22,825 views
33 votes
33 votes

Which of the following statements is FALSE?

  1. The set of rational numbers is an abelian group under addition
  2. The set of integers in an abelian group under addition
  3. The set of rational numbers form an abelian group under multiplication
  4. The set of real numbers excluding zero is an abelian group under multiplication

6 Answers

Best answer
39 votes
39 votes
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.
edited by
7 votes
7 votes

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


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.

2 votes
2 votes

 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

0 votes
0 votes
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:

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