Tag: sets and relations

Quizzes (8) Questions (20)

Questions Related to sets and relations

Multiple choice mathematics and statistics binary operations properties of binary operations discrete mathematics sets and relations

If the binary operation $*$ is defined on a set of ordered pairs of real numbers as $(a, b) * (c, d) = (a \times d + b \times c, b \times d)$ and is associative, then $(1, 2) * (3, 5) * (3, 4)$ is equal to

  1. $(74,40)$
  2. $(32,40)$
  3. $(23,11)$
  4. $(7,11)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Binary operation of odered pairs ;$(a,b)\ast(c,d)=(a\times d+b\times c,b\times d)$ is associative.

$\Rightarrow (1,2)\ast(3,5)\ast(3,4)=((1,2)\ast(3,5))\ast(3,4)$
$=(1\times 5+2\times 3,2\times 5)\ast(3,4)$
$=(11,10)\ast(3,4)$
$=(74,40)$

Multiple choice mathematics and statistics binary operations properties of binary operations discrete mathematics sets and relations

The set of all real numbers under the usual multiplication operation is not a group since

  1. multiplication is not a binary operation

  2. multiplication is not associative

  3. identity element does not exist

  4. zero has no inverse
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Set of real number i.e. $(-\infty,0)\cup(0,\infty)$ is under usual multiplication operation because $0\in R$ and zero do not have an inverse i.e. it can not give ordered airs to be included in a group.

Multiple choice mathematics and statistics binary operations properties of binary operations discrete mathematics sets and relations

If * is defined on the set R of all real numbers by $a*b=\sqrt{a^2+b^2}$, find the identity element in R with respect to *.

  1. 0

  2. 1

  3. 2

  4. 3
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let e be the identity element in R with respect to . Then,

$a*e=a=e*a$ for all $a\in R$
$a*e=a$ and $e*a=a$ for all $a\in R$
$\sqrt{a^2+e^2}=a$ and $\sqrt{e^2+a^2}=a$ for all $a\in R$
$a^2+e^2=a^2$ and $e^2+a^2=a^2$ for all $a\in R$
$e=0$
Hence, 0 is the identity element in R with respect to $$.

Multiple choice mathematics and statistics binary operations properties of binary operations discrete mathematics sets and relations

Subtraction of integers is an operation that is

  1. commutative and associative

  2. not commutative but associative

  3. neither commutative nor associative

  4. commutative but not associative.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

By property of integers, subtraction of integers is neither associative nor commutative.
Example
$9-3=6$
$3-6=-6\neq(9-3)$
$1-(4-8)=1-(-4)=5$
$(1-4)-8=-3-8=-11$
Hence $(a-b)-c\neq a-(b-c)$

Multiple choice mathematics and statistics binary operations properties of binary operations discrete mathematics sets and relations

If $\displaystyle M\cup N=N\cup R$ and $\displaystyle M\cap  N=N\cap R$  then which of the following is necessarily true?

  1. M=N

  2. N=R

  3. M=R

  4. M=N=R
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

We know that Union and Intersection of sets is commutative.
We see that if $ R $ is replaced by $ M $, then the relations show the commutative property being satisfied.

So, $ M = R $

Multiple choice mathematics and statistics binary operations properties of binary operations discrete mathematics sets and relations

A closed set with respect to some binary operation is called semi- group if

  1. $*$ is associative
  2. $*$ is commutative
  3. $*$ is anti-commutative

  4. identity element exists
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In mathematics, a semigroup is an algebraic structure consisting of a set together with an associative binary operation.

Hence, $A$ is correct.

Multiple choice mathematics and statistics binary operations properties of binary operations discrete mathematics sets and relations

If $A \ast B = A \cap B$ on $P(X)$, then identify for $\ast$ is ________$(X \neq \phi)$

  1. $\phi$
  2. $X$
  3. $U$
  4. $A$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Given: $A \ast B = A \cap B$ on $P(X)$
Now, $P(X)$ is a power set of $X$ which is the set of all subsets of $X$
So, $P(X)$ will have $\phi$ as its smallest set and $X$ will be the largest set.
Since, $A\ast B$ is defined on $P(X)$
So, $A \cap B\in P(X)$
So, $A\subseteq X$ and $B\subseteq X$
$\therefore A\cap X=A$ and $X\cap A=A$
$\therefore A\ast X=A$ and $X\ast A=A$
Hence, $X$ is the identity element for $\ast$.