A binary operation \circ on the set of real numbers \mathbb{R} is defined by a \circ b = a… — JAMB UTME

A binary operation $\circ$ on the set of real numbers $\mathbb{R}$ is defined by $a \circ b = a + b - \frac{1}{2}ab$ . What is the inverse of $3$ under this operation?

$6$ Cevap
B
$-6$
C
$-3$
D
$\frac{1}{3}$

Cevap

The inverse of

3

under the defined binary operation is

6

To find the inverse of

3

, we first find the identity element

e

using

a \circ e = a \implies a + e - \frac{1}{2}ae = a

, which gives

e = 0

. Then, setting

3 \circ x = 0

gives

3 + x - \frac{3}{2}x = 0 \implies 3 = \frac{1}{2}x \implies x = 6

Adım Adım Çözüm

Find the identity element

e

of the operation.

e = 0

By definition of identity element,

a \circ e = a

. Substituting into the operation formula gives

a + e - \frac{1}{2}ae = a \implies e\left(1 - \frac{1}{2}a\right) = 0 \implies e = 0

for all

a \neq 2

Set up the inverse equation for the element

3

3 \circ x = 0

Let

x

be the inverse of

3

. By definition of inverse element,

3 \circ x = e

, where

e = 0

Apply the binary operation definition to the left-hand side.

3 + x - \frac{1}{2}(3)(x) = 0

Substitute

a = 3

and

b = x

into

a \circ b = a + b - \frac{1}{2}ab

Solve the linear equation for

x

x = 6

Simplify the equation:

3 + x - \frac{3}{2}x = 0 \implies 3 - \frac{1}{2}x = 0 \implies \frac{1}{2}x = 3 \implies x = 6

Anahtar Kavram

Identity and Inverse Elements of Binary Operations

Soru

Cevap

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