Question

* is defined on R by a*b=a+b+ab. Is * commutative? Is it associative? Answer the same question if a*b=a-b+ab.

Answer 1

$\underline{\large{\text{Answer}}} : \\ \\ \boxed{\text{a}} \\ \\ \text{The given relation is} \\ \text{a} \ast \text{b = a + b + ab}$

$\text{i)} \: \underline {{\text{Checking for commutative property}}} : \\ \\ \text{a} \ast \text{b = a + b + ab} \\ \text{b} \ast \text{a = b + a + ba} \\ \\ \therefore \text{a} \ast \text{b} = \text{b} \ast \text{a} \: \: \: \: \{\because \text{ab = ba} \: \} \text{ and thus } (\ast)\: \text{is} \\ \text{commutative.}$

$\text{ii)} \: \underline{\text{Checking for associative property}} : \\ \\ \text{a}\ast ( \text{b} \ast \text{c}) \\ = \text{a} \ast \text{(b + c + bc)} \\ = \text{a + (b + c + bc) + a(b + c + bc)} \\ = \text{a + b + c + ab + bc + ca + abc} \\ \\ \text{and} \: (\text{a} \ast \text{b} )\ast \text{c} \\ = \text{(a + b + ab)}\ast \text{c} \\ = \text{a + b + ab + c + (a + b + ab)c} \\ = \text{a + b + c + ab + bc + ca + abc} \\ \\ \therefore \text{a} \ast (\text{b} \ast \text{c}) = ( \text{a} \ast \text{b}) \ast \text{c} \: \: \text{and thus} \: ( \ast) \: \text{is} \\ \text{associative.}$

$\boxed{\text{b}}\\ \\ \text{The given relation is} \\ \text{a} \ast \text{b = a - b + ab} \\ \\ \text{i)}\:\underline{\text{Checking for commutative property}} : \\ \\ \text{a} \ast \text{b = a - b + ab} \\ \text{b} \ast \text{a = b - a + ba} \\ \\ \therefore \text{a} \ast \text{b} \neq \text{b} \ast \text{a} \: \: \: \: \{\because { \text{a - b} \neq \text{b - a}} \: \} \text{ and thus } (\ast)\: \text{is} \\ \text{not commutative.} \\ \\ \text{ii)} \:\underline{\text{Checking for associative property}} : \\ \\ \text{a}\ast ( \text{b} \ast \text{c}) \\ = \text{a} \ast \text{(b - c + bc)} \\ = \text{a - (b - c + bc) + a(b - c + bc)} \\ = \text{a - b + c + ab - bc - ca + abc} \\ \\ \text{and} \: (\text{a} \ast \text{b} )\ast \text{c} \\ = \text{(a - b + ab)}\ast \text{c} \\ = \text{a - b + ab - c + (a - b + ab)c} \\ = \text{a - b - c + ab - bc + ca + abc} \\ \\ \because \text{a} \ast (\text{b} \ast \text{c}) \neq ( \text{a} \ast \text{b}) \ast \text{c}, \: ( \ast) \: \text{is not associative}.$

$\bigstar \underline{\large{\text{MarkAsBrainliest}}}\bigstar$

Answer 2

Answer:

\underline{\large{\text{Answer}}} : \\ \\ \boxed{\text{a}} \\ \\ \text{The given relation is} \\ \text{a} \ast \text{b = a + b + ab}

\text{i)} \: \underline {{\text{Checking for commutative property}}} : \\ \\ \text{a} \ast \text{b = a + b + ab} \\ \text{b} \ast \text{a = b + a + ba} \\ \\ \therefore \text{a} \ast \text{b} = \text{b} \ast \text{a} \: \: \: \: \{\because \text{ab = ba} \: \} \text{ and thus } (\ast)\: \text{is} \\ \text{commutative.}

\text{ii)} \: \underline{\text{Checking for associative property}} : \\ \\ \text{a}\ast ( \text{b} \ast \text{c}) \\ = \text{a} \ast \text{(b + c + bc)} \\ = \text{a + (b + c + bc) + a(b + c + bc)} \\ = \text{a + b + c + ab + bc + ca + abc} \\ \\ \text{and} \: (\text{a} \ast \text{b} )\ast \text{c} \\ = \text{(a + b + ab)}\ast \text{c} \\ = \text{a + b + ab + c + (a + b + ab)c} \\ = \text{a + b + c + ab + bc + ca + abc} \\ \\ \therefore \text{a} \ast (\text{b} \ast \text{c}) = ( \text{a} \ast \text{b}) \ast \text{c} \: \: \text{and thus} \: ( \ast) \: \text{is} \\ \text{associative.}

\boxed{\text{b}}\\ \\ \text{The given relation is} \\ \text{a} \ast \text{b = a - b + ab} \\ \\ \text{i)}\:\underline{\text{Checking for commutative property}} : \\ \\ \text{a} \ast \text{b = a - b + ab} \\ \text{b} \ast \text{a = b - a + ba} \\ \\ \therefore \text{a} \ast \text{b} \neq \text{b} \ast \text{a} \: \: \: \: \{\because { \text{a - b} \neq \text{b - a}} \: \} \text{ and thus } (\ast)\: \text{is} \\ \text{not commutative.} \\ \\ \text{ii)} \:\underline{\text{Checking for associative property}} : \\ \\ \text{a}\ast ( \text{b} \ast \text{c}) \\ = \text{a} \ast \text{(b - c + bc)} \\ = \text{a - (b - c + bc) + a(b - c + bc)} \\ = \text{a - b + c + ab - bc - ca + abc} \\ \\ \text{and} \: (\text{a} \ast \text{b} )\ast \text{c} \\ = \text{(a - b + ab)}\ast \text{c} \\ = \text{a - b + ab - c + (a - b + ab)c} \\ = \text{a - b - c + ab - bc + ca + abc} \\ \\ \because \text{a} \ast (\text{b} \ast \text{c}) \neq ( \text{a} \ast \text{b}) \ast \text{c}, \: ( \ast) \: \text{is not associative}.

\bigstar \underline{\large{\text{MarkAsBrainliest}}}\bigstar

Step-by-step explanation: