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Zorluk: ZorBinary Operations

A binary operation ∘\circ defined on the set of real numbers R∖{1}\mathbb{R} \setminus \{1\} is given by a∘b=a+b−aba \circ b = a + b - ab. What is the inverse of 33 under this operation?

  1. 32\frac{3}{2}Cevap
  2. B
    −32-\frac{3}{2}
  3. C
    11
  4. D
    13\frac{1}{3}

Cevap

The inverse of 33 under the given operation is 32\frac{3}{2}.
To find the inverse of 33 under the operation a∘b=a+b−aba \circ b = a + b - ab, we must first determine the identity element ee. Setting a∘e=aa \circ e = a yields a+e−ae=aa + e - ae = a, which simplifies to e(1−a)=0e(1 - a) = 0. For all a≠1a \neq 1, the identity element is e=0e = 0. Next, using the definition of inverse 3∘x=03 \circ x = 0, we substitute into the operational formula to obtain 3+x−3x=0  ⟹  3−2x=0  ⟹  x=323 + x - 3x = 0 \implies 3 - 2x = 0 \implies x = \frac{3}{2}. Thus, the option specifying 32\frac{3}{2} is correct.

Adım Adım Çözüm

1
Find the identity element ee of the operation ∘\circ.
e=0e = 0
By definition of identity, a∘e=aa \circ e = a. Substituting into the operational formula gives a+e−ae=a  ⟹  e(1−a)=0a + e - ae = a \implies e(1 - a) = 0. Since a≠1a \neq 1, e=0e = 0.
2
Set up the inverse equation for 33, letting xx be the inverse of 33.
3∘x=03 \circ x = 0
By definition of an inverse element, a∘a−1=ea \circ a^{-1} = e.
3
Expand 3∘x3 \circ x using the operation rule and solve for xx.
x=32x = \frac{3}{2}
3+x−3x=0  ⟹  3−2x=0  ⟹  2x=3  ⟹  x=323 + x - 3x = 0 \implies 3 - 2x = 0 \implies 2x = 3 \implies x = \frac{3}{2}.

Anahtar Kavram

Identity and Inverse Elements in Binary Operations
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Konular

Binary Operations

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