12 multiple-choice questions, progressively harder.
Let a⋆b=2a+b−1a \star b = 2a + b - 1a⋆b=2a+b−1. What is 2⋆(3⋆4)2 \star (3 \star 4)2⋆(3⋆4)?
Solution
Correct answer: D
Work from the inside out.
3⋆4=2(3)+4−1=93 \star 4 = 2(3) + 4 - 1 = 93⋆4=2(3)+4−1=9
Then 999 becomes the second input of the outer star.
2⋆9=2(2)+9−1=122 \star 9 = 2(2) + 9 - 1 = 122⋆9=2(2)+9−1=12
Let a⋆b=2a+b−1a \star b = 2a + b - 1a⋆b=2a+b−1 and a⋄b=a2−ba \diamond b = a^2 - ba⋄b=a2−b. What is (2⋄1)⋆5(2 \diamond 1) \star 5(2⋄1)⋆5?
Correct answer: C
Evaluate the inner diamond first.
2⋄1=22−1=32 \diamond 1 = 2^2 - 1 = 32⋄1=22−1=3
Then apply the star with 555.
3⋆5=2(3)+5−1=103 \star 5 = 2(3) + 5 - 1 = 103⋆5=2(3)+5−1=10
For a⋆b=2a+b−1a \star b = 2a + b - 1a⋆b=2a+b−1, you found (2⋆3)⋆4=15(2 \star 3) \star 4 = 15(2⋆3)⋆4=15 but 2⋆(3⋆4)=122 \star (3 \star 4) = 122⋆(3⋆4)=12. Is ⋆\star⋆ associative?
Correct answer: B
Associative means the grouping never changes the answer. Here it does.
(2⋆3)⋆4=15≠12=2⋆(3⋆4)(2 \star 3) \star 4 = 15 \ne 12 = 2 \star (3 \star 4)(2⋆3)⋆4=15=12=2⋆(3⋆4)
A single mismatch is enough to show ⋆\star⋆ is not associative.
A turn in a game with aaa hits and bbb misses scores a⊙b=5a−2ba \odot b = 5a - 2ba⊙b=5a−2b points. How many points is a turn with 444 hits and 333 misses?
Substitute a=4a = 4a=4 hits and b=3b = 3b=3 misses into the scoring rule.
4⊙3=5(4)−2(3)=20−6=144 \odot 3 = 5(4) - 2(3) = 20 - 6 = 144⊙3=5(4)−2(3)=20−6=14
The turn scores 141414 points.
Let a⊕b=a+b+aba \oplus b = a + b + aba⊕b=a+b+ab. Which value eee satisfies a⊕e=aa \oplus e = aa⊕e=a for every aaa?
Correct answer: A
Set up the identity condition and solve for eee.
a+e+ae=a ⟹ e(1+a)=0a + e + ae = a \;\Longrightarrow\; e(1 + a) = 0a+e+ae=a⟹e(1+a)=0
For this to hold for every aaa, the factor eee must be 000. Check: a⊕0=a+0+0=aa \oplus 0 = a + 0 + 0 = aa⊕0=a+0+0=a.
Let a⊗b=a+b−4a \otimes b = a + b - 4a⊗b=a+b−4. Is this operation commutative?
Swap the inputs in the rule.
b⊗a=b+a−4=a+b−4=a⊗bb \otimes a = b + a - 4 = a + b - 4 = a \otimes bb⊗a=b+a−4=a+b−4=a⊗b
Because addition is commutative, the expression is unchanged when the inputs trade places, so ⊗\otimes⊗ is commutative.
Let a⊕b=a+b+aba \oplus b = a + b + aba⊕b=a+b+ab. Compare (1⊕2)⊕3(1 \oplus 2) \oplus 3(1⊕2)⊕3 with 1⊕(2⊕3)1 \oplus (2 \oplus 3)1⊕(2⊕3).
The left grouping gives 232323. For the right grouping, start inside.
2⊕3=2+3+6=11,1⊕11=1+11+11=232 \oplus 3 = 2 + 3 + 6 = 11, \qquad 1 \oplus 11 = 1 + 11 + 11 = 232⊕3=2+3+6=11,1⊕11=1+11+11=23
Both groupings equal 232323, matching the fact that ⊕\oplus⊕ is associative.
Let a⊗b=a+b−4a \otimes b = a + b - 4a⊗b=a+b−4. Is this operation associative?
Group the three inputs both ways.
(a⊗b)⊗c=(a+b−4)+c−4=a+b+c−8(a \otimes b) \otimes c = (a + b - 4) + c - 4 = a + b + c - 8(a⊗b)⊗c=(a+b−4)+c−4=a+b+c−8
a⊗(b⊗c)=a+(b+c−4)−4=a+b+c−8a \otimes (b \otimes c) = a + (b + c - 4) - 4 = a + b + c - 8a⊗(b⊗c)=a+(b+c−4)−4=a+b+c−8
Both groupings give a+b+c−8a + b + c - 8a+b+c−8, so ⊗\otimes⊗ is associative.
An operation ∗\ast∗ has an identity element eee. What does that mean?
An identity element is a fixed value that leaves every input unchanged from both sides.
a∗e=a and e∗a=a for every aa \ast e = a \;\text{ and }\; e \ast a = a \;\text{ for every } aa∗e=a and e∗a=a for every a
Both sides are required because the operation need not be commutative. Having an identity is separate from being commutative or associative; an operation can have any combination of these properties.
For which operation is 111 an identity element, meaning a∗1=aa \ast 1 = aa∗1=a and 1∗a=a1 \ast a = a1∗a=a for every aaa?
Test both sides for each rule. For multiplication, a⋅1=aa \cdot 1 = aa⋅1=a and 1⋅a=a1 \cdot a = a1⋅a=a, so 111 is a two-sided identity.
a⋅1=a and 1⋅a=aa \cdot 1 = a \;\text{ and }\; 1 \cdot a = aa⋅1=a and 1⋅a=a
The others fail: a+1≠aa + 1 \ne aa+1=a, and a+1+a=2a+1≠aa + 1 + a = 2a + 1 \ne aa+1+a=2a+1=a, and a2−1≠aa^2 - 1 \ne aa2−1=a in general.
Let a⋆b=2a+b−1a \star b = 2a + b - 1a⋆b=2a+b−1. Find the value of xxx for which x⋆x=17x \star x = 17x⋆x=17.
Put xxx in both slots and simplify.
x⋆x=2x+x−1=3x−1x \star x = 2x + x - 1 = 3x - 1x⋆x=2x+x−1=3x−1
Set it equal to 171717: 3x−1=173x - 1 = 173x−1=17, so 3x=183x = 183x=18 and x=6x = 6x=6.
For a⋆b=2a+b−1a \star b = 2a + b - 1a⋆b=2a+b−1, you find (1⋆2)⋆3=8(1 \star 2) \star 3 = 8(1⋆2)⋆3=8 but 1⋆(2⋆3)=71 \star (2 \star 3) = 71⋆(2⋆3)=7. What does this show?
Associativity would require the two groupings to match.
(1⋆2)⋆3=8≠7=1⋆(2⋆3)(1 \star 2) \star 3 = 8 \ne 7 = 1 \star (2 \star 3)(1⋆2)⋆3=8=7=1⋆(2⋆3)
Because the grouping changes the result, ⋆\star⋆ is not associative.
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