Answer - 1 : - (i) Relation R in the set A={1,2,3,….13,14} defined as R={(x,y):3x – y = 0}
(ii) Relation R in the set N of naturalnumbers defined as R= {(x, y): y = x + 5 and x < 4}
(iii) Relation R in the set A={1,2,3,4,5,6} as R= {(x, y): y is divisible by x}
(iv) Relation R in the set Z of allintegers defined as R = {(x, y): x – y is an integer}
(v) Relation R in the set A of humanbeings in a town at a particular time given by
(a) R = {(x, y): x and y work at thesame place}
(b) R = {(x, y) : x and y live in thesame locality}
(c) R = {(x, y) : x is exactly 7 cmtaller than y}
(d) R={(x,y) : x is wife of y}
(e) R= {(x, y): x is father of y}

Solution

(i) Relation R in the set A = {1, 2,….,14} defined as R= {(x,y): 3x -y =0}
(a) Put y = x, 3x – x ≠ 0 => R is not reflexive.
(b) If 3x – y = 0, then 3y – x ≠ 0, R is not symmetric
(c) If 3x – y = 0, 3y – z = 0,then 3x – z ≠ 0,R is not transitive.

(ii) Relations in the set Nof natural numbers in defined by R = {(x, y): y = x + 5 and x < 4}
(a) Putting y = x, x ≠ x + 5, R is not reflexive
(b) Putting y = x + 5, then x ≠ y + 5,R is not symmetric.
(c) If y = x + 5, z = y + 5, then z ≠ x + 5 =>R is not transitive.

(iii) Relation R in the setA = {1,2,3,4,5,6} asR = {(x, y): y is divisible by x}
(a) Putting y = x, x is divisible by x => R is reflexive.
(b) If y is divisible by x, then x is not divisible by y when x ≠ y => R isnot symmetric.
(c) If y is divisible by x and z is divisible by y then z is divisible by xe.g., 2 is divisible by 1,4 is divisible by 2.
=> 4 is divisible by 1 => R is transitive.

(iv) Relation R in Z of allintegers defined as R = {(x, y): x – y is an integer}
(a) x – x=0 is an integer => R is reflexive
(b) x – y is an integer so is y – x => R is transitive.
(c) x – y is an integer, y- z is an integer and x – z is also an integer =>R is transitive.

(v) R is a set of humanbeings in a town at a particular time.

(a) R = {(x, y)} : x and y work at the same place. It is reflexive as x worksat the same place. It is symmetric since x and y or y and x work at same place.
It is transitive since X, y work at the same place and if y, z work at the sameplace,u then x and z also work at the same place.

(b) R : {(x, y) : x and y line in the same locality}
With similar reasoning as in part (a), R is reflexive, symmetrical andtransitive.

(c) R: {(x, y)}: x is exactly 7 cm taller than y it is not reflexive: x cannot7 cm taller than x. It is not symmetric: x is exactly 7 cm taller than y, ycannot be exactly 7 cm taller than x. It is not transitive: If x is exactly 7cm taller than y and if y is exactly taller than z, then x is not exactly 7 cmtaller than z.

(d) R = {(x, y): x is wife of y}
R is not reflexive: x cannot be wife of x. R is not symmetric: x is wife of ybut y is not wife of x.
R is not transitive: if x is a wife of y then y cannot be the wife of anybodyelse.

(e) R= {(x, y): x is a father of y}
It is not reflexive: x cannot be father of himself. It is not symmetric: x is afather of y but y cannot be the father of x.It is not transitive: x is a father of y and y is a father of z then x cannotbe the father of z.

Question - 2 : - Show that the relation R in the set R of realnumbers, defined as
R = {(a, b) : a ≤, b²} is neitherreflexive nor symmetric nor transitive.

Answer - 2 : -

(i) R is not reflexive, a is not less than or equal to a²for all a ∈R, e.g., is not less than

(ii) R is not symmetric since if a ≤ b² then b is not less than or equal to a²e.g. 2 < 5² but 5 is not less than 2².
(iii) R is not transitive: If a ≤ b², b ≤ c², then a is not less than c², e.g.2 < (-2)², -2 < (-1)², but 2 is not less than (-1)².

Question - 3 : - Check whether the relation R defined in the set{1,2,3,4,5,6} as
R={(a, b): b = a+1} is reflexive,symmetric or transitive.

Answer - 3 : - (i)R is not reflexive a ≠ a + 1.
(ii) R is not symmetric if b = a + 1, then a ≠ b+ 1
(iii) R is not transitive if b = a + 1, c = b + 1 then c ≠ a + 1.

Question - 4 : - Show that the relation R in R defined as R={(a, b):a ≤ b}, is reflexive and transitive but not symmetric.
R = {(a,b):a≤b}

Answer - 4 : -

(i) R is reflexive, replacing b by a, a ≤ a =>a = a istrue.
(ii) R is not symmetric, a ≤ b, and b ≤ a which is not true 2 < 3, but 3 isnot less than 2.
(iii) R is transitive, if a ≤ b and b ≤ c, then a ≤ c, e.g. 2 < 3, 3 < 4=> 2 < 4.

Question - 5 : - Check whether the relation R in R defined by R ={(a, b): a ≤ b³} is reflexive, symmetric or transitive.

Answer - 5 : -

(i) R is not reflexive.
(ii) R is not symmetric.
(iii) R is not transitive.

Question - 6 : - Show that the relation R in the set {1,2,3} given byR = {(1, 2), (2,1)} is symmetric but neither reflexive nor transitive.

Answer - 6 : -

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 thebooks in a library of a college, given by R= {(x, y): x and y have same numberof pages} is an equivalence relation.

Answer - 7 : -

(i) The number of pages in a book remain the same
=> Relation R is reflexive.
(ii) The book x has the same number of pages as the book y.
=> Book y has the same number of pages as the book x.
=> The relation R is symmetric.
(iii) Book x and y have the same number of pages. Also book y and z have thesame number of pages.
=> Books x and z also have the same number of pages.
R is transitive also Thus, R is an equivalence relation.

Question - 8 : - Show that the relation R in the set A= {1,2,3,4,5}given by R = {(a, b) : |a – b| is even}, is an equivalence relation. Show thatall 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 anyelement of {2,4}.

Answer - 8 : -

A= {1,2,3,4,5} and R= {(a,b): |a – b| is even}
R= {(1,3), (1,5), (3,5), (2,4)}
(a) (i) Let us take any element of a set A. then |a – a| = 0 which is even.
=> R is reflexive.
(ii) If |a – b| is even, then |b – a| is also even, where,
R = {(a, b) : |a – b| is even} => R is symmetric.
(iii) Further a – c = a – b + b – c
If |a – b| and |b – c| are even, then their sum |a – b + b – c| is also even.
=> |a – c| is even,
∴R is transitive. Hence R is an equivalence relation.

(b) Elements of {1,3,5} are related to each other.
Since|1 – 3| = 2, |3 – 5| = 2, |1 – 5| = 4. All are even numbers.
= Elements of {1, 3, 5} are related to each other. Similarly elements of {2,4}are related to each other. Since |2 – 4| = 2 an even number.
No element of set {1, 3, 5} is related to any element of {2,4}.

Question - 9 : - Show that each of the relation R in the set A = {x ∈ Z: 0 ≤ x ≤ 12}, given by
(i) R={(a,b) : |a – b|is a multiple of4}
(ii) R={(a, b) : a = b}
is an equivalence relation. Find theset of all elements related to 1 in each case.

Answer - 9 : -

The set A ∴{x ∈Z : 0 ≤ x ≤ 12} = {0, 1, 2,…..12}
(i) R= {(a, b): |a – b| is a multiple of 4}
|a – b| = 4k on b = a + 4k.
∴R={(1,5),(1,9),(2, 6), (2, 10), (3, 7), (3, 11) , (4,8), (4,12), (5,9), (6,10),(7,11), (8.12) ,(0,0),(1,1), (2,2),…..(12,12)
(a) (a – a) = 0 = 4k,where k = 0=>(a,a)∈R
∴R is reflexive.
(b) If |a – b| = 4k,then|b – a| = 4k i.e. (a,b) and (b, a) both belong to R. Ris symmetric.
(c) a – c = a – b + b – c
when a – b and b – c are both multiples of 4 then a – c is also a multiple of4. This shows if (a,b)(b,c) ∈R then a – c also ∈R
∴R is an equivalence relation. The sets related 1 are {(1,5), (1,9)}.

(ii) R= {(a, b):a = b}
{(0,0), (1,1), (2,2)…..(12,12)}
(a) a = a => (a, a) ∈R
R is reflexive.
(b) Again if (a, b) ∈R,then (b, a) also ∈R
Since a = b and (a, b) ∈R => R is symmetric.
(c) If(a, b) ∈R,then(b, c) ∈R=>a = b = c
∴a = c => (a, c) ∈R,Hence, R is transitive set related is {1}.

Question - 10 : - Give an example of a relation which is

Answer - 10 : - (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.

Solution

Let A = set of straight lines in a plane
(i) R: {(a, b): a is perpendicular to b} let a, b be two perpendicular lines.

(ii) Let A = set of real numbers R= {(a,b):a>b}
(a) An element is not greater than itself
∴Ris not reflexive.
(b) If a > b than b is not greater than a => R is not symmetric
(c) If a > b also b > c, then a > c thus R is transitive
Hence, R is transitive but neither reflexive nor symmetric.

(iii) The relation R in the set {1,2,3} is given by R= {(a,b) :a + b≤4}
R= {(1,1), (1,2), (2,1), (1,3), (3,1), (2,2)} (1,1), (2,2) ∈ R => R is reflexive
(1,2), (2,1), (1,3), (3,1) => R is symmetric
But it is not transitive, since (2,1) ∈R,(1,3) ∈Rbut (2,3)∉R.

(iv) The relation R in the set {1,2,3} given by R = {(a, b):a≤b} = {(1,2), (2,2), (3,3), (2,3), 0,3)}
(a) (1,1), (2,2), (3,3) ≤R => R is reflexive
(b) (1, 2) ∈R,but (2, 1) ∉R=> R is not symmetric
(c) (1,2) ∈R,(2,3) ∈R,Also(1,3) ∈R=>Ris transitive.

(v) The relation R in the set {1,2,3} given by R= {(a, b): 0< |a – b| ≤ 2} = {(1,2), (2,1), (1, 3), (3,1), (2,3), (3,2)}
(a) R is not reflexive
∵(1,1),(2,2), (3,3) do not belong to R
(b) R is symmetric
∵(1,2), (2,1), (1,3), (3,1), (2,3), (3,2) ∈R
(c) R is transitive (1,2) ∈R,(2,3) ∈R,Also (1,3) ∈R

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