excersie 1.1 continued

Question 6:

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.

solution =  

 

Let A = {1, 2, 3}.

A relation R on A is defined as R = {(1, 2), (2, 1)}.

It is seen that (1, 1), (2, 2), (3, 3) ∉R.

∴ R is not reflexive.

Now, as (1, 2) ∈ R and (2, 1) ∈ R, then R is symmetric.

Now, (1, 2) and (2, 1) ∈ R

However,

(1, 1) ∉ R

∴ R is not transitive.

Hence, R is symmetric but neither reflexive nor transitive.

Question 7:

Show that the relation R in the set A of all the books in a library of a college, given by R = {(x, y): x and y have same number of pages} is an equivalence relation.

solution  =

Set A is the set of all books in the library of a college.

R = {x, y): x and y have the same number of pages}

Now, R is reflexive since (x, x) ∈ R as x and x has the same number of pages.

Let (x, y) ∈ R ⇒ x and y have the same number of pages.

⇒ y and x have the same number of pages.

⇒ (y, x) ∈ R

∴R is symmetric.

Now, let (x, y) ∈R and (y, z) ∈ R.

⇒ x and y and have the same number of pages and y and z have the same number of pages.

⇒ x and z have the same number of pages.

⇒ (x, z) ∈ R

∴R is transitive.

Hence, R is an equivalence relation.

Question 8:

Show that the relation R in the set A = {1, 2, 3, 4, 5} given by

, is an equivalence relation. Show that all the elements of {1, 3, 5} are related to each other and all the elements of {2, 4} are related to each other. But no element of {1, 3, 5} is related to any element of {2, 4}.

solution =

A = {1, 2, 3, 4, 5}

It is clear that for any element a ∈A, we have (which is even).

∴R is reflexive.

Let (a, b) ∈ R.

∴R is symmetric.

Now, let (a, b) ∈ R and (b, c) ∈ R.

⇒ (a, c) ∈ R

∴R is transitive.

Hence, R is an equivalence relation.

Now, all elements of the set {1, 3, 5} are related to each other as all the elements of this subset are odd. Thus, the modulus of the difference between any two elements will be even.

Similarly, all elements of the set {2, 4} are related to each other as all the elements of this subset are even.

Also, no element of the subset {1, 3, 5} can be related to any element of {2, 4} as all elements of {1, 3, 5} are odd and all elements of {2, 4} are even. Thus, the modulus of the difference between the two elements (from each of these two subsets) will not be even.

Question 9:

Show that each of the relation R in the set, given by

(i)

(ii)

is an equivalence relation. Find the set of all elements related to 1 in each case.

solution=

 

(i)

For any element a ∈A, we have (a, a) ∈ R as is a multiple of 4.

∴R is reflexive.

Now, let (a, b) ∈ R ⇒ is a multiple of 4.

⇒ (b, a) ∈ R

∴R is symmetric.

Now, let (a, b), (b, c) ∈ R.

⇒ (a, c) ∈R

∴ R is transitive.

Hence, R is an equivalence relation.

The set of elements related to 1 is {1, 5, 9} since

(ii) R = {(a, b): a = b}

For any element a ∈A, we have (a, a) ∈ R, since a = a.

∴R is reflexive.

Now, let (a, b) ∈ R.

⇒ a = b

⇒ b = a

⇒ (b, a) ∈ R

∴R is symmetric.

Now, let (a, b) ∈ R and (b, c) ∈ R.

⇒ a = b and b = c

⇒ a = c

⇒ (a, c) ∈ R

∴ R is transitive.

Hence, R is an equivalence relation.

The elements in R that are related to 1 will be those elements from set A which are equal to 1.

Hence, the set of elements related to 1 is {1}.

Question 10:

Given an example of a relation. Which is

(i) Symmetric but neither reflexive nor transitive.

(ii) Transitive but neither reflexive nor symmetric.

(iii) Reflexive and symmetric but not transitive.

(iv) Reflexive and transitive but not symmetric.

(v) Symmetric and transitive but not reflexive.

solutions= 

(i) Let A = {5, 6, 7}.

Define a relation R on A as R = {(5, 6), (6, 5)}.

Relation R is not reflexive as (5, 5), (6, 6), (7, 7) ∉ R.

Now, as (5, 6) ∈ R and also (6, 5) ∈ R, R is symmetric.

(5, 6), (6, 5) ∈ R, but (5, 5) ∉ R

∴R is not transitive.

Hence, relation R is symmetric but not reflexive or transitive.

(ii)Consider a relation R in R defined as:

R = {(a, b): a < b}

For any a ∈ R, we have (a, a) ∉ R since a cannot be strictly less than a itself. In fact, a = a.

∴ R is not reflexive.

Now,

(1, 2) ∈ R (as 1 < 2)

But, 2 is not less than 1.

∴ (2, 1) ∉ R

∴ R is not symmetric.

Now, let (a, b), (b, c) ∈ R.

⇒ a < b and b < c

⇒ a < c

⇒ (a, c) ∈ R

∴R is transitive.

Hence, relation R is transitive but not reflexive and symmetric.

(iii)Let A = {4, 6, 8}.

Define a relation R on A as:

A = {(4, 4), (6, 6), (8, 8), (4, 6), (6, 4), (6, 8), (8, 6)}

Relation R is reflexive since for every a ∈ A, (a, a) ∈R i.e., (4, 4), (6, 6), (8, 8)} ∈ R.

Relation R is symmetric since (a, b) ∈ R ⇒ (b, a) ∈ R for all a, b ∈ R.

Relation R is not transitive since (4, 6), (6, 8) ∈ R, but (4, 8) ∉ R.

Hence, relation R is reflexive and symmetric but not transitive.

(iv) Define a relation R in R as:

R = {a, b): a³ ≥ b³}

Clearly (a, a) ∈ R as a³ = a³.

∴R is reflexive.

Now,

(2, 1) ∈ R (as 2³ ≥ 1³)

But,

(1, 2) ∉ R (as 1³ < 2³)

∴ R is not symmetric.

Now,

Let (a, b), (b, c) ∈ R.

⇒ a³ ≥ b³ and b³ ≥ c³

⇒ a³ ≥ c³

⇒ (a, c) ∈ R

∴R is transitive.

Hence, relation R is reflexive and transitive but not symmetric.

(v) Let A = {−5, −6}.

Define a relation R on A as:

R = {(−5, −6), (−6, −5), (−5, −5)}

Relation R is not reflexive as (−6, −6) ∉ R.

Relation R is symmetric as (−5, −6) ∈ R and (−6, −5}∈R.

It is seen that (−5, −6), (−6, −5) ∈ R. Also, (−5, −5) ∈ R.

∴The relation R is transitive.

Hence, relation R is symmetric and transitive but not reflexive.

Question 11:

Show that the relation R in the set A of points in a plane given by R = {(P, Q): distance of the point P from the origin is same as the distance of the point Q from the origin}, is an equivalence relation. Further, show that the set of all point related to a point P ≠ (0, 0) is the circle passing through P with origin as center.

solutions =

R = {(P, Q): distance of point P from the origin is the same as the distance of point Q from the origin}

Clearly, (P, P) ∈ R since the distance of point P from the origin is always the same as the distance of the same point P from the origin.

∴R is reflexive.

Now,

Let (P, Q) ∈ R.

⇒ The distance of point P from the origin is the same as the distance of point Q from the origin.

⇒ The distance of point Q from the origin is the same as the distance of point P from the origin.

⇒ (Q, P) ∈ R

∴R is symmetric.

Now,

Let (P, Q), (Q, S) ∈ R.

⇒ The distance of points P and Q from the origin is the same and also, the distance of points Q and S from the origin is the same.

⇒ The distance of points P and S from the origin is the same.

⇒ (P, S) ∈ R

∴R is transitive.

Therefore, R is an equivalence relation.

The set of all points related to P ≠ (0, 0) will be those points whose distance from the origin is the same as the distance of point P from the origin.

In other words, if O (0, 0) is the origin and OP = k, then the set of all points related to P is at a distance of k from the origin.

Hence, this set of points forms a circle with the centre as the origin and this circle passes through point P.

 

 

Question 12:

Show that the relation R defined in the set A of all triangles as R = {(T₁, T₂): T₁ is similar to T₂}, is equivalence relation. Consider three right angle triangles T₁ with sides 3, 4, 5, T₂ with sides 5, 12, 13 and T₃ with sides 6, 8, 10. Which triangles among T₁, T₂ and T₃ are related?

solutions =  

R = {(T₁, T₂): T₁ is similar to T₂}

R is reflexive since every triangle is similar to itself.

Further, if (T₁, T₂) ∈ R, then T₁ is similar to T_2.

⇒ T₂ is similar to T_1.

⇒ (T₂, T₁) ∈R

∴R is symmetric.

Now,

Let (T₁, T₂), (T₂, T₃) ∈ R.

⇒ T₁ is similar to T₂ and T₂ is similar to T_3.

⇒ T₁ is similar to T_3.

⇒ (T₁, T₃) ∈ R

∴ R is transitive.

Thus, R is an equivalence relation.

Now, we can observe that:

∴The corresponding sides of triangles T₁ and T₃ are in the same ratio.

Then, triangle T₁ is similar to triangle T_3.

Hence, T₁­ is related to T₃.

 

Question 13:

Show that the relation R defined in 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?

solutions = 

R = {(P₁, P₂): P₁ and P₂ have same the number of sides}

R is reflexive since (P₁, P₁) ∈ R as the same polygon has the same number of sides with itself.

Let (P₁, P₂) ∈ R.

⇒ P₁ and P_2­ have the same number of sides.

⇒ P₂ and P₁ have the same number of sides.

⇒ (P₂, P₁) ∈ R

∴R is symmetric.

Now,

Let (P₁, P₂), (P₂, P₃) ∈ R.

⇒ P₁ and P₂ have the same number of sides. Also, P₂ and P₃ have the same number of sides.

⇒ P₁ and P₃ have the same number of sides.

⇒ (P₁, P₃) ∈ R

∴R is transitive.

Hence, R is an equivalence relation.

The elements in A related to the right-angled triangle (T) with sides 3, 4, and 5 are those polygons which have 3 sides (since T is a polygon with 3 sides).

Hence, the set of all elements in A related to triangle T is the set of all triangles.

Question 14:

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.

solutions =

 

R = {(L₁, L₂): L₁ is parallel to L₂}

R is reflexive as any line L₁ is parallel to itself i.e., (L₁, L₁) ∈ R.

Now,

Let (L₁, L₂) ∈ R.

⇒ L₁ is parallel to L_2.

⇒ L₂ is parallel to L_1.

⇒ (L₂, L₁) ∈ R

∴ R is symmetric.

Now,

Let (L₁, L₂), (L₂, L₃) ∈R.

⇒ L₁ is parallel to L₂. Also, L₂ is parallel to L_3.

⇒ L₁ is parallel to L_3.

∴R is transitive.

Hence, R is an equivalence relation.

The set of all lines related to the line y = 2x + 4 is the set of all lines that are parallel to the line y = 2x + 4.

Slope of line y = 2x + 4 is m = 2

It is known that parallel lines have the same slopes.

The line parallel to the given line is of the form y = 2x + c, where c ∈R.

Hence, the set of all lines related to the given line is given by y = 2x + c, where c ∈ R.

 

Question 15:

Let R be the relation in the set {1, 2, 3, 4} given by R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}. Choose the correct answer.

(A) R is reflexive and symmetric but not transitive.

(B) R is reflexive and transitive but not symmetric.

(C) R is symmetric and transitive but not reflexive.

(D) R is an equivalence relation.

solutions =

 

R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}

It is seen that (a, a) ∈ R, for every a ∈{1, 2, 3, 4}.

∴ R is reflexive.

It is seen that (1, 2) ∈ R, but (2, 1) ∉ R.

∴R is not symmetric.

Also, it is observed that (a, b), (b, c) ∈ R ⇒ (a, c) ∈ R for all a, b, c ∈ {1, 2, 3, 4}.

∴ R is transitive.

Hence, R is reflexive and transitive but not symmetric.

The correct answer is B.

Question 16:

Let R be the relation in the set N given by R = {(a, b): a = b − 2, b > 6}. Choose the correct answer.

(A) (2, 4) ∈ R (B) (3, 8) ∈R (C) (6, 8) ∈R (D) (8, 7) ∈ R

solutions=

 

R = {(a, b): a = b − 2, b > 6}

Now, since b > 6, (2, 4) ∉ R

Also, as 3 ≠ 8 − 2, (3, 8) ∉ R

And, as 8 ≠ 7 − 2

 (8, 7) ∉ R

Now, consider (6, 8).

We have 8 > 6 and also, 6 = 8 − 2.

∴(6, 8) ∈ R

The correct answer is C.