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 carpenter needs to make a wooden cuboidal box, closed from all sides, which has a square base and fixed volume. Since he is short of the paint required to paint the box on completion, he wants the surface area to be minimum.
On the basis of the above information, answer the following questions :
Find a relation between \( x \) and \( y \) such that the surface area \( S \) is minimum.
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:\[ x^2 = y \quad \text{or} \quad f(x) = x^2 \]
Bittu and Chintu were partners in a firm sharing profit and losses in the ratio of 4 : 3. Their Balance Sheet as at 31st March, 2024 was as follows:
On 1st April, 2024, Diya was admitted in the firm for \( \frac{1}{7} \)th share in the profits on the following terms:
Prepare Revaluation Account and Partners' Capital Accounts.
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: