GATE Overflow Test Series | Discrete Mathematics | Test 3 | Question: 5

Question

Which of the following is NOT correct? (Mark all the appropriate choices)

• A. $(\mathbb{R^{\ast}},\times),$ the set of real numbers excluding $0$, under multiplication operation is a group.
• B. $(\mathbb{Q},\times),$ the set of rational numbers under multiplication operation is a group.
• C. $(\mathbb{Z},+),$ the set of integers under addition operation is a group.
• D. $\mathbb{Z}_{n},$ the set of integers $\{0, 1, \ldots, n-1\}$, under addition modulo $n$ operation is a group.

Comments

What is the inverse of 0 in opt D??

How it is a group??
@gatecse

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@Smrutiranjan Dalai here in option d it is addition modulo and since 0 is identity element so its inverse is 0 itself

Answer 1

$(\mathbb{Q},\times)$ is an algebraic structure because  $\mathbb{Q}$ is closed under multiplication operation.
 
It is a semi group because $'\times'$ is associative on $\mathbb{Q}$.

It is a monoid because identity element $e=1\in \mathbb{Q}$.

A monoid $(\mathbb{Q},\times)$ with identity element $'e'$ is called a group if for each element $a\in \mathbb{Q},$ there exists an element $b\in \mathbb{Q},$ such that $(a \times b) = (b \times a) = e.$

Then $'b'$ is called inverse of an element $'a',$ denoted by $a^{-1}$.

Now, $a\times b = 1$

$\implies \boxed{b = \dfrac{1}{a}}$

Inverse of $'a'$ is $\dfrac{1}{a}$

Inverse of $'0'$ is $\dfrac{1}{0}\notin \mathbb{Q}$

It is not a group.

So, the correct answer is $(B)$. For $A, C$ and $D,$ both identity as well as inverse exists.

Comments

@gatecse Sir in option D inverse of 0 is n. Then how it is a group?

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@Abhinandan1 here in option d it is addition modulo and since 0 is identity element so its inverse is 0 itself not n.

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For 1 the inverse will be n-1, for 2 it will be n-2 and so on