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Defining New Operations: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Advanced (beyond the core course) Advanced. This problem set goes beyond core Algebra I. You can skip it.

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Problem 1 of 10

Problem 1 Two signed inputs

Define $a\oslash b=(a-b)/(a+b)$ when $a+b\ne0$ . Find $(-3)\oslash1$ .
1. Hint 1
  The two input slots keep their order in both parts of the fraction.
2. Hint 2
  Compute the numerator and denominator separately before dividing.
Answer

$2$ .

Full solution

The denominator is $-3+1=-2$ , so the operation is defined.
Substitution gives
$(-3)\oslash1=\frac{-3-1}{-3+1}.$
The numerator is $-4$ , and dividing by $-2$ gives $2$ .
Multiplying $2$ by the denominator $-2$ returns the numerator $-4$ .

Answer

$2$ .

Key idea

A defined operation may have restrictions on the pair of inputs as well as a rule for its output.
Problem 1: Mark complete
Problem 2 Two expression slots

For all real inputs, define $a\boxplus b=(a+b)^2-ab$ . Write $(t+1)\boxplus(t-1)$ as an expanded polynomial.
1. Hint 1
  Substitute each whole expression wherever its input appears.
2. Hint 2
  Simplify the sum of the inputs and their product before combining the results.
Answer

$3t^2+1$ .

Full solution

The sum of the inputs is $2t$ and their product is $t^2-1$ .
Therefore the operation gives
$(2t)^2-(t^2-1).$
Expanding and collecting gives
$3t^2+1.$
At $t=0$ , the inputs are $1$ and $-1$ , producing $1$ , as the polynomial also does.

Answer

$3t^2+1$ .

Key idea

Both slots can contain whole expressions, and each must stay intact during substitution.
Problem 2: Mark complete
Problem 3 Related input expressions

For all real inputs, define $a\circleddash b=a^2-ab$ . Find every real $x$ satisfying $(x+1)\circleddash(x-1)=8$ .
1. Hint 1
  Replace the custom symbol with its rule before solving.
2. Hint 2
  Both terms share the full first input as a factor, which can simplify the equation.
Answer

$x=3$ .

Full solution

Substitution gives
$(x+1)^2-(x+1)(x-1)=8.$
Factor the shared expression without dividing by it:
$(x+1)((x+1)-(x-1))=8.$
Thus $2(x+1)=8$ , which gives $x=3$ .

Check with inputs $4$ and $2$ .
$4^2-4(2)=8.$
The simplification was valid for every real $x$ , so no other solutions were lost.

Answer

$x=3$ .

Key idea

Replacing a custom symbol can reveal cancellation that turns an apparent quadratic into a linear equation.
Problem 3: Mark complete
Problem 4 An unknown inside the inner pair

Define $a\boxtimes b=a^2+b^2$ for real inputs. Find every real $x$ with $(x\boxtimes1)\boxtimes2=29$ .
1. Hint 1
  The inner pair must become a single expression before the outer rule can be applied.
2. Hint 2
  After substituting, a square of an expression equals a number, so ask which values that expression is allowed to take before solving for $x$ .
Answer

$x=2$ or $x=-2$ .

Full solution

The inner pair gives $x\boxtimes1=x^2+1$ , and that expression fills the first outer slot, so the equation becomes
$(x^2+1)^2+2^2=29.$
Subtracting $4$ from both sides leaves
$(x^2+1)^2=25.$

A square equal to $25$ leaves two branches, $x^2+1=5$ and $x^2+1=-5$ .
The second is impossible, since $x^2+1$ is at least $1$ for every real $x$ .
The first gives $x^2=4$ , so $x=2$ or $x=-2$ .
Both check: $2\boxtimes1=5$ and $(-2)\boxtimes1=5$ , and
$5\boxtimes2=29.$

Answer

$x=2$ or $x=-2$ .

Key idea

A square equal to a positive number opens two branches, so test each one and keep every real solution it allows, negative values included.
Problem 4: Mark complete
Problem 5 A rule from records

An operation on real numbers has rule $a\boxdot b=ma+nb$ . Its records include $1\boxdot2=8$ and $2\boxdot1=7$ . Find $m$ and $n$ , then find $(2\boxdot3)-(3\boxdot2)$ .
1. Hint 1
  The two records form equations for the two fixed coefficients.
2. Hint 2
  Solve those equations, then use the recovered rule in both orders.
Answer

$m=2$ , $n=3$ ; $(2\boxdot3)-(3\boxdot2)=1$ .

Full solution

The records give $m+2n=8$ and $2m+n=7$ .
Double the first equation and subtract the second to get
$3n=9.$
Therefore $n=3$ and $m=2$ .

The completed rule gives $2\boxdot3=13$ and $3\boxdot2=12$ , so their difference is $1$ .
The records check as $2+6=8$ and $4+3=7$ .

Answer

$m=2$ , $n=3$ ; $(2\boxdot3)-(3\boxdot2)=1$ .

Key idea

Multiple records can determine how a custom rule weights its two input slots.
Problem 5: Mark complete
Problem 6 Checking a classmate's expansion

For all real inputs, define $a\circledast b=a^2-3b$ . Asked for $(m-2)\circledast5$ , Rowan writes $m^2-2^2-3(5)$ and simplifies it to $m^2-19$ . Name the first step that is wrong, and give the correct expanded expression.
1. Hint 1
  The first input here is a whole expression, and the rule squares whatever sits in that slot.
2. Hint 2
  Substitute $m-2$ into the squared slot as one block, expand that square, then combine it with the second term.
Answer

The first wrong step is replacing $(m-2)^2$ with $m^2-2^2$ . The correct expression is $m^2-4m-11$ .

Full solution

The rule squares the entire first input, so substituting the two inputs gives
$(m-2)^2-3(5).$
Rowan's second term is right, since $3(5)$ is $15$ .

His first step is the wrong one: $(m-2)^2$ is not $m^2-2^2$ , because squaring a difference produces the middle term $-4m$ .
Expanding correctly gives $m^2-4m+4$ , so the operation gives
$m^2-4m-11.$
At $m=0$ the rule gives $-11$ , while Rowan's expression gives $-19$ , so the two disagree.

Answer

The first wrong step is replacing $(m-2)^2$ with $m^2-2^2$ . The correct expression is $m^2-4m-11$ .

Key idea

An expression input enters its slot as one block, so a squared slot is expanded rather than squared piece by piece.
Problem 6: Mark complete
Problem 7 A prescribed identity

For all real inputs, define $a\circledcirc b=kab$ , where $k$ is fixed. Choose $k$ so that $3$ is an identity element, and verify the identity from both sides.
1. Hint 1
  An identity must return every input when placed in either slot.
2. Hint 2
  Write the two identity conditions and compare the coefficient of the unchanged input.
Answer

$k=\frac13$ ; $a\circledcirc3=a$ and $3\circledcirc a=a$ for every real $a$ .

Full solution

The condition $a\circledcirc3=a$ says $3ka=a$ for every $a$ .
Taking $a=1$ gives $3k=1$ , hence
$k=\frac13.$
For this value,
$a\circledcirc3=\frac{3a}{3}$
equals $a$ , and
$3\circledcirc a=\frac{3a}{3}$
also equals $a$ .
Both required conditions hold for all real inputs.

Answer

$k=\frac13$ ; $a\circledcirc3=a$ and $3\circledcirc a=a$ for every real $a$ .

Key idea

A specified identity can determine an unknown operation, but it must work in both input positions.
Problem 7: Mark complete
Problem 8 Order and grouping

Define $a\barwedge b=|a-b|$ for all real inputs. Kai says this operation is both commutative and associative. Decide each part of his claim, and justify each decision either for every real input or with a specific counterexample.
1. Hint 1
  Commutativity asks whether the rule is unchanged when the two inputs trade places; associativity asks whether the two groupings of three inputs agree. Decide each on its own evidence.
2. Hint 2
  Compare $|a-b|$ with $|b-a|$ in general, then compute both groupings of the triple $1$ , $2$ , and $4$ .
Answer

Commutative: yes. Associative: no; for example $(1\barwedge2)\barwedge4=3$ but $1\barwedge(2\barwedge4)=1$ .

Full solution

Swapping inputs gives
$b-a=-(a-b).$
Opposite numbers have the same absolute value, so
$|b-a|=|a-b|.$
This proves commutativity for every pair.

One triple settles the other half.
For the left grouping, $1\barwedge2=1$ , followed by
$1\barwedge4=3.$
For the right grouping, $2\barwedge4=2$ , followed by
$1\barwedge2=1.$
Since $3\ne1$ , associativity fails.

Answer

Commutative: yes. Associative: no; for example $(1\barwedge2)\barwedge4=3$ but $1\barwedge(2\barwedge4)=1$ .

Key idea

An operation may ignore input order while still depending on grouping.
Problem 8: Mark complete
Problem 9 A claim about the first input

Define $a\dagger b=a+b-ab$ for all real inputs. Lila claims at least two values of the first input make the output the same for every possible second input. Decide whether Lila is correct, and name every first input with that property together with the constant output it produces.
1. Hint 1
  Keep the first input fixed and collect the terms involving the second input.
2. Hint 2
  Compare the outputs at second inputs zero and one to find a necessary condition, then check it generally.
Answer

No; exactly one first input has that property: $a=1$ , with constant output $1$ .

Full solution

The rule can be written
$a\dagger b=a+(1-a)b.$
In particular, the outputs at $b=0$ and $b=1$ are $a$ and $1$ .
If all outputs agree, these two must agree, so $a=1$ .

For this value,
$1\dagger b=1+b-b,$
which equals $1$ for every real $b$ .
No other first input can work, so exactly one value has the property and Lila is wrong to claim two or more.

Answer

No; exactly one first input has that property: $a=1$ , with constant output $1$ .

Key idea

Comparing selected inputs can identify a necessary condition, which must then be checked for all inputs.
Problem 9: Mark complete
Problem 10 Two properties of a root rule

Define $a\veebar b=\sqrt{a^2+b^2}$ for inputs that are zero or positive. Decide whether this operation has an identity element and whether it is associative, and justify both conclusions for every allowed input.
1. Hint 1
  The nonnegative input restriction matters when a square and a square root cancel.
2. Hint 2
  Test the identity condition at the input $0$ to narrow the candidates to one, check that candidate in both orders, then compare the expressions the two groupings produce.
Answer

Identity: $0$ . The operation is associative on the stated input set.

Full solution

If $e$ is an identity, using input $0$ requires $\sqrt{e^2}=0$ , hence $e=0$ .
For every allowed $a$ , the two orders with zero both give
$\sqrt{a^2}=a.$
Thus zero is an identity.

For allowed $a$ , $b$ , and $c$ , each inner output is nonnegative, so it is itself an allowed input.
The left grouping gives
$(a\veebar b)\veebar c=\sqrt{(\sqrt{a^2+b^2})^2+c^2},$
and the right grouping gives
$a\veebar(b\veebar c)=\sqrt{a^2+(\sqrt{b^2+c^2})^2}.$
Each inner radicand is nonnegative, so each square undoes its own root, and both expressions reduce to
$\sqrt{a^2+b^2+c^2}.$
The two groupings therefore agree for every allowed triple, so the operation is associative.

Answer

Identity: $0$ . The operation is associative on the stated input set.

Key idea

A domain restriction can make simplifications valid and support a general identity or associativity proof.
Problem 10: Mark complete

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