression for $a \oplus b$ . Since the two are identical for every $a$ and $b$ , the operation $\oplus$ is commutative. Notice what did the work: the new operation inherits its symmetry straight from the symmetry of the $+$ and $\times$ hiding inside its rule. A rule built symmetrically in its two inputs stays the same when they trade places.

Check your understanding

For the operation $a \diamond b = a^2 - b$ , which statement is true?

3 \diamond 1 = 1 \diamond 3

3 \diamond 1 \ne 1 \diamond 3

\diamond

is commutative D the order of the inputs never matters

Nested operations: work from the inside out

Once you can combine two inputs, you can combine three by applying the operation twice. The catch is that the two inputs of the second step must themselves be single values, so a three-input expression needs parentheses to say which pair goes first. Face a grouped expression the same way you face nested parentheses in arithmetic or an inside-out composition. Evaluate the innermost parentheses first, turn that result into a single number, then apply the operation again.

Watch what the grouping does to the answer.

Worked example 4 Two groupings, two different answers

Let $a \star b = 2a + b - 1$ . Compute $(1 \star 2) \star 3$ and $1 \star (2 \star 3)$ .

For $(1 \star 2) \star 3$ , the parentheses on the left go first. Evaluate the inner star, then use its result as the first input of the outer star:

1 \star 2 = 2(1) + 2 - 1 = 3, \qquad 3 \star 3 = 2(3) + 3 - 1 = 8.

So $(1 \star 2) \star 3 = 8$ . Now regroup to the right. This time the inner star on $2$ and $3$ goes first, and its result becomes the second input of the outer star:

2 \star 3 = 2(2) + 3 - 1 = 6, \qquad 1 \star 6 = 2(1) + 6 - 1 = 7.

So $1 \star (2 \star 3) = 7$ . The two groupings give $8$ and $7$ , which are not equal.

The two answers disagreeing tells you something important. An operation is called associative when the grouping never matters, the way $(2 + 3) + 4$ and $2 + (3 + 4)$ are both $9$ . The star fails that test, since $(1 \star 2) \star 3 = 8$ but $1 \star (2 \star 3) = 7$ , so the star is not associative. Just like commutativity, associativity is a property to check rather than assume, and when it fails the parentheses are not optional. Without them, " $1 \star 2 \star 3$ " would be ambiguous, because the two readings genuinely produce different values.

Looking for an identity element

Ordinary addition has a special number, $0$ , that changes nothing: $a + 0 = a$ for every $a$ . Ordinary multiplication has $1$ , since $a \times 1 = a$ . A value that leaves every input untouched from both sides is called an identity element for the operation. Untouched from both sides means the value gives back $a$ whether it is combined on the left or the right of $a$ . Requiring both sides matters here, because these operations need not be commutative, so leaving an input unchanged from one side alone is not enough. It is natural to ask whether an invented operation has such a value, and the question is answered by solving an equation.

Worked example 5 Finding an identity element

Does $a \oplus b = a + b + ab$ have an identity element? If so, find it.

An identity is a fixed value $e$ that leaves every input unchanged from both sides, meaning $a \oplus e = a$ and $e \oplus a = a$ no matter what $a$ is. Start with the first condition, writing it out with the rule and solving for $e$ :

a \oplus e = a + e + ae = a.

Subtract $a$ from both sides to isolate the terms with $e$ :

e + ae = 0 \quad\Longrightarrow\quad e(1 + a) = 0.

For this to hold for every $a$ , the factor $e$ itself must be $0$ , since $1 + a$ is not always zero. So the only candidate is $e = 0$ . Now check that it works from both sides:

a \oplus 0 = a + 0 + a\cdot 0 = a, \qquad 0 \oplus a = 0 + a + 0\cdot a = a.

Both orders return $a$ , which they must here, since $\oplus$ is commutative. The identity element of $\oplus$ is $0$ . Not every operation is so lucky. For the star, the condition $a \star e = 2a + e - 1 = a$ already forces $e = 1 - a$ , a value that depends on $a$ , so no single number leaves every input unchanged even from that one side. With nothing able to satisfy the condition, the star has no identity element. Whether an identity exists is one more property you discover by testing, not by assuming.

Solving an equation that hides a custom operation

Because a defined operation is only a rule, an equation that contains one is only an ordinary equation in disguise. Substitute the definition to strip the invented symbol away, and you are left with the linear or simple quadratic equations you already solve. The unknown can sit in either slot.

Worked example 6 Solving for an unknown input

Let $a \star b = 2a + b - 1$ . Solve $x \star 4 = 11$ , then $3 \star x = 10$ . Finally, with $a \diamond b = a^2 - b$ , solve $x \diamond 3 = 22$ .

For $x \star 4 = 11$ , substitute $a = x$ and $b = 4$ into the rule to turn the starred equation into a plain one:

2x + 4 - 1 = 11 \quad\Longrightarrow\quad 2x + 3 = 11 \quad\Longrightarrow\quad 2x = 8 \quad\Longrightarrow\quad x = 4.

For $3 \star x = 10$ , the unknown is the second input, so substitute $a = 3$ and $b = x$ :

2(3) + x - 1 = 10 \quad\Longrightarrow\quad x + 5 = 10 \quad\Longrightarrow\quad x = 5.

For $x \diamond 3 = 22$ , the rule squares the first input, so substituting gives a simple quadratic:

x^2 - 3 = 22 \quad\Longrightarrow\quad x^2 = 25 \quad\Longrightarrow\quad x = 5 \ \text{ or } \ x = -5.

Squaring produces two solutions, both valid, since $5^2$ and $(-5)^2$ each equal $25$ . In every case the starred or diamonded equation became an equation you have solved since your first algebra lesson, once the definition was substituted in.

Check your understanding

With $a \star b = 2a + b - 1$ , solve $x \star 5 = 16$ for $x$ .

5

11

6

7

Common mistakes

Treating the symbol as a mystery. A star, diamond, or circle-plus is not new arithmetic. It is a rule written with the usual operations, so read the definition and substitute the inputs into it.
Ignoring the order of operations inside the rule. In $a \star b = 2a + b - 1$ you must compute $2a$ before adding. Substituting the numbers does not suspend the order of operations.
Assuming the operation is commutative. Most invented operations are not, so $a \star b$ and $b \star a$ can differ. Check the two orders instead of assuming they match.
Assuming the operation is associative. The grouping usually matters too, so $(a \star b) \star c$ can differ from $a \star (b \star c)$ . Evaluate the innermost parentheses first, and do not drop them.
Substituting only part of an expression input. When a slot holds an expression like $k + 1$ , the whole expression goes in. For $a \diamond b = a^2 - b$ that means $(k + 1)^2$ , not $k^2 + 1$ .
Expecting an identity element to exist. Some operations have one, like $0$ for $\oplus$ , and some have none, like the star. Solve $a \star e = a$ to find out rather than guessing.

Key takeaways

A custom operation is a binary operation defined by a formula, a function of two inputs written with familiar operations. The symbol is the only new thing.
Evaluate it by substituting the inputs into the defining rule and simplifying with the ordinary order of operations. A slot may hold a whole expression, which drops in as a single block.
Most custom operations are not commutative: $a \star b$ and $b \star a$ can differ, because the rule treats its two slots differently. Commutativity is a property to check, as with composition.
Most custom operations are not associative: $(a \star b) \star c$ and $a \star (b \star c)$ can differ. Work nested operations from the inside out, evaluating the innermost parentheses first.
An identity element is a value $e$ with $a \star e = a$ and $e \star a = a$ for every $a$ (both sides, since these operations need not be commutative). Find it, or rule it out, by solving that condition; some operations have one and some do not.
Solve an equation containing a custom operation by substituting the definition, which turns it into an ordinary linear or simple quadratic equation.

Defining New Operations

Learning goals

A made-up symbol is only a rule

When an input is an expression

The order of the two inputs usually matters

Nested operations: work from the inside out

Looking for an identity element

Solving an equation that hides a custom operation

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Free Response Questions (FRQ)