4 Marks

Question 14 Marks

m is said to be related to n if m and n are integers and m - n is divisible by 13. Does this define an equivalence relation?

Answer

We observe the following properties of relation R.

$\text{Let R} = \big\{(\text{m, n}): \text{(m, n)}\in\text{Z}:\text{m}-\text{n}\ \text{is divisible by} 13\big\}$$$$$

Relexivity: Let m be an arbitrary element of Z. Then, $\text{m}\in\text{R}$

⇒ m - m = 0 = 0 × 13

⇒ m - m is divisible by 13

⇒ (m, m) is reflexive on Z.

Symmetry: Let $(\text{m, n})\in\text{R.}$ Then,

⇒ m - n is divisible by 13

⇒ m - n = 13p

Here, $\text{p}\in\text{Z}$

⇒ n - m = 13(-p)

Here, $-\text{p}\in\text{Z}$

⇒ n - m is divisible by 13

$\Rightarrow\ (\text{n, m})\in\text{R}$ for all $\text{m, n}\in\text{Z}$

So, R is symmetric on Z.

Transitivity: Let (m, n) and (n, o) $\in\text{R}$

⇒ m - n and n - o are divisible by 13

⇒ m - n = 13p and n - o = 13q for some $\text{p, q}\in\text{Z}$

Adding the above two, we get

⇒ m - n + n - o = 13p + 13q

⇒ m - o = 13(p + q)

Here, $\text{p}+\text{q}\in\text{Z}$

⇒ m - o is divisible by 13

$\Rightarrow\ (\text{m, o})\in\text{R}$ for all $\text{m, o}\in\text{Z}$

So, R is transitive on Z.

Hence, R is an equivalence relation on Z.

View full question & answer→

Question 24 Marks

Show that the relation R, defined on the set A of all polygons as R = {(P₁, P₂): P₁ and P₂ have same number of sides}, is an equivalence relation. What is the set of all elements in A related to the right angle triangle T with sides 3, 4 and 5?

Answer

We observe the following properties on R.
Reflexivity: Consider P₁ be an arbitrary element of A.
Then, polygon P₁ and P₁ have the same number of sides.
Since they are one and the same.
Implies that$\text{P}_1, \text{P}_1\in\text{R}$ $$ for all $\text{P}_1\in\text{A.}$
So,R is reflexive on A.
Symmetry: Consider $\text{P}_1,\text{ P}_2\in\text{R}$
Implies that P₁ and P₂ have the same number of sides.
Implies that P₂ and P₁ have the same number of sides.
Implies that $\text{P}_2,\text{ P}_1\in\text{R}$ for all $\text{P}_1,\text{ P}_2\in\text{A}$
So, R is symmetric on A.
Transitivity: Consider $\text{P}_1, \text{P}_2, \text{P}_3\in\text{R}$ $$
Implies that P₁ and P₂ have the same number of sides and P₂ and P₃ have the same number of sides
Implies that P₁, P₂ and P₃ have the same number of sides.
Implies that P₁ and P₃ have the same number of sides.
Implies that $\text{P}_1,\text{ P}_3\in\text{R}$ for all $\text{P}_1,\text{ P}_3\in\text{A.}$
So, R is transitive on A.
Hence, R is an equivalence relation on the set A. Also, the set of all the triangles $\in\text{A}$ is related to the right angle triangle T with the sides 3, 4, 5.

View full question & answer→

Question 34 Marks

Let R be the relation defined on the set A = {1, 2, 3, 4, 5, 6, 7} by R = {(a, b): both a and b are either odd or even}. Show that R is an equivalence relation. Further, show that all the elements of the subset {1, 3, 5, 7} are related to each other and all the elements of the subset {2, 4, 6} are related to each other, but no element of the subset {1, 3, 5, 7} is related to any element of the subset {2, 4, 6}.

Answer

We observe the following properties of R.
Reflexivity: Let a be an arbitrary element of R. Then,
$\text{a}\in\text{R}$
$\Rightarrow\ (\text{a, a})\in\text{R}$ for all $\text{a}\in\text{A}$
So, R is reflexive on A.
Symmetry: Let $(\text{a, b})\in\text{R}$
⇒ Both a and b are either even or odd.
⇒ Both b and a are either even or odd.
$\Rightarrow\ (\text{b, a})\in\text{R}$ for all $\text{a, b}\in\text{A}$
So, R is symmetric on A.
Transitivity: Let (a, b) and (b, c) $\in\text{R}$
⇒ Both a and b are either even or odd and both b and c are either even or odd.
⇒ a, b and c are either even or odd.
⇒ a and c both are either even or odd.
$\Rightarrow\ (\text{a, c})\in\text{R}$ for all $\text{a, c}\in\text{A}$
So, R is transitive on A.
Thus, R is an equivalence relation on A.
We observe that all the elements of the subset {1, 3, 5, 7} are odd. Thus, they are related to each other.
This is because the relation R on A is an equivalence relation.
Similarly, the elements of the subset {2, 4, 6} are even. Thus, they are related to each other because every element is even.
Hence proved.

View full question & answer→

Question 44 Marks

Show that the relation R on the set Z of integers, given by R = {(a, b): 2 divides a - b}, is an equivalence relation.

Answer

We have,

R = {(a, b): a - b is divisible by 2; a, b $\in\text{Z}$}

To prove: R is an equivalence relation.

Proof:

Reflexivity: Let $\text{a}\in\text{Z}$

⇒ a - a = 0

⇒ a - a is divisible by 2

$\Rightarrow\ (\text{a, a})\in\text{R}$

⇒ R is reflexive.

Symmetric: Let $\text{a, b}\in\text{Z}$ and $(\text{a, b})\in\text{R}$

⇒ a - b is divisible by 2

⇒ a - b = 2p For some $\text{p}\in\text{Z}$

⇒ b - a = 2 × (-p)

$\Rightarrow\ \text{b}-\text{a}\in\text{R}$

⇒ R is symmetric.

Transitive: Let $\text{a, b, c}\in\text{Z}$ and such that $(\text{a, b})\in\text{R}$ and $(\text{b, c})\in\text{R}$

⇒ a - b = 2p and b - c = q For some $\text{p, q}\in\text{Z}$

⇒ a - c = 2(p + q)

⇒ a - c is divisible by 2

$\Rightarrow\ (\text{a, c})\in\text{R}$

⇒ R is transitive.

View full question & answer→

Question 54 Marks

Let Z be the set of all integers and Z₀ be the set of all non-zero integers. Let a relation R on Z × Z₀ be defined as (a, b)R(c, d) ⇔ ad = bc for all (a, b), (c, d) ∈ Z × Z₀, Prove that R is an equivalence relation on Z × Z₀.

Answer

We have, Z be the set of integers and Z₀ be the set of non-zero integers.
R = {(a, b), (c, d): ad = bc} be a relation on Z × Z₀
Now,
Reflexivity: (a, b) ∈ Z × Z₀
⇒ ab = ba
⇒ {(a, b), (a, b)} ∈ R
⇒ R is reflexive.
Symmetric: Let {(a, b), (c, d)} ∈ R
⇒ ad = bc
⇒ cd = da
⇒ {(c, d), (a, b)} ∈ R
⇒ R is symmetric.
Transitive: Let (a, b), (c, d) ∈ R and (c, d), (e, f) ∈ R
⇒ ad = bc and cf = de
$\Rightarrow\ \frac{\text{a}}{\text{b}}=\frac{\text{c}}{\text{d}}$ and $\frac{\text{c}}{\text{d}}=\frac{\text{e}}{\text{f}}$
$\Rightarrow\ \frac{\text{a}}{\text{b}}=\frac{\text{e}}{\text{f}}$
$\Rightarrow\ \text{af}=\text{be}$
$\Rightarrow\ (\text{a, b})(\text{e, f})\in\text{R}$
⇒ R is transitive.
Hence, R is an equivalence relation on Z × Z₀

View full question & answer→

Question 64 Marks

Prove that the relation R on Z defined by $(\text{a, b})\in\text{R}\Leftrightarrow\ \text{a}-\text{b}$ is divisible by 5 is an equivalence relation on Z.

Answer

We observe the following properties of relation R.
Reflexivity: Let a be an arbitrary element of R. Then,
⇒ a - a = 0 = 0 × 5
⇒ a - a is divisible by 5
$\Rightarrow\ (\text{a, a})\in\text{R}$ for all $\text{a}\in\text{Z}$
So, R is reflexive on Z.
Symmetry: Let $(\text{a, b})\in\text{R}$
⇒ a - b is divisible by 5
⇒ a - b = 5p for some $\text{p}\in\text{Z}$
⇒ b - a = 5(-p)
Here, $-\text{p}\in\text{Z}$ [Since $\text{p}\in\text{Z}$]
⇒ b - a is divisible by 5
$\Rightarrow\ (\text{b, a})\in\text{R}$ for all $\text{a, b}\in\text{Z}$
So, R is symmetric on Z.
Transitivity: Let a, b and $\text{b},\text{c}\in\text{R}$
⇒ a - b is divisible by 5
⇒ a - b = 5p for some Z
Also, b - c is divisible by 5
⇒ b - c = 5q for some Z
Adding the above two, we get
a - b + b - c = 5p + 5q
⇒ a - c = 5(p + q)
⇒ a - c is divisible by 5
Here, $\text{p}+\text{q}\in\text{Z}$
$\Rightarrow\ \text{a, c}\in\text{R}$ for all $\text{a, c}\in\text{Z}$
So, R is transitive on Z.
Hence, R is an equivalence relation on Z.

View full question & answer→

Question 74 Marks

Let Z be the set of integers. Show that the relation R = {(a, b): a, b ∈ Z and a + b is even} is an equivalence relation on Z.

Answer

We have, Z be set of integers and R = {(a, b): a, b ∈ Z and a + b is even} be a relation on Z.

We want to prove that R is an equivalence relation on Z.

Now,

Reflexivity: Let $\text{a}\in\text{Z}$

⇒ a + a is even

[if a is even ⇒ a + a is even, if a is odd ⇒ a + a is even]

$\Rightarrow\ (\text{a, a})\in\text{R}$

⇒ R is reflexive.

Symmetric: Let $\text{a, b}\in\text{Z}$ and $(\text{a, b})\in\text{R}$

⇒ a + b is even

⇒ b + a is even

$\Rightarrow\ (\text{b, a})\in\text{R}$

⇒ R is symmetric.

Transitivity: Let $(\text{a, b})\in\text{R}$ and $(\text{b, c})\in\text{R}$ For some $\text{a, b, c}\in\text{Z}$

⇒ a + b is even and b + c is even

[if b is odd, then a and c must be odd ⇒ a + c is even, if b is even, then a and c must be even ⇒ a + c is even]

⇒ a + c is even

$\Rightarrow\ (\text{a, c})\in\text{R}$

⇒ R is transitive.

Hence, R is an equivalence relation on Z.

View full question & answer→

Question 84 Marks

Let R be a relation on the set A of ordered pair of integers defined by (x, y)R(u, v) if xv = yu. Show that R is an equivalence relation.

Answer

We observe the following properties of R.
Reflexivity: Consider a, b be an arbitrary element of the set A.
Then $\text{a, b}\in\text{A}$
Implies that ab = ba
Implies that a, bRa, b
Thus, R is reflexive on A.
Symmetry: Consider x, y and u, v $\in\text{A}$ such that x, yRu, v.
Then xv = yu
Implies that vx = uy
Implies that uy = vx
Implies that u, vRx, y.
So, R is symmetric on A.
Transitivity: Let x, y, u, v and p, q $\in\text{R}$ such that x, yRu, v and u, vRp, q.
Implies that xv = yu and uq = vp.
Multiplying the corresponding sides,
We get xv × uq = yu × vp
Implies that xq = yp
Implies that x, yRp, q.
So, R is transitive on A.
Hence, R is an equivalence relation on A.

View full question & answer→

Question 94 Marks

Let R be a relation defined on the set of natural numbers N as,
R = {(x, y): x, y ∈ N, 2x + y = 41}
Find the domain and range of R. Also, verify whether R is:

Reflexive.
Symmetric.
Transitive.

Answer

{(x, y): x, y ∈ N, 2x + y = 41}
Domain of R = {1, 2, 3, 4, ....., 20}
Then Domain of R is x ∈ N such that
2x + y = 41
$\text{x}=\frac{41-\text{y}}{2}$
Therefore, Domain of R is:
R = {1, 2, 3, 4, ..., 20}
Range of R is y ∈ N
Such that range of R = {1, 3, 5, ...., 37, 39}
Let x be an arbitrary element of R.
Since, $(2,2)\notin\text{R,}$ R is not reflexive.
Hence, R is not symmetric.
Finally, since, $(15,11)\notin\text{R}$ and $(11,19)\notin\text{R}$ but $(15,19)\notin\text{R}$
Thus, R is not transitive.

View full question & answer→

Question 104 Marks

Let L be the set of all lines in XY-plane and R be the relation in L defined as R = {(L₁, L₂): L₁ is parallel to L₂}. Show that R is an equivalence relation. Find the set of all lines related to the line y = 2x + 4.

Answer

We observe the following properties of R.

Reflexivity: Let L₁ be an arbitrary element of the set L. Then,

$\text{L}_1\in\text{L}$

⇒ L₁ is parallel to L₁ [Every line is parallel to itself]

$\Rightarrow\ (\text{L}_1,\text{ L}_1)\in\text{R}$ for all $\text{L}_1\in\text{L}$

So, R is reflexive on L.

Symmetry: Let $(\text{L}_1,\text{ L}_2)\in\text{R}$

⇒ L₁ is parallel to L₂

⇒ L₂ is parallel to L₁

$\Rightarrow\ (\text{L}_2,\text{ L}_1)\in\text{R}\text{ for } \text{ all } \text{L}_1 \text{ and } \text{L}_2\in\text{L}$ $$ $$

So, R is symmetric on L.

Transitivity: $\text{ Let }(\text{L}_1, \text{L}_2) \text{ and } (\text{L}_2, \text{L}_3) \in\text{R}$$$

⇒ L₁ is parallel to L₂ and L₂ is parallel to L₃

⇒ L₁, L₂ and L₃ are all parallel to each other

⇒ L₁ is parallel to L₃

$\Rightarrow\ (\text{L}_1,\text{ L}_3)\in\text{R}$

So, R is transitive on L.

Hence, R is an equivalence relation on L.

Set of all the lines related to

y = 2x + 4 = L' = {(x, y): y = 2x + c, where c ∈ R}

View full question & answer→

Question 114 Marks

Show that the relation R defined by R = {(a, b): a - b is divisible by 3; a, b ∈ Z} is an equivalence relation.

Answer

We observe the following relations of relation R.
Reflexivity: Let a be an arbitrary element of R. Then,
a - a = 0 = 0 × 3
⇒ a - a is divisible by 3
$\Rightarrow\ (\text{a, a})\in\text{R}$ for all $\text{a}\in\text{Z}$
So, R is reflexive on Z.
Symmetry: Let $(\text{a, b})\in\text{R}$
⇒ a - b is divisible by 3
⇒ a - b = 3p for some $\text{p}\in\text{Z}$
⇒ b - a = 3(-p)
Here, $-\text{p}\in\text{Z}$
⇒ b - a is divisible by 3
⇒ (b, a)∈R for all a, $\text{b}\in\text{Z}$
So, R is symmetric on Z.
Transitivity: Let a, b and b, c $\in\text{R}$
⇒ a - b and b - c are divisible by 3
⇒ a - b = 3p for some $\text{p}\in\text{Z}$
and b - c = 3q for some $\text{q}\in\text{Z}$
Adding the above two, we get
a - b + b - c = 3p + 3q
⇒ a - c = 3(p + q)
Here, p + q $\in\text{Z}$
⇒ a - c is divisible by 3
$ \Rightarrow \text{a, c} \in \text{R}$ for all a, c $\in\text{Z}$
So, R is transitive on Z.
Hence, R is an equivalence relation on Z.

View full question & answer→

Question 124 Marks

Let S be a relation on the set R of all real numbers defined by S = {(a, b) ∈ R × R: a² + b² = 1} prove that S is not an equivalence relation on R.

Answer

We observe the following properties of S.
Reflexivity: Consider a be an arbitrary element of R.
Then, $\text{a}\in\text{R}$
Implies that $\text{a}^2+\text{a}^2\neq1\ \forall\ \text{a}\in\text{R}$
Implies that $\text{a, a}\notin\text{S.}$
So, S is not reflexive on R.
Symmetry: Consider $\text{a, b}\in\text{R}$
Implies that a² + b² = 1
Implies that b² + a²= 1
Implies that $\text{b, a}\in\text{S}$ for all $\text{a, b}\in\text{R}$
So, S is symmetric on R.
Transitivity: Let a, b and $\text{b, c}\in\text{R}$
Implies that a² + b² = 1 and b² + c² = 1
Adding the above two, we get $\text{a}^2+\text{c}^2=2-2\text{b}^2\neq1$ for all $\text{a, b, c}\in\text{R}.$
So, S is not transitive on R.
Hence, S is not an equivalence relation on R.

View full question & answer→

Question 134 Marks

Show that the relation R on the set A = {x ∈ Z; 0 ≤ x ≤ 12}, given by R = {(a, b): a = b}, is an equivalence relation. Find the set of all elements related to 1.

Answer

We have, A = {x ∈ Z; 0 ≤ x ≤ 12} be a set and R = {(a, b): a = b} be a relation on A

Now,

Reflexivity: Let $\text{a}\in\text{A}$

⇒ a = a

$\Rightarrow\ (\text{a, a})\in\text{R}$

⇒ R is reflexive.

Symmetric: Let $\text{a, b}\in\text{A}$ and $(\text{a, b})\in\text{R}$

⇒ a = b

⇒ b = a

$\Rightarrow\ (\text{b, a})\in\text{R}$

⇒ R is symmetric.

Transitive:$\text{Let a, b & c}\in\text{A}$ $$

and Let $(\text{a, b})\in\text{R}\ \text{ and }\ (\text{b, c})\in\text{R}$ $$

⇒ a = b and b = c

⇒ a = c

$\Rightarrow\ (\text{a, c})\in\text{R}$

⇒ R is transitive.

Since R is being reflexive, symmetric and transitive, so R is an equivalance relation.

Also, we need to find the set of all elements related to 1.

Since the relation is given by, R = {(a, b): a = b}, and 1 is an element of A,

R = {(1, 1): 1 = 1}

Thus, the set of all elements related to 1 is 1.

View full question & answer→

Question 144 Marks

Let A be the set of all human beings in a town at a particular time. Determine whether the following relations are reflexive, symmetric and transitive:
R = {(x, y): x and y live in the same locality}

Answer

A be the set of human beings.
R = {(x, y): x and y live in the same locality}
Reflexive: Since x and x lives in the same locality.
$\Rightarrow\ (\text{x, x})\in\text{R}$
⇒ R is Reflexive.
Symmetric: Let $(\text{x, y})\in\text{R}$
⇒ x and y lives in the same locality.
⇒ y and x lives in the same locality.
$\Rightarrow\ (\text{y, x})\in\text{R}$
Transitive: Let $(\text{x, y})\in\text{R}$ and $(\text{y, z})\in\text{R}$
$(\text{x, y})\in\text{R}$
⇒ x and y lives in the same locality and $(\text{y, z})\in\text{R}$
⇒ y and z lives in the same locality.
⇒ x and z lives in the same locality.
$\Rightarrow\ (\text{x, z})\in\text{R}$
⇒ R is transitive.

View full question & answer→

Question 154 Marks

Let A be the set of all human beings in a town at a particular time. Determine whether the following relations are reflexive, symmetric and transitive:
R = {(x, y): x is wife of y}

Answer

Reflexivity: Let x be an element of R. Then,
x is wife of x cannot be true.
$\Rightarrow\ (\text{x, x})\notin\text{R}$ so, R is not a reflexive relation.

View full question & answer→

Question 164 Marks

Let A be the set of all human beings in a town at a particular time. Determine whether the following relations are reflexive, symmetric and transitive:
R = {(x, y): x and y work at the same place}

Answer

Reflexivity: Consider x be an arbitrary element of R .
Then, $\text{x}\in\text{R}$
Implies that, x and x work at the same place is true since they are the same.
Implies that x, $\text{x}\in\text{R},$ therefore
R is a reflexive relation.
Symmetry: Consider $\text{x},\text{y}\in\text{R}$
Implies that x and y work at the same place
Implies that $\text{x},\text{y}\in\text{R}$
R is a symmetric relation.
Transitivity: Consider $\text{x},\text{y}\in\text{R}$ and $\text{y},\text{z}\in\text{R}.$
Then x and y works at the same place.
y and z also work at the same place
Implies that, x, y and z all work at the same place
Implies that x and z work at the same place.
Implies that $\text{x},\text{z}\in\text{R}$
Therefore, R is a transitive relation.

View full question & answer→

Question 174 Marks

Let O be the origin. We define a relation between two points P and Q in a plane if OP = OQ. Show that the relation, so defined is an equivalence relation.

Answer

Let A be set of points on plane.

Let R = {(P, Q): OP = OQ} be a relation on A where O is the origin.

To prove R is an equivalence relation, we need to show that R is reflexive, symmetric and transitive on A.

Now,

Reflexivity: Let $\text{P}\in\text{A}$

Since OP = OP

$\Rightarrow\ (\text{P, P})\in\text{R}$

⇒ R is reflexive.

Symmetric: Let $(\text{P, Q})\in\text{R}$ for $\text{P, Q}\in\text{A}$

Then OP = OQ

⇒ OQ = OP

$\Rightarrow\ (\text{Q, P})\in\text{R}$

⇒ R is symmetric.

Transitive: Let $(\text{P, Q})\in\text{R}$ and $(\text{Q, S})\in\text{R}$

⇒ OP = OQ and OQ = OS

⇒ OP = OS

$\Rightarrow\ (\text{P, S})\in\text{R}$

⇒ R is transitive.

Thus, R is an equivalence relation on A.

View full question & answer→

Question 184 Marks

Let n be a fixed positive integer. Define a relation R on Z as follows:
$(\text{a, b})\in\text{R}\Leftrightarrow\ \text{a}-\text{b}$ is divisible by n. Show that R is an equivalence relation on Z.

Answer

We observe the following properties of R.

Reflexivity: Consider $\text{a}\in\text{N}$

Here, a - a = 0 = 0 × n

Implies that a - a is divisible by n

Implies that $\text{a, a}\in\text{R}$

Implies that $\text{a, a}\in\text{R}$ for all $\text{a}\in\text{Z}.$

So, R is reflexive on Z.

Symmetry: Consider $\text{a, b}\in\text{R}$

Here a - b is divisible by n

Implies that a - b = np for some $\text{p}\in\text{Z}$

Implies that b - a = n - p.

Implies that b - a is divisible by n $\big[\text{p}\in\text{Z}$ implies that $-\text{p}\in\text{Z}\big]$

implies that $\text{b, a}\in\text{R}$

So, R is symmetric on Z.

Transitivity: Consider a, b and b, c $\in\text{R}$

Here, a - b is divisible by n and b - c is divisible by n.

implies that a - b = np for some $\text{p}\in\text{Z}$ and b - c = nq for some $\text{q}\in\text{Z}$

Adding the above two

we get a - b + b - c = np + nq

Implies that a - c = n(p + q).

Here, $\text{p}+\text{q}\in\text{Z}$

Implies that $\text{a, c}\in\text{R}$ for all $\text{a, c}\in\text{Z.}$

So, R is transitive on Z.

Hence, R is an equivalence relation on Z.

View full question & answer→

Question 194 Marks

Let C be the set of all complex numbers and C₀be the set of all no-zero complex numbers. Let a relation R on C₀be defined as $\text{z}_1\text{R z}_2\Leftrightarrow\frac{\text{z}_1-\text{z}_2}{\text{z}_1+\text{z}_2}$ is real for all $\text{z}_1,\ \text{z}_2\in\text{C}_0.$ Show that R is an equivalence relation.

Answer

Test for reflexivity: Since, $\frac{\text{z}_1-\text{z}_1}{\text{z}_1+\text{z}_1}=0,$ which is a real number.

So, $(\text{z}_1,\text{ z}_1)\in\text{R}$

Hence, R is relexive relation.

Test for symmetric: Let $(\text{z}_1,\text{ z}_2)\in\text{R.}$

Then, $\frac{\text{z}_1-\text{z}_2}{\text{z}_1+\text{z}_2}=\text{x,}$ where x is real

$\Rightarrow\ -\Big(\frac{\text{z}_1-\text{ z}_2}{\text{z}_1+\text{ z}_2}\Big)=-\text{x}$

$\Rightarrow\ \Big(\frac{\text{z}_2-\text{ z}_1}{\text{z}_2+\text{ z}_1}\Big)=-\text{x},$ is also a real number

So, $(\text{z}_2,\text{ z}_1)\in\text{R}$

Hence, R is symmetric relation.

Test for transivity: Let $(\text{z}_1,\text{ z}_2)\in\text{R}$ and $(\text{z}_2,\text{ z}_3)\in\text{R}.$

Then,

$\frac{\text{z}_1-\text{z}_2}{\text{z}_1+\text{z}_2}=\text{x},$ where x is a real number.

$\Rightarrow\ \text{z}_1-\text{z}_2=\text{xz}_1+\text{xz}_2$

$\Rightarrow\ \text{z}_1-\text{xz}_1=\text{z}_2+\text{xz}_2$

$\Rightarrow\ \text{z}_1(1-\text{x})=\text{z}_2(1+\text{x})$

$\Rightarrow\ \frac{\text{z}_1}{\text{z}_2}=\frac{(1+\text{x})}{(1-\text{x})}\ .....(1)$

Also,

$\frac{\text{z}_2-\text{z}_3}{\text{z}_2+\text{z}_3}=\text{y},$ where y is a real number.

$\Rightarrow\ \text{z}_2-\text{z}_3=\text{yz}_2+\text{yz}_3$

$\Rightarrow\ \text{z}_2-\text{yz}_2=\text{z}_3+\text{yz}_3$

$\Rightarrow\ \text{z}_2(1-\text{y})=\text{z}_3(1+\text{y})$

$\Rightarrow\ \frac{\text{z}_2}{\text{z}_3}=\frac{(1+\text{y})}{(1-\text{y})}\ .....(2)$

Dividing (1) and (2), we get

$\frac{\text{z}_1}{\text{z}_3}=\Big(\frac{1+\text{x}}{1-\text{x}}\Big)\times\Big(\frac{1-\text{y}}{1+\text{y}}\Big)=\text{z,}$ where z is a real number.

$\Rightarrow\ \frac{\text{z}_1-\text{z}_3}{\text{z}_1+\text{z}_3}=\frac{\text{z}-1}{\text{z}+1},$ which is real

$\Rightarrow\ (\text{z}_1,\text{ z}_3)\in\text{R}$

Hence, R is transitive relation.

From (i), (ii) and (iii), R is an equivalenve relation.

View full question & answer→

Question 204 Marks

If R and S are relations on a set A, then prove that:

R and S are symmetric $\Rightarrow\ \text{R}\cap\text{S}$ and $\text{R}\cup\text{S}$ are symmetric
R is reflexive and S is any relation $\Rightarrow\ \text{R}\cup\text{S}$ is reflexive.

Answer

R and S are symmetric relations on the set A.

$\Rightarrow\ \text{R}\subset\text{A}\times\text{A}$ and $\text{S}\subset\text{A}\times\text{A}$

$\Rightarrow\ \text{R}\cap\text{S}\subset\text{A}\times\text{A}$

Thus, $\text{R}\cap\text{S}$ is a relation on A.

Let $\text{a, b}\in\text{A}$ such that $(\text{a, b})\in\text{R}\cap\text{S.}$ Then,

$(\text{a, b})\in\text{R}\cap\text{S}$

$\Rightarrow\ (\text{a, b})\in\text{R}$ and $(\text{a, b})\in\text{S}$

$\Rightarrow\ (\text{b, a})\in\text{R}$ and $(\text{b, a})\in\text{S}$ [Since R and S are symmetric]

$\Rightarrow\ (\text{b, a})\in\text{R}\cap\text{S}$

Thus,

$(\text{a, b})\in\text{R}\cap\text{S}$

$\Rightarrow\ (\text{b, a})\in\text{R}\cap\text{S}$ for all $\text{a, b}\in\text{A}$

So, $\text{R}\cap\text{S}$ is symmetric on A.

Also,

Let $\text{a, b}\in\text{A}$ such that $(\text{a, b})\in\text{R}\cup\text{S}$

$\Rightarrow\ (\text{a, b})\in\text{R}$ or $(\text{a, b})\in\text{S}$

$\Rightarrow\ (\text{b, a})\in\text{R}$ or $(\text{b, a})\in\text{S}$ [Since R and S are symmetric]

$\Rightarrow\ (\text{b, a})\in\text{R}\cup\text{S}$

So, $\text{R}\cup\text{S}$ is symmetric on A.

R is reflexive and S is any relation.

Suppose $\text{a}\in\text{A.}$ Then,

$(\text{a, a})\in\text{R}$ [Since R is reflexive]

$\Rightarrow\ (\text{a, a})\in\text{R}\cup\text{S}$

$\Rightarrow\ \text{R}\cup\text{S}$ is reflexive on A.

View full question & answer→

Question 214 Marks

Test whether the following relations R₃are:

Reflexive.
Symmetric.
Transitive.

R₃ on R defined by $(\text{a, b})\in\text{R}_3\Leftrightarrow\ \text{a}^2-4\text{ab}+3\text{b}^2=0$

Answer

Reflexivity: Consider a be an arbitrary element of R₃

Then, $\text{a}\in\text{R}_3$

Implies that a₂ - 4a₂ + 3a₂ = 0

So, R₃ is reflexive.

Symmetry: Consider, $\text{a, b}\in\text{R}_3$

Implies that a₂ - 4a₂b₂ + 3b₂ = 0

But $\text{b}_2-4\text{b}_2\text{a}_2+3\text{a}_2\neq0$ for all $\text{a, b}\in\text{R}$

So, R₃ is not symmetric.

Transitivity: $1,2\in\text{R}_3$ and $2,3\in\text{R}_3$

Implies that 1 - 8 + 6 = 0 and 4 - 24 + 27 = 0

But $1 - 12 + 9 \neq0$

So, R₃ is not transitive.

View full question & answer→

Question 224 Marks

Test whether the following relations R₁are:

Reflexive.
Symmetric.
Transitive.

R₁ on Q₀ defined by $(\text{a, b})\in\text{R}_1\Leftrightarrow\ \text{a}=\frac{1}{\text{b}}.$

Answer

$\text{R}_1=\Big\{(\text{x, y}),\ \text{x, y}\in\text{Q}_0, \text{x}=\frac{1}{\text{y}}\Big\}$

Reflexivity: Let, $\text{x}\in\text{Q}_0$

$\Rightarrow\ \text{x}\neq\frac{1}{\text{x}}$

$\Rightarrow\ (\text{x, x})\in\text{R}_1$

$\therefore$ R₁ is not reflexive.

Symmetric: Let, $(\text{x, y})\in\text{R}_1$

$\Rightarrow\ \text{x}=\frac{1}{\text{y}}$

$\Rightarrow\ \text{y}=\frac{1}{\text{x}}$

$\Rightarrow\ (\text{y, x})\in\text{R}_1$

$\therefore$ R₁ is Symmetric.

Transitive: Let, $(\text{x, y})\in\text{R}_1$ and $(\text{y, z})\in\text{R}_1$

$\Rightarrow\ \text{x}=\frac{1}{\text{y}}$ and $\text{y}=\frac{1}{\text{z}}$

$\Rightarrow\ \text{x}=\text{z}$

$\Rightarrow\ (\text{x, z})\notin\text{R}_1$

$\therefore$ R₁ is not Transitive.

View full question & answer→

Maths 4 Marks

Test yourself on this topic