Question

Difficulty: MediumBinary Operations

A binary operation \circ on the set of real numbers R\mathbb{R} is defined by ab=a+b+2aba \circ b = a + b + 2ab. If the identity element of the operation is ee, what is the value of xx such that the inverse of xx under \circ is equal to 22?

Answer: -0.4

Answer

The value of xx is 0.4-0.4.
First find the identity element ee by solving ae=aa \circ e = a, which yields a+e+2ae=a    e(1+2a)=0    e=0a + e + 2ae = a \implies e(1 + 2a) = 0 \implies e = 0. Next, by definition of an inverse, xx1=ex \circ x^{-1} = e. Substituting x1=2x^{-1} = 2 and e=0e = 0 gives x2=0x \circ 2 = 0. Expanding this using the binary operation rule yields x+2+2(x)(2)=0    5x+2=0    x=0.4x + 2 + 2(x)(2) = 0 \implies 5x + 2 = 0 \implies x = -0.4.

Step-by-Step Solution

1
Find the identity element ee of the operation \circ
e=0e = 0
By definition of identity element, ae=a    a+e+2ae=aa \circ e = a \implies a + e + 2ae = a, which simplifies to e(1+2a)=0e(1 + 2a) = 0, giving e=0e = 0.
2
Set up the inverse equation using x1=2x^{-1} = 2
x2=0x \circ 2 = 0
The inverse of xx satisfies xx1=ex \circ x^{-1} = e. Since x1=2x^{-1} = 2 and e=0e = 0, x2=0x \circ 2 = 0.
3
Solve for xx
x=0.4x = -0.4
Expanding x2=0x \circ 2 = 0 gives x+2+4x=0    5x=2    x=0.4x + 2 + 4x = 0 \implies 5x = -2 \implies x = -0.4.

Key Concept

Identity and Inverse Elements in Binary Operations
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