1-1 function

One-To-One Function or Injective function

A function is said to be a One-to-One Function, if for each element of range, there is a unique domain.

More About One to One Function

It satisfies both vertical line test as well as horizontal line test.
It is also Known as injective function.

Example of One to One Function

In the given figure, every element of range has unique domain.

So, the given function is one-to-one function.

                                                 Or

A function for which every element of the range of the function corresponds to exactly one element of the domain. One-to-one is often written as 1-1.

Note: y = f(x) is a function if it passes the vertical line test. It is a 1-1 function if it passes both the vertical line test and the horizontal line test. 

   Ques. is g(x)=3x-2 is a one one function?

   

   Sol:       let g(a) = g(b)

             =>  3a-2 = 3b-2

             =>  3a = 3b

             =>  a = b

            Hence g(x) is one to one function

Equivalence Relation on a given set

Equivalence Relation

            A relation R, defined on a set P is an equivalence relation if 

          and only if

       (i) R is reflexive, i.e. p R p for all p ∈ P.

       (ii) R is symmetric, i.e. p R q ⇒ q R p for all p,    q ∈ P.

        (iii) R is transitive, i.e. p R q and q R r ⇒ p R r    for all p, q, r ∈ P.

        The relation defined by “x is equal to y” on the    set P of Integers  is an equivalence relation.

  Solved examples on equivalence              relation                     

      Q 1.  Let P be a set of triangles in a plane. The 

       relation R is defined as “m is                                                            

       similar to n, m, n ∈ P”.

 Solution:

 We see that R is;

     (i) Reflexive, for, every triangle is similar to  

     itself.

     (ii) Symmetric, for, if m is similar to n, then n

     is also similar to m.

     (iii) Transitive, for, if m be similar to n and n be

     similar to r, then m is also similar to r.

  Hence R is an equivalence relation.

        Q 2.   A relation R is defined on the set Z by “p 

        R q if p – q is divisible by 2” for

        p, q ∈ Z. Examine if R is an equivalence 

        relation on Z.

           Solution:

       (i) Let p ∈ Z. Then p – p is divisible by 2. 

     Therefore p R p holds for all p in Z and R is 

     reflexive.

  (ii) Let p, q ∈ Z and pRq hold. Then p – q is 

  divisible by 2 and therefore

  q – p is also divisible by 2.

  Thus, p R q ⇒ q R p and therefore R is 

  symmetric.

(iii) Let p, q, r ∈ Z and p R q, q R r both hold. Then p – q and q – r are both divisible by 2.

Therefore p – r = (p – q) + (q – r) is divisible by 2.

Thus, p R q and q R r  ⇒ p R r and therefore R is transitive.

Since R is reflexive, symmetric and transitive so, R is an equivalence relation on Z.

      Q.3.   Let x be a positive integer. A relation R is 

      defined on the set Z by “a R b

      if and only if p – q is divisible by x” for p, q ∈ Z. 

      Show that R is an

      equivalence relation on set Z.

Solution:

(i) Let p ∈ Z. Then p – p = 0, which is divisible by x

Therefore, p R p holds for all p ∈ Z.

Hence, R is reflexive.

(ii) Let p, q ∈ Z and p R q holds. Then p – q is divisible by x and therefore, q – p is also divisible by x.

Thus, p R q ⇒ q R p.

Hence, R is symmetric.

(iii) Let p, q, r ∈ Z and p R q, q R r both hold. Then p – q is divisible by x and q – r is also divisible by x. Therefore, p – r = (p – q) + (q – r) is divisible by x.

Thus,  p R q and q R r ⇒ p R r

Therefore, R is transitive.

Hence R is an equivalence relation on set Z

          

      Q.4. Let S be the set of all lines in 3 dimensional 

      space. A relation ρ is defined

      on S by “l ρ m if and only if l lies on the plane of 

       m” for l, m ∈ S.

      Examine if ρ is (i) reflexive, (ii) symmetric, 

    (iii) transitive

Solution:

(i) Reflexive: Let l ∈ S. Then l is coplanar with itself.

Therefore, lρl holds for all l in S.

Hence, ρ is reflexive

(ii)  Symmetric: Let l, m ∈ S and l ρ m holds. Then l lies on the plane of m.

Therefore, m lies on the plane of l. Thus, l ρ m ⇒  m ρ l and therefore ρ is symmetric.

(iii) Transitive: Let l, m, p ∈ S and l ρ m, m ρ p both hold. Then l lies on the plane of m and m lies on the plane of p. This does not always implies that l lies on the plane of p.

That is, l ρ m and m ρ p do not necessarily imply l ρ p.

Therefore, ρ is not transitive.

Since, R is reflexive and symmetric but not transitive so, R is not an equivalence relation on set Z

Transitive relation

TRANSITIVE RELATION

Let P be a set on which the relation R is defined.

R is said to be transitive, if

(p, q) ∈ R and (q, r) ∈ R ⇒ (p, r) ∈ R,

That is pRq and qRr ⇒ pRr where p, q, r ∈ P.

The relation is said to be non-transitive, if

(p, q) ∈ R and (q, r) ∈ R do not imply (p, r ) ∈ R.

For example, in the set P of whole numbers if the relation R be defined by ‘x less than y’ then

p < q and q < r imply p < r, that is, pRq and qRr ⇒ pRr.

Hence this  is a transitive relation.

Solved examples of transitive relation on given set:

1. Let L be given positive integer.

Let R = {(p, q) : p, q  ∈ Z and (p – q) is divisible by L}.

Show that R is transitive relation.

Solution:

Given R = {(p, q) : p, q ∈ Z, and (p – q) is divisible by L}.

Let p, q, r ∈ R such that (p, q) ∈ R and (q, r) ∈ R. Then

   ⇒ (p – q) is divisible by L,and (q – r) is divisible by L.

   ⇒ {(p – q) + (q – r)} is divisible by L.

   ⇒ (p – r) is divisible by L.

   ⇒ (p, r) ∈ R.

Therefore, (p, q) ∈ R and (q, r) ∈ R   ⇒ (p, r) ∈ R.

So, R is a transitive relation.

2. A relation ρ on the set N is given by “ρ = {(x, y) ∈ N × N : x is divisor of y}”. Examine whether ρ is transitive or not transitive relation on set N.

Solution:

Given ρ = {(x, y) ∈ N × N : x is divisor of y}.

Let p, q, r ∈ N and (p, q) ∈ ρ and  (q, r ) ∈ ρ. Then

  (p, q) ∈ ρ and  (q, r ) ∈ ρ  ⇒ p is divisor of q and q is divisor of r.

⇒ p is divisor of r.

 ⇒ (p, r) ∈ ρ

Therefore, (p, q) ∈ ρ and q, r) ∈ ρ ⇒ (p, r) ∈ ρ.

Hence, R is a transitive relation.

Symmetric Relation

SYMMETRIC RELATION

Let P be a set on which the relation R is defined. Then R is said to be a symmetric relation, if (p, q) ∈ R ⇒ (q, p) ∈ R, that is, pRq ⇒ qRp for all (p, q) ∈ R.

e.g- the set P of natural numbers. If a relation P be defined by “x + y = 5”, then this relation is symmetric in P, for

p + q = 5 ⇒ q + p = 5

But in the set P of natural numbers if the relation R be defined as ‘x is a divisor of y’, then the relation R is not symmetric as 4R8 does not imply 8R4  for, 4 divides 8 but 8 does not divide 4.

Solved example on symmetric relation on set:

1. A relation R is defined on the set Z by {p R q if p – q is divisible by 5} for p, q ∈ Z. Examine if R is a symmetric relation on Z.

Solution:

Let p, q ∈ Z such that pRq hold. Then p – q is divisible by 5, i.e. p – q =5z and therefore q – p =5z hence q - p is divisible by 5.

Thus, pRq ⇒ qRp and therefore R is symmetric.

2. A relation R is defined on the set Z (set of all integers) by {pRq if and only if 2p + 3q is divisible by 5}, for all p, q ∈ Z. Examine if R is a symmetric relation on Z.

Solution:

Let p, q ∈ Z such that pRq holds i.e., 2p + 3q = 5z, which is divisible by 5. Now, 2q + 3p = 5p – 2p + 5q – 3q = 5(p + q) – (2p + 3q) is also divisible by 5.

Therefore pRq implies qRp for all p,q in Z i.e. R is symmetric.

3. Let R be a relation on Q, defined by R = {(p, q) : p, q ∈ Q and p – q ∈ Z}. Show that R is Symmetric relation.

Solution:

Given R = {(p, q) : p, q ∈ Q, and p – q ∈ Z}.

Let (p,q) ∈ R ⇒ (p – q) ∈ Z, i.e. (p – q) is an integer.

               ⇒ -(p – q) is an integer

               ⇒ (q – p) is an integer

               ⇒ (q, p) ∈ R

Thus, (p, q) ∈ R ⇒ (q, p) ∈ R

Therefore, R is symmetric.

4. Let m be given fixed positive integer.

Let R = {(p, q) : p, q  ∈ Z and (p – q) is divisible by m}.

Show that R is symmetric relation.

Solution:

Given R = {(p, q) : p, q ∈ Z, and (p – q) is divisible by m}.

Let (p, q) ∈ R . Then,

                ⇒ (p – q) is divisible by m

               ⇒ -(p – q) is divisible by m

               ⇒ (q – p) is divisible by m

               ⇒ (q, p) ∈ R

Thus, (p, q) ∈ R ⇒ (q, p) ∈ R

Therefore, R is symmetric relation on set Z.

Reflexive Relation

Reflexive Relation

It is a binary element in which every element is related to itself.

Let A be a set and R be the relation defined on it.

R is set to be reflexive, if (p, p) ∈ R for all p ∈ A that is, every element of A is R-related to itself, in other words pRp for every p ∈ A.

A relation R on a set A is not reflexive if there is at least one element p ∈ A such that (p, p) ∉ R.

Consider, for example, a set A = {l,m,n,o}.

The relation R1

But the relation R2

Solved example of reflexive relation on set:

1. A relation R is defined on the set Z (set of all integers) by “aRb if and only if 2a + 3b is divisible by 5”, for all a, b ∈ Z. Examine if R is a reflexive relation on Z.

Solution:

Let b ∈ Z. Now 2b + 3b = 5b, which is divisible by 5. Therefore bRb holds for all b in Z i.e. R is reflexive.

2. A relation R is defined on the set Z by “aRb if a – b is divisible by 5” for a, b ∈ Z. Examine if R is a reflexive relation on Z.

Solution:

Let p ∈ Z. Then p – p is divisible by 5. Therefore pRp holds for all p in Z i.e. R is reflexive.

3. Consider the set Z in which a relation R is defined by ‘aRb if and only if a + 3b is divisible by 4, for a, b ∈ Z. Show that R is a reflexive relation on on setZ.

Solution:

Let a ∈ Z. Now a + 3a = 4a, which is divisible by 4. Therefore aRa holds for all a in Z i.e. R is reflexive.

4. A relation ρ is defined on the set of all real numbers R by ‘xρy’ if and only if |x – y| ≤ y, for x, y ∈ R. Show that the ρ is not reflexive relation.

Solution:

The relation ρ is not reflexive as x = -2 ∈ R but |x – x| = 0 which is not less than -2(= x).