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Question
Let * be a binary operation on the set Q of rational numbers as follows:
(i) a * b = a − b
(ii) a * b = a2 + b2
(iii) a * b = a + ab
(iv) a * b = (a − b)2
(v) a * b = ab/4
(vi) a * b = ab2
Find which of the binary operations are commutative and which are associative.
Solution
(i) On Q, the operation * is defined as a * b = a − b.
It can be observed that:
`1/2 * 1/3 = 1/2 - 1/3 = (3-2)/6 = 1/6` " and " `1/3 * 1/2 = 1/3- 1/2 = (2-3)/6 = (-1)/6`
`:. 1/2 * 1/3 != 1/3 * 1/2` where `1/2, 1/3 in Q`
Thus, the operation * is not commutative.
It can also be observed that:
`(1/2 * 1/3) * 1/4 = (1/2 - 1/3) * 1/4 = 1/6 * 1/4 = 1/6 - 1/4 = (2 -3)/12 = (-1)/12`
`1/2 * (1/3 * 1/4) = 1/2 * (1/3 - 1/4) = 1/2 * 1/12 = 1/2 - 1/12 = (6 -1)/12 = 5/12`
`:. (1/2 * 1/3) * 1/4 != 1/2 * (1/3 * 1/4)` where `1/2, 1/3, 1/4 in Q`
Thus, the operation * is not associative.
(ii) On Q, the operation * is defined as a * b = a2 + b2.
For a, b ∈ Q, we have:
`a* b = a^2 + b^2 = b^2 + a^2 = b * a`
∴a * b = b * a
Thus, the operation * is commutative.
It can be observed that:
(1*2)*3 =(12 + 22)*3 = (1+4)*3 = 5*3 = 52+32= 25 + 9 = 34
1*(2*3)=1*(22+32) = 1*(4+9) = 1*13 = 12+ 132 = 1 + 169 = 170
∴ (1*2)*3 ≠ 1*(2*3) , where 1, 2, 3 ∈ Q
Thus, the operation * is not associative.
(iii) On Q, the operation * is defined as a * b = a + ab.
It can be observed that:
`1 *2 = 1 + 1 xx 2 = 1 + 2 = 3`
`2 * 1 = 2 + 2 xx 1 = 2 + 2 =4`
`:. 1 * 2 != 2 *1 where 1, 2 in Q`
Thus, the operation * is not commutative.
It can also be observed that:
`(1 * 2)*3 = (1 + 1 xx 2) * 3 = 3 * 3 = 3 + 3 xx 3 = 3 + 9 = 12`
`1 * (2 * 3) = 1 *(2+2xx3) = 1 + 1 xx 8 = 9`
`:.(1 * 2)*3 != 1 *(2 * 3)` where 1, 2. 3 ∈ Q
Thus, the operation * is not associative.
(iv) On Q, the operation * is defined by a * b = (a − b)2.
For a, b ∈ Q, we have:
a * b = (a − b)2
b * a = (b − a)2 = [− (a − b)]2 = (a − b)2
∴ a * b = b * a
Thus, the operation * is commutative.
It can be observed that:
`(1 * 2)*3 = (1 - 2)^2 * 3 = (-1)^2 * 3 = 1 * 3 = (1-3)^2 = (-2)^2 = 4`
`1 * (2 * 3) = 1 * (2 - 3)^2 = 1*(-1)^2 = 1*1 = (1 - 1)^2 = 0`
`:. (1 * 2) * 3 != 1 *(2 * 3)` where 1,2,3 ∈ Q
Thus, the operation * is not associative.
On Q, the operation * is defined as `a * b = "ab"/4`
For a, b ∈ Q, we have:
`a * b = "ab"/4 = ba/4 = b * a`
∴ a * b = b * a
Thus, the operation * is commutative.
For a, b, c ∈ Q, we have:
`(a * b) * c = "ab"/4 * c = ("ab"/4 . c)/4 = "abc"/16`
`a * (b * c) = a * bc/4= (a . "bc"/4)/4 = "abc"/16`
= ∴(a * b) * c = a * (b * c)
Thus, the operation * is associative.
(vi) On Q, the operation * is defined as a * b = ab2
It can be observed that:
`1/2 * 1/3 = 1/2 . (1/3)^2 = 1/2 . 1/9 = 1/18`
`1/3 * 1/2 = 1/3 . (1/2)^2 = 1/3 . 1/4 = 1/12`
`:. 1/2 * 1/3 != 1/3 * 1/2` where `1/2, 1/3 in Q`
Thus, the operation * is not commutative.
It can also be observed that:
`(1/2 * 1/3) * 1/4 = [1/2.(1/3)^2]* 1/4 = 1/18 * 1/4 = 1/18 . (1/4)^2 = 1/(18xx16)`
`1/2 * (1/3 * 1/4) = 1/2 * [1/3 . (1/4)^2] = 1/2 * 1/48 = 1/2 . (1/48)^2 = 1/(2 xx (48)^2)`
`:. (1/2 * 1/3) * 1/4 != 1/2 (1/3 * 1/4)` where `1/2, 1/3, 1.4 in Q`
Thus, the operation * is not associative.
Hence, the operations defined in (ii), (iv), (v) are commutative and the operation defined in (v) is associative.
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