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A binary operation Ηon a set of integers is defined as x⊙y=x2+y2+2xy. Which one of the following statements is true aboutΗ?
- a)Commutative but not associative
- b)Both commutative and associative
- c)Associative but not commutative
- d)Neither commutative nor associative
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
A binary operation Ηon a set of integers is defined as x⊙y=x2+y2+2xy....
1.Check for commutative
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(x⊙y)=(y⊙x)
x2 + y2 + 2xy = y2 + x2 + 2yx
Since LHS = RHS
So ⊙ is commutative.
2. Check for associative:
(x⊙y)⊙z=x⊙(y⊙z)
(x2 + y2 + 2xy) Η z = x Η ( y2 + z2 + 2yz)
x4 + y. + 4x2y2 + 2x2y2 + 4xy3 + 4x3y + z2 + 2x2z + 2y2z + 4xyz ≠ x2 + y4 + z4 + 4y2z2 + 2y2z2 + 4y3z + 4yz3 + 2xy2 + 4xyz
So A is not associative.
Most Upvoted Answer
A binary operation Ηon a set of integers is defined as x⊙y=x2+y2+2xy....
To determine whether the binary operation ⊙ is commutative or associative, we need to examine its properties.
Commutative Property:
A binary operation ⊙ is commutative if for any two elements x and y in the set, x⊙y = y⊙x. In other words, the order of the operands does not affect the result.
Associative Property:
A binary operation ⊙ is associative if for any three elements x, y, and z in the set, (x⊙y)⊙z = x⊙(y⊙z). In other words, the grouping of the operands does not affect the result.
Let's analyze the given operation ⊙ to determine whether it satisfies these properties.
Commutative Property:
To check if ⊙ is commutative, we need to verify if x⊙y = y⊙x holds true for any x and y in the set.
Let's take arbitrary values for x and y and evaluate both sides of the equation:
x⊙y = x^2 * y^2 / (2xy)
y⊙x = y^2 * x^2 / (2yx)
If we simplify both expressions, we get:
x⊙y = y⊙x = xy
Since xy is not always equal to x⊙y, the operation ⊙ is not commutative.
Associative Property:
To check if ⊙ is associative, we need to verify if (x⊙y)⊙z = x⊙(y⊙z) holds true for any x, y, and z in the set.
Let's take arbitrary values for x, y, and z and evaluate both sides of the equation:
(x⊙y)⊙z = ((x^2 * y^2) / (2xy))⊙z = (((x^2 * y^2) / (2xy))^2 * z^2) / (2((x^2 * y^2) / (2xy))z)
x⊙(y⊙z) = x⊙((y^2 * z^2) / (2yz)) = (x^2 * ((y^2 * z^2) / (2yz))^2) / (2xy((y^2 * z^2) / (2yz)))
If we simplify both expressions, we get:
(x⊙y)⊙z = x⊙(y⊙z) = xyz
Since xyz is not always equal to (x⊙y)⊙z, the operation ⊙ is not associative.
Therefore, the correct answer is option (A) - Commutative but not associative.
Commutative Property:
A binary operation ⊙ is commutative if for any two elements x and y in the set, x⊙y = y⊙x. In other words, the order of the operands does not affect the result.
Associative Property:
A binary operation ⊙ is associative if for any three elements x, y, and z in the set, (x⊙y)⊙z = x⊙(y⊙z). In other words, the grouping of the operands does not affect the result.
Let's analyze the given operation ⊙ to determine whether it satisfies these properties.
Commutative Property:
To check if ⊙ is commutative, we need to verify if x⊙y = y⊙x holds true for any x and y in the set.
Let's take arbitrary values for x and y and evaluate both sides of the equation:
x⊙y = x^2 * y^2 / (2xy)
y⊙x = y^2 * x^2 / (2yx)
If we simplify both expressions, we get:
x⊙y = y⊙x = xy
Since xy is not always equal to x⊙y, the operation ⊙ is not commutative.
Associative Property:
To check if ⊙ is associative, we need to verify if (x⊙y)⊙z = x⊙(y⊙z) holds true for any x, y, and z in the set.
Let's take arbitrary values for x, y, and z and evaluate both sides of the equation:
(x⊙y)⊙z = ((x^2 * y^2) / (2xy))⊙z = (((x^2 * y^2) / (2xy))^2 * z^2) / (2((x^2 * y^2) / (2xy))z)
x⊙(y⊙z) = x⊙((y^2 * z^2) / (2yz)) = (x^2 * ((y^2 * z^2) / (2yz))^2) / (2xy((y^2 * z^2) / (2yz)))
If we simplify both expressions, we get:
(x⊙y)⊙z = x⊙(y⊙z) = xyz
Since xyz is not always equal to (x⊙y)⊙z, the operation ⊙ is not associative.
Therefore, the correct answer is option (A) - Commutative but not associative.
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