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 = ab / 2
(iv) On Z + , define a * b = 2 ^ (ab)
(v) On Z + , define a * b = a^b
(vi) On R − {−1}, define a * b = a / (b+1)
Answers
here a,b ∈ R then, (a - b) ∈ R
so, * is binary operation.
we can see that 1*2 = 1 - 2 = -1
and 2*1 = 2 - 1 = 1
e.g., 1*2 ≠ 2*1 , where 1,2∈R, therefore, * is not commutative.
Also, we get,
(1 * 2) * 3 = (1 – 2) * 3 = -1 * 3 = -1 -3 = -4
1 * (2 * 3) = 1 * (2 – 3) = 1 * -1 = 1 – (-1) = 2
e.g., (1*2)*3 ≠ 1*(2*3) , where 1,2,3∈ R
therefore, the operation * is not associative.
(ii) On Q, define a*b = ab + 1
here a , b ∈ Q so, (ab + 1) ∈ Q
therefore, * is binary operation.
now we can see that 1*2 = 1 × 2 + 1 = 3
2*1 = 2 × 1 + 1 = 3
e.g., 1*2 = 2*1 = 3 , where 1,2 ∈ Q
therefore , * is commutative.
also we get,
(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)
⇒ the operation * is not associative.
(iii) On Q, define a*b = ab/2
we can see that 1*2 = 1 × 2/2 = 2/2 = 1
2*1 = 2 × 1/2 = 2/2 = 1
e.g., 1*2 = 2*1 = 1 , where 1,2 ∈ Q.
therefore, * is commutative.
also we get,
(1*2)*3 = (1 × 2/2)*3 = 1*3 = 1 × 3/2 = 3/2
1*(2*3) = 1*(2×3/2) = 1*3 = 1×3/2 = 3/2
e.g., (1*2)*3 = 1*(2*3) , where 1,2,3∈ Q
therefore, * is associative.
(iv) It is given that on Z⁺, define a ∗ b = 2^ab
2^ab ∈ Z⁺, so operation * is binary
We know that ab = ba for a,b ϵ Z⁺
⇒ 2^ab = 2^ab for a,b ∈ Z⁺
⇒ a * b = a * b for a, b ∈ Z⁺
⇒ The operation * is commutative.
Also, we get,
(1 * 2) * 3 = 2⁽¹×²⁾ *3 = 4 * 3 = 2⁽⁴׳⁾= 2¹²
1 * (2 * 3) = 1 * 2⁽²×³⁾ = 1 * 2⁶ = 2⁽¹×⁶⁴⁾ =2⁶⁴
⇒ (1 * 2) * 3 ≠ 1 * (2 * 3)
⇒ The operation * is not associative
(v) is given that On Z⁺, define a ∗ b = a^b
a^b Z⁺, so operation * is binary.
We know that ab = ba for a, b ϵ Z⁺
⇒ 1 * 2 =1² and 2 * 1 = 2¹ =2
⇒ 1 * 2 ≠ 2 *1, where 1,2 ϵ Z⁺
⇒ The operation * is not commutative.
Also, we get,
(1 * 2) * 3 = 2³ *3 = 8 * 3 = 24
1 * (2 * 3) = 1 * 3² = 1 * 9 = 9
⇒ (1 * 2) * 3 ≠ 1 * (2 * 3), where 1,2,3 ϵ Z⁺
⇒ The operation * is not associative.
(vi) On R-{-1}, define a*b = a/(b + 1)
we can see that 1*2 = 1/(2 + 1) = 1/3
2*1 = 2/(1 + 1) = 1
e.g., 1*2 ≠ 2*1 where 1,2 belongs to R -{-1}
therefore, * is not commutative.
also we get.
(1*2)*3 = [1/(2 + 1) ]*3 = 1/3*3 = 1/3/(3 + 1) = 1/12
1*(2*3) = 1*[2/(3 +1)] = 1*2/4 = 1/(2/4 + 1) = 2/3
e.g., (1*2)*3 ≠ 1*(2*3) , where 1,2,3 belongs to R-{-1}
therefore , * is not associative.
Answer:
(i) On Z, define a* b = a - b
here a,b ∈ R then, (a - b) ∈ R
so, * is binary operation.
we can see that 1*2 = 1 - 2 = -1
and 2*1 = 2 - 1 = 1
e.g., 1*2 ≠ 2*1 , where 1,2∈R, therefore, * is not commutative.
Also, we get,
(1 * 2) * 3 = (1 – 2) * 3 = -1 * 3 = -1 -3 = -4
1 * (2 * 3) = 1 * (2 – 3) = 1 * -1 = 1 – (-1) = 2
e.g., (1*2)*3 ≠ 1*(2*3) , where 1,2,3∈ R
therefore, the operation * is not associative.
(ii) On Q, define a*b = ab + 1
here a , b ∈ Q so, (ab + 1) ∈ Q
therefore, * is binary operation.
now we can see that 1*2 = 1 × 2 + 1 = 3
2*1 = 2 × 1 + 1 = 3
e.g., 1*2 = 2*1 = 3 , where 1,2 ∈ Q
therefore , * is commutative.
also we get,
(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)
⇒ the operation * is not associative.
(iii) On Q, define a*b = ab/2
we can see that 1*2 = 1 × 2/2 = 2/2 = 1
2*1 = 2 × 1/2 = 2/2 = 1
e.g., 1*2 = 2*1 = 1 , where 1,2 ∈ Q.
therefore, * is commutative.
also we get,
(1*2)*3 = (1 × 2/2)*3 = 1*3 = 1 × 3/2 = 3/2
1*(2*3) = 1*(2×3/2) = 1*3 = 1×3/2 = 3/2
e.g., (1*2)*3 = 1*(2*3) , where 1,2,3∈ Q
therefore, * is associative.
(iv) It is given that on Z⁺, define a ∗ b = 2^ab
2^ab ∈ Z⁺, so operation * is binary
We know that ab = ba for a,b ϵ Z⁺
⇒ 2^ab = 2^ab for a,b ∈ Z⁺
⇒ a * b = a * b for a, b ∈ Z⁺
⇒ The operation * is commutative.
Also, we get,
(1 * 2) * 3 = 2⁽¹×²⁾ *3 = 4 * 3 = 2⁽⁴׳⁾= 2¹²
1 * (2 * 3) = 1 * 2⁽²×³⁾ = 1 * 2⁶ = 2⁽¹×⁶⁴⁾ =2⁶⁴
⇒ (1 * 2) * 3 ≠ 1 * (2 * 3)
⇒ The operation * is not associative
(v) is given that On Z⁺, define a ∗ b = a^b
a^b Z⁺, so operation * is binary.
We know that ab = ba for a, b ϵ Z⁺
⇒ 1 * 2 =1² and 2 * 1 = 2¹ =2
⇒ 1 * 2 ≠ 2 *1, where 1,2 ϵ Z⁺
⇒ The operation * is not commutative.
Also, we get,
(1 * 2) * 3 = 2³ *3 = 8 * 3 = 24
1 * (2 * 3) = 1 * 3² = 1 * 9 = 9
⇒ (1 * 2) * 3 ≠ 1 * (2 * 3), where 1,2,3 ϵ Z⁺
⇒ The operation * is not associative.
(vi) On R-{-1}, define a*b = a/(b + 1)
we can see that 1*2 = 1/(2 + 1) = 1/3
2*1 = 2/(1 + 1) = 1
e.g., 1*2 ≠ 2*1 where 1,2 belongs to R -{-1}
therefore, * is not commutative.
also we get.
(1*2)*3 = [1/(2 + 1) ]*3 = 1/3*3 = 1/3/(3 + 1) = 1/12
1*(2*3) = 1*[2/(3 +1)] = 1*2/4 = 1/(2/4 + 1) = 2/3
e.g., (1*2)*3 ≠ 1*(2*3) , where 1,2,3 belongs to R-{-1}
therefore , * is not associative.
Step-by-step explanation: