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.
LIST I | LIST II | ||
A. | Range of y=cosec-1x | I. | R-(-1, 1) |
B. | Domain of sec-1x | II. | (0, π) |
C. | Domain of sin-1x | III. | [-1, 1] |
D. | Range of y=cot-1x | IV. | \([\frac{-π}{2},\frac{π}{2}]\)-{0} |
What is the Planning Process?
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.
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There are four main types of binary operations which are: