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

@Deepak Poonia

Sir why D is not false.

If I take two irrational number that is root(2) and pie, then multiplication isn't even defined. We could say we would get a real number from multiplying two irrationals and the structure is closed. But for following associatvity we need to have a proper real value that terminates.

 

Say a*(b*c) = (a*b)*c

Then if a: root(2) and b: pie and c:1

The values for checking associativity is never possible.

 

Please resolve my doubts.

 

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@Tushar Rana

a : √2 = 1.41421356237 == 1.4 ( approximate )
b : π = 3.14 
c : 1

now

a * ( b * c ) = 1.4 * ( 3.14 * 1 ) = 1.4 * 3.14 = 4.396

similarly 

( a * b ) * c  = ( 1.4 * 3.14 ) * 1 = 1.4 * 3.14 = 4.396

so associativity holds, for this real number under multiplication 

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Go classes : https://youtu.be/9VTNQtBUefQ?list=PLIPZ2_p3RNHgsZOgtGQ6Yg_KR7sedZOgT&t=14225

Answer 1

Answer: C

Rational numbers will include 0. As the group should be under multiplication we will not have any inverse element for 0. Thus, not even satisfying the group property.

Comments

Actually, the identity element exists for (Q,*) is 1 but there is no inverse for 0. That's why (Q,*) is not a group.

Answer 2

The identity element of any structure operated on multiplication is 1.

ie, $x*1=x$

But if you have 0 in your set, then

0 multiplied by what = 1? There's no answer to that.

So, having 0 in your set, if it is defined under multiplication would fail to have an inverse.

 

No inverse => Not a group => Not Abelian.

Option C

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Why is D True? Because it excludes 0.

Why are A and B True? Because they're defined on addition. Means their identity is 0.

ie, $x+0=x$

They'll have closure, associativity, identity, inverse and commutativity.

Answer 3

 The identity element exists for (Q, ⨉) is 1 but there is no inverse for 0. That's why (Q, ⨉) is not a group.So, it is not an abelian group.

Option (c) is false.

The correct answer is, (c)The set of rational numbers form an abelian group under multiplication

Answer 4

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 5

For a set to be abelian group:

1. It should be group [closed,associative,identity,inverse]

2.for every 2 pair of elements of the group it shoud be commutative.

option c is not a group because identity element (e=1) and o which is a rational no. fails to have inverse as its multiplication with invere shoud be equal to 1.

Answer 6

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.