2 Marks

Question 12 Marks

Given a non empty set X, consider P (X) which is the set of all subsets of X.
Define the relation R in P (X) as follows:
For subsets A, B in P (X), ARB if and only if A $\subset$ B. Is R an equivalence relation on P (X)? Justify your answer.

Answer

A $\subset$ A $\therefore$ R is reflexive.
If A $\subset$ B then B $\subset$ A is not true $\therefore$ R is not symmetric.
If A $\subset$ B, B $\subset$ C, then A $\subset$ C $\therefore$ R is transitive.

Therefore, R is not equivalent relation.

View full question & answer→

Question 22 Marks

Show that the function f : R $\rightarrow$ {x $\in$ R : -1 < x < 1} defined by $f(x) = \frac{x}{{1 + |x|}}$, x $\in$ R is one-one and onto function.

Answer

f is one-one: For any x, y $\in$ R, we have f(x) : f(y)
$\Rightarrow \frac{x}{{1 + |x|}} = \frac{y}{{|y| + 1}}$
$\Rightarrow$ xy + x = xy + y
$\Rightarrow$ x = y
Therefore, f is one-one function.
If f is one-one, let y = R – {1}, then f(x) = y
$\Rightarrow \frac{x}{{x + 1}} = y$
$\Rightarrow x = \frac{y}{{1 - y}}$
It is clear that x $\in$ R for all y = R – {1}, also x $\ne$=-1
Because x = -1
$\Rightarrow \frac{y}{{1 - y}} = - 1$
$\Rightarrow$ y = -1 + y which is not possible.
Thus for each R – {1} there exists $x = \frac{y}{{1 - y}} \in$ R – {1} such that
$f(x) = \frac{x}{{x + 1}} = \frac{{\frac{y}{{1 - y}}}}{{\frac{y}{{1 - y}} + 1}} = y$
Therefore f is onto function.

View full question & answer→

Question 32 Marks

Let A and B be sets. Show that f : A $\times $ B $\rightarrow$ B $\times $ A such that f(a, b) = (b, a) is a bijective function.

Answer

Injectivity: Let (a₁, b₁) and (a₂, b₂) $\in$ A $\times$ B such that f(a₁, b₁) = f(a₂, b₂)
$\Rightarrow$ (b₁, a₁) = (b₂, a₂)
$\Rightarrow$ b₁ = b₂ and a₁ = a₂
$\Rightarrow$ (a₁, b₁) = (a₂, b₂) for all (a₁, b₁), (a₂, b₂) $\in$ A $\times$ B
So, f is injective.
Surjectivity: Let (b, a) be an arbitrary element of B $\times$ A. Then b $\in$ B and a $\in$ A.
$\Rightarrow$ (a, b) $\in$ A $\times$ B
Thus, for all (b, a) $\in$ B $\times$ A, there exists (a, b) $\in$ A $\times$ B such that f(a, b) = (b, a)
So, f : A $\times$ B $\rightarrow$ B $\times$ A is an onto function, therefore f is bijective.

View full question & answer→

Question 42 Marks

State whether the function is one-one, onto or bijective. Justify your answer. f: R $\rightarrow$ R defined by f(x) = 1+ x²

Answer

Let x₁, x₂ $\in$ R
If f(x₁) = f(x₂)
$1 + x_1^2 = 1 + x_1^2$
$x_1^2 = x_1^2$
${x_1} = \pm {x_2}$
Hence not one - one
y = 1 + x²
$x = \pm \left( {\sqrt {1 - y} } \right)$
$f\left( {\sqrt {1 - y} } \right) = 1 + (1 - y) = 2 - y \ne y$
Therefore, f is not onto.

View full question & answer→

Question 52 Marks

State whether the function is one-one, onto or bijective. Justify your answer. f: R $\rightarrow$ R defined by f(x) = 3 - 4x.

Answer

Let (x$_1$, x₂ ) $\in$ R such that
f(x₁) = f(x₂)
3 - 4x₁ = 3 - 4x₂
x₁ = x₂
Hence one–one
Y = 3 - 4x
$x = \left( {\frac{{3 - y}}{4}} \right)$
$f\left( {\frac{{3 - y}}{4}} \right) = 3 - 4\left( {\frac{{3 - y}}{4}} \right)$
f(x) = y
= y
Hence onto also.

View full question & answer→

Question 62 Marks

Show that the Signum Function $f : R \rightarrow R$, given by $f(x)=\left\{\begin{array}{l}1, \text { if } x>0 \\ 0, \text { if } x=0 \text { is neither one-one nor } \\ 1, \text { if } x<0\end{array}\right.$ onto.

Answer

Signum Function f : R $\rightarrow$ R, given by $f(x) = \left\{ {\begin{array}{*{20}{c}} {1,\;if\;x > 0} \\ {0,\;if\;x = 0} \\ { - 1,\;if\;x < 0} \end{array}} \right.$
f(1) = f(2) = 1
Two distinct elements have same image.
$\therefore$ f is not one-one.
Except -1, 0, 1 no other members of co-domain of f has any pre-image its domain.
$\therefore$ f is not onto.
Therefore, f is neither one-one nor onto.

View full question & answer→

Question 72 Marks

Show that the Modulus Function f : R $\rightarrow$ R, given by f(x) = |x|, is neither one-one nor onto, where |x| is x, if x is positive or 0 and |x| is -x, if x is negative.

Answer

Modulus Function f : R $\rightarrow$ R, given by f(x) = |x|
One-one: f(1) = |1| = 1 and f(2) = |2| = 2,so distinct elements have same image.So, f is not one-one.
onto: f takes only positive values, so range(f) = set of positive real numbers $\neq$ R, codomain. So, f is not onto.

View full question & answer→

Question 82 Marks

Prove that the Greatest integer Function f : R $\rightarrow$ R, given by f(x) = [x], is neither one-one nor onto, where [x] denotes the greatest integer less than or equal to x.

Answer

Function f : R $\rightarrow$ R, given by f(x) = [x]
$\because \;1 \leqslant x \leqslant 2$, f(x) = 1
$\therefore$ f(1) = 1 and f(1.1) = 1
$\therefore$ f is not one-one.
f takes only integer values, therefore range(f) = set of integers,which is not equal to R, codomain.
Therefore, f is not onto.

View full question & answer→

Question 92 Marks

Check the injectivity and surjectivity of the below function:
f : Z $\rightarrow$ Z given by f(x) = x³

Answer

It is given that f : Z $\rightarrow$ Z given by f (x) = x³
We can see that for x, y $\in$ N,
f(x) = f(y)
$\Rightarrow$ x³ = y³
$\Rightarrow$ x = y
$\Rightarrow$ f is injective.
Now, let 2 $\in$ Z. But, we can see that there does not exists any x in Z such that
f(x) = x³ = 2
$\Rightarrow$ f is not surjective.
Therefore, function f is injective but not surjective.

View full question & answer→

Question 102 Marks

Check the injectivity and surjectivity of the below function:
f : N $\rightarrow$ N given by f(x) = x³

Answer

The function f : N $\rightarrow$ N is given by f (x) = x³
Clearly for x, y $\in$ N,
f(x) = f(y)
$\Rightarrow$ x³ = y³
$\Rightarrow$ x = y
$\Rightarrow$ f is injective.
Now, let 2 $\in$ N. But, we can see that there does not exists any x in N such that
f(x) = x³ = 2
$\Rightarrow$ f is not surjective.
Therefore, function f is injective but not surjective.

View full question & answer→

Question 112 Marks

Check the injectivity and surjectivity of the below function:
f : R $\rightarrow$ R given by f(x) = x²

Answer

f : R $\rightarrow$ R given by f(x) = x²
As f(-1) = f(1) = 1
$\Rightarrow$ -1 and 1 have same image. $\therefore$ f is not injective.
e.g. -2 $\in$ co-domain, but $\sqrt { - 2} \notin$ R=domain of f. $\therefore$ f is not surjective.

View full question & answer→

Question 122 Marks

Check the injectivity and surjectivity of the below function:
f : Z $\rightarrow$ Z given by f(x) = x²

Answer

f : Z $\rightarrow$ Z given by f(x) = x²
Since, $z = \left\{ {0,\; \pm 1,\; \pm 2,\; \pm 3,\;....} \right\}$ therefore, f(-1) = f(1) = 1
$\Rightarrow$ -1 and 1 have same image. $\therefore$ f is not injective.
There are such numbers of co-domain which have no image in domain Z.
e.g. 3 $\in$ co-domain, but $\sqrt 3 \notin$ domain of f. $\therefore$ f is not surjective.

View full question & answer→

Question 132 Marks

Check the injectivity and surjectivity of the below function:
f : N $\rightarrow$ N given by f(x) = x²

Answer

f : N $\rightarrow$ N given by f(x) = x²
If f(x₁) = f(x₂) then $x_1^2 = x_2^2$
$\Rightarrow$ x₁ = x₂
$\therefore$ f is injective.
There are such numbers of co-domain which have no image in domain N.
e.g. 3 $\in$ co-domain N, but there is no pre-image in domain of f.
therefore f is not onto.
$\therefore$ f is not surjective.

View full question & answer→

Question 142 Marks

Show that the function f : R₊ $\rightarrow$ R₊ defined by $f(x) = \frac{1}{x}$ is one-one and onto, where R₊ is the set of all non-zero real numbers. Is the result true, if the domain R₊ is replaced by N with co-domain being same as R₊?

Answer

$f(x) = \frac{1}{x},f:{R_ * } \to {R_ * }$
Part I: $f({x_1}) = \frac{1}{{{x_1}}}$ and $f({x_2}) = \frac{1}{{{x_2}}}$
If f(x₁) = f(x₂) then $\frac{1}{{{x_1}}} = \frac{1}{{{x_2}}}$
$\Rightarrow$ x₁ = x₂
$\therefore$ f is one-one.
$f(x) = \frac{1}{x}$
$\Rightarrow y = \frac{1}{x}$
$\Rightarrow x = \frac{1}{y}$
$\Rightarrow f\left( {\frac{1}{y}} \right) = y$ $\therefore$ f is onto.
Part II: When domain R is replaced by N, co-domain remaining the same, then, f : N$\rightarrow$R$_*$
If f(x₁) = f(x₂)
$\Rightarrow \frac{1}{{{n_1}}} = \frac{1}{{{n_2}}}$
$\Rightarrow$ n₁ = n₂ where n₁, n₂ $\in$ N
$\therefore$ f is one-one.
But, every real number belonging to co-domain may not have a pre-image in N.
e.g. 2 in codomain R_* does not have pre-image in N as if
x in N be pre-image of 2,then f(x) = 2 $\Rightarrow$ $\frac{1}{x}$ = 2 $\Rightarrow$ x = $\frac{1}{2} \notin N$.
$\therefore$ f is not onto.

View full question & answer→

Question 152 Marks

Show that the relation R in the set {1, 2, 3} given by R = {(1, 2), (2, 1)} is symmetric but neither reflexive nor transitive.

Answer

R = {(1, 2), (2, 1)}, so for (a, a), (1, 1) $\notin$ R. $\therefore$R is not reflexive.
Also if (a, b) $\in$R then (b, a) $\in$ R $\therefore$ R is symmetric.
Now (a, b) $\in$ R and (b, c) $\in$ R ,then does not imply (a, c) $\notin$ R as (1,2) $\in$ R and (2,1) $\in$ R but (1,1) $\notin$ R $\therefore$ R is not transitive.
Therefore, R is symmetric but neither reflexive nor transitive.

View full question & answer→

Question 162 Marks

Determine whether the relation is reflexive, symmetric and transitive:
Relation R in the set A of human beings in a town at a particular time given by
R = {(x, y) : x is father of y}

Answer

It is given that R = {(x, y) : x is father of y}
$\Rightarrow$ (x, x) $\notin$ R as x cannot be the father of himself.
$\Rightarrow$ R is not reflexive.
Now, if (x, y) $\in$ R, then x is the father of y.
$\Rightarrow$ But y is not father of x.
$\Rightarrow$ (y, x) $\notin$ R
$\Rightarrow$ R is not symmetric.
Now, let (x, y), (y, z) $\in$ R
$\Rightarrow$ x is the father of y and y is the father of z.
$\Rightarrow$ x is not the father of z.
$\Rightarrow$ Indeed x is the grandfather of z.
$\Rightarrow$ (x, z) $\notin$ R
$\Rightarrow$ R is not transitive.
Therefore, R is neither reflexive, nor symmetric, nor transitive.

View full question & answer→

Question 172 Marks

Determine whether the relation is reflexive, symmetric and transitive:
Relation R in the set A of human beings in a town at a particular time given by
R = {(x, y) : x is wife of y}

Answer

It is given that R = {(x, y) : x is wife of y}
Clearly, (x,x) $\notin$ R as x cannot be the wife of herself.
⇒ R is not reflexive.
Now, if (x,y) $\in$ R, then x is the wife of y.
$\Rightarrow$ But y is not wife of x.
$\Rightarrow$ (y,x) $\notin$ R
$\Rightarrow$ R is not symmetric.
Further, let (x,y), (y,z) $\in$ R
$\Rightarrow$ x is the wife of y and y is the wife of z.
$\Rightarrow$ This is not possible.
$\Rightarrow$ (x,z) $\notin$ R
$\Rightarrow$ R is not transitive.
Therefore, R is neither reflexive, nor symmetric, nor transitive.

View full question & answer→

Question 182 Marks

Determine whether the relation is reflexive, symmetric and transitive:
Relation R in the set A of human beings in a town at a particular time given by
R = {(x, y) : x is exactly 7 cm taller than y}

Answer

It is given that R = {(x, y) : x is exactly 7 cm taller than y}
Clearly, (x,x) $\notin$ R as a human being x cannot be taller than himself.
$\Rightarrow$ R is not reflexive.
Now, if (x,y) $\in$ R, then x is exactly 7 cm taller than y.
$\Rightarrow$ But y is not taller than x.
$\Rightarrow$ (y,x) $\notin$ R
$\Rightarrow$ R is not symmetric.
Further, let (x,y), (y,z) $\in$ R
$\Rightarrow$ x is exactly 7 cm taller than y and y is exactly 7 cm taller than z.
$\Rightarrow$ x is exactly 14 cm taller than z.
$\Rightarrow$ (x,z) $\notin$ R
$\Rightarrow$ R is not transitive.
Therefore, R is neither reflexive, nor symmetric, nor transitive.

View full question & answer→

Question 192 Marks

Determine whether the below relation is reflexive, symmetric and transitive:
Relation R in the set A of human beings in a town at a particular time given by
R = {(x, y) : x and y live in the same locality}

Answer

Given that R = {(x, y) : x and y live in the same locality}
Clearly, (x, x) $\in$ R as x and x live in the same locality.
$\Rightarrow$ R is reflexive.
Now, if (x, y) $\in$ R, then x and y live in the same locality.
$\Rightarrow$ y and x live in the same locality.
$\Rightarrow$ (y, x) $\in$ R
$\Rightarrow$ R is symmetric.
Further, let (x, y), (y, z) $\in$ R
$\Rightarrow$ x and y live in the same locality and y and z live in the same locality.
$\Rightarrow$ x and z live in the same locality
$\Rightarrow$ (x, z) $\in$ R
$\Rightarrow$ R is transitive.
Therefore, R is reflexive, symmetric and transitive.

View full question & answer→

Question 202 Marks

Determine whether the below relation is reflexive, symmetric and transitive:
Relation R in the set A of human beings in a town at a particular time given by
R = {(x, y) : x and y work at the same place}

Answer

It is given that R = {(x, y) : x and y work at the same place}
Clearly, (x, x) $\in$ R, as we can say x and x work at the same place.
$\Rightarrow$ R is reflexive.
Now, if (x, y) $\in$ R, then x and y work on the same place.
$\Rightarrow$ y and x work at the same place.
$\Rightarrow$ (y, x) $\in$ R
$\Rightarrow$ R is symmetric.
Further, let (x, y), (y, z) $\in$ R
$\Rightarrow$ x and y work at the same place and y and z work at the same place.
$\Rightarrow$ x and z work at the same place
$\Rightarrow$ (x, z) $\in$ R
$\Rightarrow$ R is transitive.
Therefore, R is reflexive, symmetric and transitive.

View full question & answer→

Question 212 Marks

Give an example of a relation which is symmetric and transitive but not reflexive

Answer

Let A = {-7, -8}
Define a relation R on A as:
R = { (-7, -7)}
Relation R is not reflexive as (a, a) $\notin$ R
Relation R is symmetric as (-7, -7) $\in$ R and (-7, -7) $\in$ R
Clearly R is transitive.
Therefore, relation R is symmetric and transitive but not reflexive.

View full question & answer→

Question 222 Marks

Give an example of a relation which is reflexive and transitive but not symmetric.

Answer

Let us define a relation R in R as
R = {(a,b) : a³ $\ge$ b³}
It is clear that (a,a) $\in$ R as a³ = a³
$\Rightarrow$ R is reflexive.
Now, (2,1) $\in$ R, but (1,2) $\notin$ R
$\Rightarrow$ R is not symmetric.
Now, let (a,b) (b,c) $\in$ R
$\Rightarrow$ a³$\ge$ b³ and b³$\ge$ c³
$\Rightarrow$ a³$\ge$ c³
$\Rightarrow$ (a,c) $\in$ R
$\Rightarrow$ R is transitive.
Therefore, relation R is reflexive and transitive but not symmetric.

View full question & answer→

Question 232 Marks

Give an example of a relation which is reflexive and symmetric but not transitive.

Answer

Let us take A = {2,4,6}
Define a relation R on A as:
A = {(2,2), (4,4), (6,6), (2,4), (4,2), (4,6), (6,4)}
Relation of R is reflexive as for every a $\in$ A,
(a,a) $\in$ R
$\Rightarrow$ (2,2), (4,4), (6,6) $\in$ R,
Relation R is symmetric as (a,b) $\in$ R
$\Rightarrow$ (b,a) $\in$ R for all a ,b $\in$ R
And Relation R is not transitive as (2,4), (4,6) $\in$ R,
but (2,6) $\notin$ R
Therefore, relation R is reflexive and symmetric but not transitive.

View full question & answer→

Question 242 Marks

Give an example of a relation which is transitive but neither reflexive nor symmetric.

Answer

Let a relation R is defined as:
R = {(a,b): a < b}
For any a $\in$ R, we have (a,a) $\notin$ R as a cannot be strictly less than itself.
In fact, a = a,
Therefore, R is not reflexive.
Now, (1,2) $\in$ R but 2 > 1
$\Rightarrow$ (2,1)) $\notin$ R.
$\Rightarrow$ R is not symmetric.
Now, let (a,b), (b,c) $\in$ R
$\Rightarrow$ a < b and b < c
$\Rightarrow$ a < c
$\Rightarrow$ (a,c) $\in$ R
$\Rightarrow$ R is transitive.
Therefore, relation R is transitive but not reflexive and symmetric.

View full question & answer→

Question 252 Marks

Give an example of a relation which is symmetric but neither reflexive nor transitive.

Answer

Let A = {3,4,5}
Define a relation R on A as R = {(3,4), (4,3)}
Relation R is not reflexive as (3,3), (4,4) ,(5,5) $\notin$ R.
Now, as (3,4) $\in$ R and (4,3) $\in$ R,
R is symmetric.
Further, (3,4),(4,3) $\in$ R, but (3,3) $\notin$ R
$\Rightarrow$ R is not transitive.
Therefore, relation R is symmetric but not reflexive or transitive.

View full question & answer→

Maths 2 Marks

Test yourself on this topic