Consider a binary operation *on N defined as \(a*b=a^3+b^3.\)Choose the correct answer.
Is * both associative and commutative?
Is * commutative but not associative?
Is * associative but not commutative?
Is * neither commutative nor associative?
On N, the operation * is defined as \(a*b=a^3+b^3.\)
For, a, b, ∈ N, we have: \(a*b=a^3+b^3=b^3+a^3=b*a\) [Addition is commutative in N]
Therefore, the operation * is commutative. It can be observed that:
\((1*2)*3=(1^3+2^3)*3=9*3\)= \(9^3+3^3=756\)
\(1*(2*3)=1*(2^3+3^3)=1*(8+27)=1*35=1^3+35^3=1+35^3=42876\)
∴ (1 * 2) * 3 ≠ 1 * (2 * 3) ; where 1, 2, 3 ∈ N
Therefore, the operation * is not associative.
Hence, the operation * is commutative, but not associative. Thus, the correct answer is B.
A school is organizing a debate competition with participants as speakers and judges. $ S = \{S_1, S_2, S_3, S_4\} $ where $ S = \{S_1, S_2, S_3, S_4\} $ represents the set of speakers. The judges are represented by the set: $ J = \{J_1, J_2, J_3\} $ where $ J = \{J_1, J_2, J_3\} $ represents the set of judges. Each speaker can be assigned only one judge. Let $ R $ be a relation from set $ S $ to $ J $ defined as: $ R = \{(x, y) : \text{speaker } x \text{ is judged by judge } y, x \in S, y \in J\} $.
During the festival season, a mela was organized by the Resident Welfare Association at a park near the society. The main attraction of the mela was a huge swing, which traced the path of a parabola given by the equation:\[ x^2 = y \quad \text{or} \quad f(x) = x^2 \]
A binary operation can be understood as a function f (x, y) that applies to two elements of the same set S, such that the result will also be an element of the set S. Examples of binary operations are the addition of integers, multiplication of whole numbers, etc. A binary operation is a rule that is applied on two elements of a set and the resultant element also belongs to the same set.
Read More: Truth Table
There are four main types of binary operations which are: