For each binary operation * defined below, determine whether * is commutative or associative.
(i) On Z, define a * b=a−b
(ii) On Q, define a * b=ab+1
(iii) On Q, define a * b= \(\frac {ab}{2}\).
(iv) On Z+, define a * b=2ab
(v) On Z+, define a * b=ab
(vi) On R−{−1},define a * b= \(\frac {a}{b+1}\)
(i) On Z, * is defined by a * b = a − b.
It can be observed that 1 * 2 = 1 − 2 = 1 and 2 * 1 = 2 − 1 = 1.
∴1 * 2 ≠ 2 * 1; where 1, 2 ∈ Z
Hence, the operation * is not commutative.
Also we have:
(1 * 2) * 3 = (1 − 2) * 3 = −1 * 3 = −1 − 3 = −4
1 * (2 * 3) = 1 * (2 − 3) = 1 * −1 = 1 − (−1) = 2
∴(1 * 2) * 3 ≠ 1 * (2 * 3) ;
where 1, 2, 3 ∈ Z
Hence, the operation * is not associative.
(ii) On Q, * is defined by a * b = ab + 1.
It is known that: ab = ba ∀ a, b ∈ Q ⇒ ab + 1 = ba + 1∀ a, b ∈ Q ⇒ a * b = a * b ∀a, b ∈ Q
Therefore, the operation * is commutative.
It can be observed that:
(1 * 2) * 3 = (1 × 2 + 1) * 3 = 3 * 3 = 3 × 3 + 1 = 10
1 * (2 * 3) = 1 * (2 × 3 + 1) = 1 * 7 = 1 × 7 + 1 = 8
∴(1 * 2) * 3 ≠ 1 * (2 * 3) ;
where 1, 2, 3 ∈ Q
Therefore, the operation * is not associative.
(iii) On Q, * is defined by a * b = \(\frac{ab}{2}.\)
It is known that:
ab = ba ∀ a, b ∈ Q \(\frac {ab}{2}=\frac{ba}{2}\) ∀ a, b ∈ Q
⇒ a * b = b * a ∀ a, b ∈ Q
Therefore, the operation * is commutative.
For all a, b, c ∈ Q,
we have:
\((a*b)*c=(\frac{ab}{2})*c=\frac{(\frac{ab}{2})}{\frac{c}{2}}=\frac{abc}{4}\).
\(a*(b*c)=a*(\frac{bc}{2})=\frac{a(\frac{bc}{2})}{2}=\frac{abc}{4}\).
Therefore (a * b) * c =a * (b * c)
Therefore, the operation * is associative.
(iv) On Z+, * is defined by a * b = 2ab.
It is known that:
ab = ba ∀ a, b ∈ Z+
⇒ 2ab = 2ba ∀ a, b ∈ Z+
⇒ a * b = b * a ∀ a, b ∈ Z+
Therefore, the operation * is commutative.
It can be observed that:
(1*2)*3=2(1*2)*3 = 4*3=2(4*3)=212
1*(2*3)=1*2x3=1*2 6=1*64=2 64.
∴(1 * 2) * 3 ≠ 1 * (2 * 3) ; where 1, 2, 3 ∈ Z+
Therefore, the operation * is not associative.
(v) On Z++ * is defined by a * b = ab.
It can be observed that:
1*2=12 =1 and 2*1=21=2
∴ 1 * 2 ≠ 2 * 1 ; where 1, 2 ∈ Z+
Therefore, the operation * is not commutative.
It can also be observed that:
\((2*3)*4=2^3*4=8*4=8^4=2^{12}\)
\(2(3*4)=2*3^4=2*81=2^{81}\)
∴(2 * 3) * 4 ≠ 2 * (3 * 4) ; where 2, 3, 4 ∈ Z+
Therefore, the operation * is not associative.
(vi) On R, * − {−1} is defined by \(a*b=\frac {a}{b+1}.\)
It can be observed that \(1*2=\frac{1}{2+1}=\frac{1}{3} and 2*1=\frac{2}{1+1}=\frac{2}{2}=1\).
∴1 * 2 ≠ 2 * 1 ; where 1, 2 ∈ R − {−1}
Therefore, the operation * is not commutative.
It can also be observed that:
\((1*2)*3=\frac{\frac{1}{3}}{3+1}=\frac{1}{12}.\)
\(1*(2*3)=1*\frac{2}{3+1}=1*\frac{2}{4}=\frac{1}{\frac{1}{2+1}}=\frac{1}{3}=\frac{2}{3}\)
∴ (1 * 2) * 3 ≠ 1 * (2 * 3) ; where 1, 2, 3 ∈ R − {−1}
Therefore, the operation * is not associative.
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\} $.
The correct IUPAC name of \([ \text{Pt}(\text{NH}_3)_2\text{Cl}_2 ]^{2+} \) is:
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: