Tag: sets and relations

Questions Related to 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. $(7,11)$


Correct Option: A
Explanation:

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

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

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


Correct Option: D
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.

$P\wedge (q \vee r)=(p\wedge q)\vee r$. This law is known as 

  1. Commutative law

  2. Associative law

  3. De-Morgan's law

  4. Distributive law


Correct Option: A

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


Correct Option: A
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 $$.

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.


Correct Option: C
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 $(a-b)-c\neq a-(b-c)$

If $A={6,7,8,9}$ and $B={0,1,3,4,5}$, then $A \cap B$ is

  1. $\phi$

  2. ${6,7,8,9}$

  3. ${1,2,3,4,5,6,7,8,9}$

  4. None of the above


Correct Option: A
Explanation:

$A\cap B= B \cap A= $ the elements which are common in both A and B. 

Hence $A\cap B= B \cap A$ $=\phi$
Since no element is common in A and B

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


Correct Option: C
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 $

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


Correct Option: A
Explanation:

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

Hence, $A$ is correct.

$"*"$ is said to be commutative in $A$ for all $a,b  \epsilon  A$

  1. $a+b=b+a$

  2. $a*b=b*a$

  3. $a-b=b-a$

  4. $a*b\neq b*a$


Correct Option: B
Explanation:

This is the definition of commutativeness for all binary operators.

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. $A$


Correct Option: B
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 $\ast$.