Tag: sets and relations

Quizzes (8) Questions (20)

Questions Related to sets and relations

Multiple choice mathematics and statistics sets and relations de morgan's law for set theory complement of sets different sets de morgan's law

Let the universal set, $\xi$ = {$x : 1 \leq  x \leq  15$ and x is an  integer} set H = {x : x is a multiple of 3} and set K = {x : x is an even number}. Find $n(H' \cap K)$.

  1. $2$
  2. $5$
  3. $7$
  4. $13$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

H' includes all numbers from 1 to 15 that are not multiples of 3.


$H'\cap K ={2,4,8,10,14}$

$n(H'\cap K) =5$

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

Let $\ast$ be a binary operation on the set $Q$ of rational numbers as follows:
(i) $a\ast b = a - b$ (ii) $a\ast b = a^{2} + b^{2}$
(iii) $a\ast b = a + ab$ (iv) $a\ast b = (a - b)^{2}$
(v) $a\ast b = \dfrac {ab}{4}$ (vi) $a\ast b = ab^{2}$
Find which of the binary operations are commutative and which are associative

  1. $ii, iv, v$ are commutative and $v$ associative
  2. $ii, iv, v$ are not commutative and $v$ associative
  3. $iii, iv, v$are commutative and $v$ associative
  4. $vi, iv, v$are commutative and $v$ associative
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$(i)$  $a\ast b=a-b$

Check commutative is
$a\ast b=b\ast a$
$a\ast b=a-b$
$b\ast a=b-a$
Since, $a\ast b\neq b\ast a$
$\ast$ is not commutative.
Check associative
$\ast$ is associative if
$(a\ast b)\ast c=a\ast (b\ast c)\ (a\ast b)\ast c={ (a-b) }^{ \ast  }c=(a-b)-c=a-b-c\ a\ast (b\ast c)=a\ast (b-c)=a-(b-c)=a-b+c$
Since $ (a\ast b)\ast c\neq a\ast (b\ast c)$
$\ast$ is not an associative binary operation.
$(ii)$  $a\ast b={ a }^{ 2 }+{ b }^{ 2 }$
Check commutative
$\ast$ is commutative if $a\ast b=b\ast a$
$a\ast b={ a }^{ 2 }+{ b }^{ 2 }\ b\ast a={ b }^{ 2 }+{ a }^{ 2 }={ a }^{ 2 }+{ b }^{ 2 }$
Since $ a\ast b=b\ast a\quad \forall\quad a,b\epsilon Q$
$\ast$ is commutative.
Check associative
$\ast$ is associative if
$(a\ast b)\ast c=a\ast (b\ast c)\ (a\ast b)\ast c=({ a }^{ 2 }+{ b }^{ 2 })\ast c={ ({ a }^{ 2 }+{ b }^{ 2 }) }^{ 2 }+{ c }^{ 2 }\ a\ast (b\ast c)=a\ast ({ b }^{ 2 }+{ c }^{ 2 })={ a }^{ 2 }+{ ({ b }^{ 2 }+{ c }^{ 2 }) }^{ 2 }$
Since $ (a\ast b)\ast c\neq a\ast (b\ast c)$
$\ast$ is not an associative binary operation.
$(iii)$ $a\ast b=a+b$
Check commutative
$\ast$ is commutative is $a\ast b=b\ast a$
$ a\ast b=a+ab;\quad b\ast a=b+ba$
Since $ a\ast b\neq b\ast a$
$\ast$ is not commutative.
$(iv)$ $a\ast b={ (a-b) }^{ 2 }$
Check commutative
$\ast$ is commutative if $a\ast b=b\ast a$
$ a\ast b={ (a-b) }^{ 2 }\quad ;\quad b\ast a={ (b-a) }^{ 2 }={ (a-b) }^{ 2 }$
Since $ a\ast b=b\ast a\quad \forall\quad a,b\epsilon Q$
$\ast$ is commutative.
Check associative
$\ast$ if
$(a\ast b)\ast c=a\ast (b\ast c)\ (a\ast b)\ast c={ (a-b) }^{ 2 }\ast c={ [{ (a-b) }^{ 2 }-c] }^{ 2 }\ a\ast (b\ast c)=a\ast { (b-c) }^{ 2 }={ [a-{ (b-c) }^{ 2 }] }^{ 2 }$
Since $ (a\ast b)\ast c\neq a\ast (b\ast c)$
$\ast$ is not an associative binary operation.
$(v)$ $a\ast b=\cfrac { ab }{ 4 } $
Check commutative.
$\ast$ is commutative if $a\ast b=b\ast a$
$ a\ast b=\cfrac { ab }{ 4 } \quad ;\quad b\ast a=\cfrac { ba }{ 4 } =\cfrac { ab }{ 4 } $
Since $ a\ast b=b\ast a\quad \forall\quad a,b\epsilon Q$
$\ast$ is commutative.
Check associative.
$\ast$ is association if $(a\ast b)\ast c=a\ast (b\ast c)$
$(a\ast b)\ast c=(\cfrac { \cfrac { ab }{ 4 } \ast c }{ 4 } )=\cfrac { abc }{ 16 } \ a\ast (b\ast c)=a\ast (\cfrac { bc }{ 4 } )=\cfrac { a\times \cfrac { bc }{ 4 }  }{ 4 } =\cfrac { abc }{ 16 } $
Since $ (a\ast b)\ast c=a\ast (b\ast c)\quad \forall\quad a,b,c\epsilon Q$
$\ast$ is an associative binary operation.
$(vi)$ $a\ast b={ ab }^{ 2 }$
check commutative.
$\ast$ is commutative if $a\ast b=b\ast a$
$ a\ast b={ ab }^{ 2 }\quad ;\quad b\ast a={ ba }^{ 2 }$
Since $ a\ast b\neq b\ast a$
$\ast$ is not commutative.
Check associative 
$\ast$ is associative if $(a\ast b)\ast c=a\ast (b\ast c)$
$(a\ast b)\ast c={ ab }^{ 2 }\ast c=({ ab }^{ 2 }){ c }^{ 2 }=a{ b }^{ 2 }{ c }^{ 2 }.\ a\ast (b\ast c)=a\ast { bc }^{ 2 }=a{ ({ bc }^{ 2 }) }^{ 2 }=a{ b }^{ 2 }{ c }^{ 4 }$
Since $ (a\ast b)\ast c\neq a\ast (b\ast c)$
$\ast$ is not an associate binary operation.

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

State whether the following statements are true of false. Justify.
(i) For an arbitrary binary operation $\ast$ on as set $N, a\ast a = a\forall a \epsilon N$
(ii) If $\ast$ is a commutative binary operation on $N$, then $a\ast (b\ast c) = (c\ast b) \ast a$

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

i)Given $a\ast a=a\quad \forall a\epsilon N$ on a set $N$ for an arbitary binary operation $\ast$.

The binary operation can be defined as,$a\times b={ a }^{ 2 }$
Clearly ${ a }^{ 2 }\neq a\quad \forall\quad a\epsilon A$.Therefore the statement is false.
ii)Given binary operator $\ast$ on set $N$.We need to prove that $a\ast (b\ast a)=(c\ast b)\ast a$
An operation $\ast$ on $A$  is commutative 
if $a\ast b=b\ast a\quad \forall\quad a,b\epsilon A$
$\therefore a\ast (b\ast a)=(c\ast b)\ast a$
 since $\ast$ is commutative.
$\therefore $the statement is true.

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

Consider a binary operation $\ast$ on $N$ defined as $a\ast b = a^{3} + b^{3}$. Choose the correct answer

  1. Is $\ast$ both associative and commutative?
  2. Is $\ast$ commutative but not associative?
  3. s $\ast$ associative but not commutative?
  4. Is $\ast$ neither commutative nor associative?
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Check commutative 

$\ast$ is commuatative if
$a \ast b=b\ast a$
$a\ast b={ a }^{ 3 }+{ b }^{ 3 }\quad ;\quad b\ast a={ b }^{ 3 }+{ a }^{ 3 }={ a }^{ 3 }+{ b }^{ 3 }$
Since ,$ a\ast b=b\ast a,\forall\quad a,b\epsilon N$
$\ast$ is commutative
Check associative
$\ast$ is associative if $(a\ast b)\ast c=a\ast (b\ast c)$
$(a\ast b)\ast c={ ({ a }^{ 3 }+{ b }^{ 3 }) }^{ \ast  }c={ ({ a }^{ 3 }+{ b }^{ 3 }) }^{ 3 }+{ c }^{ 3 }\ a\ast (b\ast c)={ a }^{ 3 }\ast ({ b }^{ 3 }+{ c }^{ 3 })={ { a }^{ 3 } }+{ ({ b }^{ 3 }+{ c }^{ 3 }) }^{ 3 }$
Since, $ (a\ast b)\ast c\neq a\ast (b\ast c),\forall\quad a,b,c\epsilon N$
$\ast$ is not associative.
$\therefore $Is $\ast$ commutative but not associative.

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

Consider the following statements for non empty sets A, B and C
1 $\displaystyle A-\left ( B-C \right )=\left ( A-B \right )\cup C $
2 $\displaystyle A-\left ( B\cup C \right )=\left ( A-B \right )- C $
which of the statements given above is/are correct?

  1. 1 only

  2. 2 only

  3. Both 1 and 2

  4. Neither 1 nor 2
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

1. $A- (B - C) = \displaystyle A-(B\cap C')$
$\displaystyle =A\cap (B\cap C')'$
$\displaystyle =A\cap (B'\cup (C'))$
$\displaystyle = A\cap (B'\cup C)$
Thus, $\displaystyle A-(B-C)\neq (A-B)\cup C$


2. A- $\displaystyle (B\cup C)=A\cap (B\cap C)'$
$\displaystyle = A\cap (B'\cap C)'$
$\displaystyle (A-B)-C=(A\cap B')-C$
$\displaystyle =A\cap B'\cap C'$
$\displaystyle \Rightarrow A-(B\cup C)=(A-B)-C$
Associative property. 

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

The set of integers $Z$ with the binary operation $*$ defined as $a * b = a + b+ 1$ for $a, b, Z$ is a group. The identity element of this group is

  1. $0$
  2. $1$
  3. $-1$
  4. $15$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$a\ast b=a+b+1$   (a,b,z is a group)

at $a=-1 \Rightarrow a\ast b=-1+b+1=b$
at $b=-1  \Rightarrow a\ast b=a-1+1=a$
$\Rightarrow a\ast 0=a+0+1$
$\Rightarrow$ identity element is $-1$.