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Binary Operations — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsBinary Operations5 Marks
Question
On R − {1}, a binary operation * is defined by a * b = a + b − ab. Prove that * is commutative and associative. Find the identity element for * on R − {1}. Also, prove that every element of R − {1} is invertible.
✓

Answer

Commutativity:Let, $\text{a},\text{b}\in\text{R}-\{1\}.$ Then,
a * b = a + b - ab
= b + a - ba
= b * a
Therefore,
a * b = b * a, $\forall\ \text{a},\text{b}\in\text{R}-\{1\}$
Thus, * is commutative on R - {1}.
Associativity:
Let, $\text{a},\text{b}\in\text{R}-\{1\}.$ Then,
a * (b * c) = a * (b + c - bc)
= a + b + c - bc - a(b + c - bc)
= a + b + c - bc - ab - ac - abc
(a * b) * c = (a + b - ab) * c
= a + b - ab + c - (a + b - ab)c
= a + b + c - ab - ac - bc + abc
Therefore,
a * (b * c) = (a * b) * c, $\forall\text{ a},\text{b},\text{c}\in\text{R}-\{1\}$
Thus, * is associative on R - {1}.
Finding identity element:
Let e be the element in R - {1} with respect to * such that
a * e = a = e * a, $\forall\text{ a}\in\text{R}-\{1\}$
a * e = a and e * a = a, $\forall\text{ a}\in\text{R}-\{1\}$
⇒ a + e - ae = a and e + a - ea = a, $\forall\text{ a}\in\text{R}-\{1\}$
e(1 - a) = 0, $\forall\text{ a}\in\text{R}-\{1\}$
$\text{e}=0\in\forall\text{ a}\in\text{R}-\{1\},\forall\text{ a}\in\text{R}-\{1\}$ $[\because\ \text{a}\neq1]$
Thus, 0 is the identity element in R - {1} with respect to *.
Finding inverse:
Let $\text{a}\in\text{R}-\{1\}$ and $\text{b}\in\text{R}-\{1\}$ be the inverse of a. Then,
a * b = e = b * a
a * b = e and b * a = e
⇒ a + b - ab = 0 and b + a - ba = 0
⇒ a = ab - b
⇒ a = b(a - 1)
$\Rightarrow\text{b}=\frac{\text{a}}{\text{a}-1}$
Thus, $\frac{\text{a}}{\text{a}-1}$ is inverse of $\text{a}\in\text{R}-\{1\}.$

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