Examkin

Soru

Zorluk: OrtaFunctions 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)Cevap
  2. a⋆a=0a \star a = 0Cevap
  3. (a⋆b)2=a2b2+b2a2−2(a \star b)^2 = \frac{a^2}{b^2} + \frac{b^2}{a^2} - 2Cevap
  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)

Cevap

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.

Adım Adım Çözüm

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.

Anahtar Kavram

Evaluating algebraic properties and identity statements for custom defined binary operations.
Bu soruyu puanla

Konular

Functions and Custom Symbol Operations

Bu konudan daha fazla soru