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Set of rational number of the form m/2n (.m, n integers) is a group under
- a)addition
- b)subtraction
- c)multiplication
- d)division
Correct answer is option 'A'. Can you explain this answer?
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Set of rational number of the form m/2n(.m, n integers) is a group und...
Group under Addition:
To show that the set of rational numbers of the form m/2n is a group under addition, we need to prove the following properties:
1. Closure:
For any two rational numbers m/2n and p/2q, their sum is (mq + np)/(2nq), which is also in the form m/2n. Therefore, the set is closed under addition.
2. Associativity:
For any three rational numbers m/2n, p/2q, and r/2s, the sum (m/2n + p/2q) + r/2s is equal to (mq + np)/(2nq) + r/2s, which can be simplified to (2nsq(mq + np) + 2nqrs)/(4nsq), and further simplified to (2n^2sqmq + 2n^2snp + 2nqrs)/(4nsq). Similarly, m/2n + (p/2q + r/2s) simplifies to (2nqrs(mq + np) + 2n^2snp + 2nqrs)/(4nsq). Since addition is associative for integers, the two expressions are equal. Therefore, the set is associative under addition.
3. Identity Element:
The identity element in this set is 0/1, since for any rational number m/2n, m/2n + 0/1 = (mq + 0)/(2n) = m/2n. Therefore, the set has an identity element.
4. Inverse Element:
For any rational number m/2n, its inverse is -m/2n, since m/2n + (-m/2n) = (mq + (-mq))/(2n) = 0/2n = 0/1, which is the identity element. Therefore, every element in the set has an inverse.
Since the set satisfies all the properties of a group under addition, we can conclude that the set of rational numbers of the form m/2n is a group under addition.
Non-Applicability to Other Operations:
The set of rational numbers of the form m/2n does not form a group under subtraction, multiplication, or division. This is because not all elements have inverses under these operations, which violates the property of having an inverse. For example, if we consider division, the element 1/2 does not have an inverse in the set, as there is no rational number m/2n such that (m/2n) * (1/2) = 1/1. Similarly, the set does not satisfy the closure property for subtraction and multiplication, as the result of these operations may not be in the form m/2n.
Therefore, the correct answer is option 'A' - the set of rational numbers of the form m/2n is a group under addition.
To show that the set of rational numbers of the form m/2n is a group under addition, we need to prove the following properties:
1. Closure:
For any two rational numbers m/2n and p/2q, their sum is (mq + np)/(2nq), which is also in the form m/2n. Therefore, the set is closed under addition.
2. Associativity:
For any three rational numbers m/2n, p/2q, and r/2s, the sum (m/2n + p/2q) + r/2s is equal to (mq + np)/(2nq) + r/2s, which can be simplified to (2nsq(mq + np) + 2nqrs)/(4nsq), and further simplified to (2n^2sqmq + 2n^2snp + 2nqrs)/(4nsq). Similarly, m/2n + (p/2q + r/2s) simplifies to (2nqrs(mq + np) + 2n^2snp + 2nqrs)/(4nsq). Since addition is associative for integers, the two expressions are equal. Therefore, the set is associative under addition.
3. Identity Element:
The identity element in this set is 0/1, since for any rational number m/2n, m/2n + 0/1 = (mq + 0)/(2n) = m/2n. Therefore, the set has an identity element.
4. Inverse Element:
For any rational number m/2n, its inverse is -m/2n, since m/2n + (-m/2n) = (mq + (-mq))/(2n) = 0/2n = 0/1, which is the identity element. Therefore, every element in the set has an inverse.
Since the set satisfies all the properties of a group under addition, we can conclude that the set of rational numbers of the form m/2n is a group under addition.
Non-Applicability to Other Operations:
The set of rational numbers of the form m/2n does not form a group under subtraction, multiplication, or division. This is because not all elements have inverses under these operations, which violates the property of having an inverse. For example, if we consider division, the element 1/2 does not have an inverse in the set, as there is no rational number m/2n such that (m/2n) * (1/2) = 1/1. Similarly, the set does not satisfy the closure property for subtraction and multiplication, as the result of these operations may not be in the form m/2n.
Therefore, the correct answer is option 'A' - the set of rational numbers of the form m/2n is a group under addition.
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Community Answer
Set of rational number of the form m/2n(.m, n integers) is a group und...
Satisfy 4 conditions
1.Closer
2.Associative
3. Identity
4. Inverse
1.Closer
2.Associative
3. Identity
4. Inverse
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Set of rational number of the form m/2n(.m, n integers) is a group undera)additionb)subtractionc)multiplicationd)divisionCorrect answer is option 'A'. Can you explain this answer?
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