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Difficulty: MediumFunctions and Custom Symbol Operations

For all non-zero real numbers xx and yy, the custom binary operation ⋆\star is defined by x⋆y=xy−yxx \star y = \frac{x}{y} - \frac{y}{x}. Which of the following statements must be true for all non-zero real numbers aa, bb, and cc? Select all that apply.

  1. a⋆b=−(b⋆a)a \star b = -(b \star a)Answer
  2. a⋆a=0a \star a = 0Answer
  3. (a⋆b)2=a2b2+b2a2−2(a \star b)^2 = \frac{a^2}{b^2} + \frac{b^2}{a^2} - 2Answer
  4. D
    a⋆(b⋆c)=(a⋆b)⋆ca \star (b \star c) = (a \star b) \star c
  5. E
    a⋆(b+c)=(a⋆b)+(a⋆c)a \star (b + c) = (a \star b) + (a \star c)

Answer

The statements a⋆b=−(b⋆a)a \star b = -(b \star a), a⋆a=0a \star a = 0, and (a⋆b)2=a2b2+b2a2−2(a \star b)^2 = \frac{a^2}{b^2} + \frac{b^2}{a^2} - 2 must be true for all non-zero real numbers aa, bb, and cc.
The statements showing anti-commutativity, zero self-operation, and the expanded squared expression are all mathematically valid for all non-zero real numbers upon expanding their definitions using algebra.

Step-by-Step Solution

1
Evaluate a⋆b=−(b⋆a)a \star b = -(b \star a)
a⋆b=ab−baa \star b = \frac{a}{b} - \frac{b}{a} and −(b⋆a)=−(ba−ab)=ab−ba-(b \star a) = -\left(\frac{b}{a} - \frac{a}{b}\right) = \frac{a}{b} - \frac{b}{a}.
Verify anti-commutativity property by direct substitution.
2
Evaluate a⋆a=0a \star a = 0
a⋆a=aa−aa=1−1=0a \star a = \frac{a}{a} - \frac{a}{a} = 1 - 1 = 0.
Verify identity property for identical inputs.
3
Evaluate (a⋆b)2(a \star b)^2
(a⋆b)2=(ab−ba)2=(ab)2−2(ab)(ba)+(ba)2=a2b2−2(1)+b2a2=a2b2+b2a2−2(a \star b)^2 = \left(\frac{a}{b} - \frac{b}{a}\right)^2 = \left(\frac{a}{b}\right)^2 - 2\left(\frac{a}{b}\right)\left(\frac{b}{a}\right) + \left(\frac{b}{a}\right)^2 = \frac{a^2}{b^2} - 2(1) + \frac{b^2}{a^2} = \frac{a^2}{b^2} + \frac{b^2}{a^2} - 2.
Apply binomial expansion to the squared custom operation.
4
Test associativity statement a⋆(b⋆c)=(a⋆b)⋆ca \star (b \star c) = (a \star b) \star c with a counterexample
Let a=4,b=2,c=1a = 4, b = 2, c = 1. Then b⋆c=21−12=32b \star c = \frac{2}{1} - \frac{1}{2} = \frac{3}{2}. a⋆(b⋆c)=4⋆32=43/2−3/24=83−38=5524a \star (b \star c) = 4 \star \frac{3}{2} = \frac{4}{3/2} - \frac{3/2}{4} = \frac{8}{3} - \frac{3}{8} = \frac{55}{24}. Meanwhile, a⋆b=4⋆2=42−24=32a \star b = 4 \star 2 = \frac{4}{2} - \frac{2}{4} = \frac{3}{2}, and (a⋆b)⋆c=32⋆1=3/21−13/2=32−23=56(a \star b) \star c = \frac{3}{2} \star 1 = \frac{3/2}{1} - \frac{1}{3/2} = \frac{3}{2} - \frac{2}{3} = \frac{5}{6}. Since 5524≠56\frac{55}{24} \neq \frac{5}{6}, associativity fails.
A single counterexample disproves a general identity statement.
5
Test distributivity statement a⋆(b+c)=(a⋆b)+(a⋆c)a \star (b + c) = (a \star b) + (a \star c) with a counterexample
Let a=1,b=1,c=1a = 1, b = 1, c = 1. Then a⋆(b+c)=1⋆2=12−21=−32a \star (b + c) = 1 \star 2 = \frac{1}{2} - \frac{2}{1} = -\frac{3}{2}. Meanwhile, (a⋆b)+(a⋆c)=(1⋆1)+(1⋆1)=0+0=0(a \star b) + (a \star c) = (1 \star 1) + (1 \star 1) = 0 + 0 = 0. Since −32≠0-\frac{3}{2} \neq 0, distributivity fails.
A single counterexample disproves distributivity over addition.

Key Concept

Evaluating algebraic properties and identity statements for custom defined binary operations.
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Topics

Functions and Custom Symbol Operations

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