A new operation may look familiar but obey unfamiliar laws. Swap the inputs, regroup three inputs, search for an identity, and use one counterexample to stop an unsafe assumption.
For a defined operation a ◆ b, test each claim separately. One successful example is only evidence. One failed example is a complete counterexample.
Swapping the two inputs does not change the output.
Regrouping three inputs does not change the output.
One legal value leaves every input unchanged on both sides.
Legal inputs always produce an output that is still in the stated domain.
Swapping, regrouping, keeping a value unchanged, and staying in the domain are different tests.
The outputs match only when a=b. A commutative law must work for every legal pair, so this rule is not commutative.
Subtract the swapped expressions; the difference simplifies to a−b.
The explorer accepts integer inputs. Zero is not allowed in the fraction rule. This mission compares formulas; Mission 7 separately checks whether outputs stay in a chosen domain.
Every cell compares a◆b with b◆a. A cell marked ≠ supplies a counterexample. Rows give a and columns give b.
A single legal mismatch disproves commutativity. Matching examples still need a general explanation.
The inner values match by symmetry. Combine those equal nonzero outputs.
Expand each route, solve the ordinary equation, and reject a non-integer result.
Adding 1 changes the custom operation into ordinary multiplication. Both groupings, after adding 1, give (a+1)(b+1)(c+1). Subtracting 1 gives equal original outputs. Swapping a and b also leaves (a+1)(b+1) unchanged. This proves both laws.
Add 1 to the operation result to reveal (a+1)(b+1).
The identity must belong to the domain. For max on positive integers, e=0 is illegal even though max(a,0)=a for positive a. The legal identity is 1. A successful test for one a is evidence only; the explanation must work for every legal a.
A formula may be symmetric yet fail to define a binary operation on the stated domain. Closure is checked before other laws can be used safely.
| Rule and domain | Commutative? | Associative? | Closed? | Identity | Useful witness |
|---|
Check that e is legal before testing it, and require both directions for every legal input.
Here α (alpha), β (beta), and γ (gamma) are fixed integer coefficients: positive, zero, or negative. The first two tell you how much of each input to use; γ is the constant added.
For a◆b=αa+βb+γ:
(a◆b)◆c = α²a + αβb + βc + γ(α+1).
a◆(b◆c) = αa + αβb + β²c + γ(β+1).
For equality at every a,b,c, match the coefficients of a and c and the constants. This gives α²=α, β²=β, and γ(α−β)=0.
| α, β | Associative rule | Reason |
|---|---|---|
| 0, 0 | a◆b=γ, any γ | Both routes return γ. |
| 1, 0 | a◆b=a, γ=0 | Both routes return the first input. |
| 0, 1 | a◆b=b, γ=0 | Both routes return the last input. |
| 1, 1 | a◆b=a+b+γ, any γ | Both routes return a+b+c+2γ. |
There are no other integer solutions of α²=α or β²=β, because t(t−1)=0 forces t=0 or t=1.
Define, on all integers,
a ◆ b = αa + βb + γ
For this guided example, define:
x*y=mx+ny, x△y=kxy
with positive integers m,n,k, and gives:
1*2=5, (2*3)△4=64
m=3, n=1
2*3=9, so 36k=64.
Reject: k is not an integerm=1, n=2
2*3=8, so 32k=64.
Accept: k=2(1△2)*3 = 10
Use m+2n=5 to list positive branches. The second condition must also give an integer k.
Rule reminders: a★b=3a+2b−2 for the input-order question; a○b=a/b+b/a+1/2 for the symmetry question (nonzero inputs); a⊙b=ab+a+b on nonnegative integers; and the associativity equation uses a★b=a+2b−1 on integers. Each question names the rule it needs.
Correct all eight questions to complete the workshop.
Answers: commutative; 50; 1/2; 5/2; 0; 59; No; 10. The equation 2m+9=4m+8 gives m=1/2. The equal inner circle outputs give 3○3=5/2. For the shifted rule, 3×4×5−1=59. Closure fails because 1◆1=−1. Finally 1△2=4, then 4*3=10.
Rule reminders: a★b=3a+2b−2 for the input-order question; a○b=a/b+b/a+1/2 for the symmetry question (nonzero inputs); a⊙b=ab+a+b on nonnegative integers; and the associativity equation uses a★b=a+2b−1 on integers. Each question names the rule it needs.
For integers, define a◇b=2a+b. Find 1◇2 and 2◇1. Does this pair disprove commutativity?
Answers: No; 3; No; 0; α=β. A single unequal swapped pair disproves commutativity; the circle sum is 2+1/2+1/2=3; the grouping equation requires the non-integer 1/2. Substituting 0 on either side proves the shifted identity. Equal input coefficients prove symmetry.
This certifies that the learner can test input order, symmetry, associativity, identity, and closure; use counterexamples correctly; and respect the stated domain of a newly defined operation.