GATE CSE 1996 | Question: 1.4

Question

Which of the following statements is FALSE?

• A. The set of rational numbers is an abelian group under addition
• B. The set of integers in an abelian group under addition
• C. The set of rational numbers form an abelian group under multiplication
• D. The set of real numbers excluding zero is an abelian group under multiplication

Comments

Note: $\frac{22}{7}$ is rational while $\pi$ is irrational.

So, $\frac{22}{1} \cdot \frac{1}{7} = \frac{22}{7} \in \mathbb{Q}$

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bro pi is not equal to 22/7. We take approximate value of pi as 22/7

Answer 1

A Set to qualify as a Abelian Group must satisfy 5 properties.

1. Algebraic Structure (Closure)
2. Semi Group(Associativity)
3. Monoid(Identity Element)
4. Inverse Element must exist for every element in set
5. Commutative Group

Option A, B, D are true they qualify to be an Abelian Group.

Option C: (Rational Number, *)

1. Algebraic Structure (SATISFIES) as closure property satisfies.
2. Semi Group (SATISFIED) as multiplication is associative.
3. Monoid(SATISFIED), Identity element is 1, a*1 = a.
4. Inverse(NOT SATISFIED), this property says that for every element an inverse should exist, such that when we perform the operation with the inverse element we must get back the identity element since this is multiplication reciprocal will be inverse in every case, except 0. (0 is a rational number). Only for 0 this conditions fails
5. Commutative(SATISFIED) Multiplication is commutative as well.

Hence if we could remove 0 from this set it will qualify as abelian Group.

 

Answer 2

A Set to qualify as a Abelian Group must satisfy 5 properties.

1. Algebraic Structure (SATISFIES) 
2. Semi Group (SATISFIED) 
3. Monoid(SATISFIED),
4. Inverse(NOT SATISFIED)
5. Commutative(SATISFIED) 

Hence if we could remove 0 from this set it will qualify as abelian Group

Therefore, The set of rational numbers form an abelian group under multiplication.