Question

Difficulty: HardBinary Operations

A binary operation ⋆\star on the set of real numbers R∖{1}\mathbb{R} \setminus \{1\} is defined by a⋆b=a+b−aba \star b = a + b - ab. If y−1y^{-1} denotes the inverse of an element yy under ⋆\star, find the value of xx such that (x⋆3)−1=2(x \star 3)^{-1} = 2.

Answer: 0.5

Answer

The value of xx is 0.50.5.
To solve for xx, first calculate the identity element e=0e = 0 using a⋆e=aa \star e = a. Next, use y⋆y−1=0y \star y^{-1} = 0 to get the inverse formula y−1=yy−1y^{-1} = \frac{y}{y - 1}. Substituting y=x⋆3=3−2xy = x \star 3 = 3 - 2x into the inverse formula yields 3−2x2−2x=2\frac{3 - 2x}{2 - 2x} = 2, which simplifies to 2x=12x = 1, giving x=0.5x = 0.5.

Step-by-Step Solution

1
Find the identity element ee of the operation ⋆\star.
e=0e = 0
By definition of the identity element, a⋆e=a  ⟹  a+e−ae=a  ⟹  e(1−a)=0a \star e = a \implies a + e - ae = a \implies e(1 - a) = 0, so e=0e = 0 for all a≠1a \neq 1.
2
Derive the general expression for the inverse element y−1y^{-1} of yy.
y−1=yy−1y^{-1} = \frac{y}{y - 1}
By definition of inverse element, y⋆y−1=e  ⟹  y+y−1−yy−1=0  ⟹  y−1(1−y)=−y  ⟹  y−1=yy−1y \star y^{-1} = e \implies y + y^{-1} - y y^{-1} = 0 \implies y^{-1}(1 - y) = -y \implies y^{-1} = \frac{y}{y - 1}.
3
Evaluate the inner operation x⋆3x \star 3.
x⋆3=3−2xx \star 3 = 3 - 2x
Using the operational rule a⋆b=a+b−aba \star b = a + b - ab, we obtain x⋆3=x+3−3x=3−2xx \star 3 = x + 3 - 3x = 3 - 2x.
4
Set up and solve the equation (x⋆3)−1=2(x \star 3)^{-1} = 2.
x=0.5x = 0.5
Letting y=3−2xy = 3 - 2x, its inverse is y−1=3−2x(3−2x)−1=3−2x2−2xy^{-1} = \frac{3 - 2x}{(3 - 2x) - 1} = \frac{3 - 2x}{2 - 2x}. Equating this to 22 gives 3−2x2−2x=2  ⟹  3−2x=4−4x  ⟹  2x=1  ⟹  x=0.5\frac{3 - 2x}{2 - 2x} = 2 \implies 3 - 2x = 4 - 4x \implies 2x = 1 \implies x = 0.5.

Key Concept

Identity and Inverse Elements in Binary Operations
Rate this question