Functions - Part 2 Supartha Podder One-to-One and Onto Functions A - PowerPoint PPT Presentation

Functions - Part 2 Supartha Podder

One-to-One and Onto Functions A function f : A → B is said to be one-to-one, or an injunction, if and only if f ( a ) = f ( b ) implies that a = b for all a , b ∈ A . A function is said to be injective if it is one-to-one. ∀ a ∈ A , ∀ b ∈ A ( f ( a ) = f ( b ) → a = b ) 1/15

Onto Functions Onto A function f from A to B is called onto, or a surjection, if and only if for every element b ∈ B there is an element a ∈ A with f ( a ) = b . A function f is called surjective if it is onto. ∀ y ∈ B ∃ x ∈ A ( f ( x ) = y ) 2/15

Example 3/15

Example of Bijective Function Let f : ( R + × R + ) → ( R + × R + ) be the function defined by f ( r , s ) = (2 r , rs ). Proof that f is bijective. Proof of Injectivity 4/15

Example of Bijective Function Let f : ( R + × R + ) → ( R + × R + ) be the function defined by f ( r , s ) = (2 r , rs ). Proof that f is bijective. Proof of Injectivity Let ( a , b ) , ( c , d ) ∈ R + × R + be arbitrary elements of f ’s domain. 4/15

Example of Bijective Function Let f : ( R + × R + ) → ( R + × R + ) be the function defined by f ( r , s ) = (2 r , rs ). Proof that f is bijective. Proof of Injectivity Let ( a , b ) , ( c , d ) ∈ R + × R + be arbitrary elements of f ’s domain. Assume f ( a , b ) = f ( c , d ) (Goal is to prove ( a , b ) = ( c , d ).) 4/15

Example of Bijective Function Let f : ( R + × R + ) → ( R + × R + ) be the function defined by f ( r , s ) = (2 r , rs ). Proof that f is bijective. Proof of Injectivity Let ( a , b ) , ( c , d ) ∈ R + × R + be arbitrary elements of f ’s domain. Assume f ( a , b ) = f ( c , d ) (Goal is to prove ( a , b ) = ( c , d ).) Then, (2 a , ab ) = (2 c , cd ) ⇒ 2 a = 2 c and ab = cd . 4/15

Example of Bijective Function Let f : ( R + × R + ) → ( R + × R + ) be the function defined by f ( r , s ) = (2 r , rs ). Proof that f is bijective. Proof of Injectivity Let ( a , b ) , ( c , d ) ∈ R + × R + be arbitrary elements of f ’s domain. Assume f ( a , b ) = f ( c , d ) (Goal is to prove ( a , b ) = ( c , d ).) Then, (2 a , ab ) = (2 c , cd ) ⇒ 2 a = 2 c and ab = cd . For two ordered pairs to be equal, both their 1st and 2nd coordinates must be equal. 4/15

Example of Bijective Function Let f : ( R + × R + ) → ( R + × R + ) be the function defined by f ( r , s ) = (2 r , rs ). Proof that f is bijective. Proof of Injectivity Let ( a , b ) , ( c , d ) ∈ R + × R + be arbitrary elements of f ’s domain. Assume f ( a , b ) = f ( c , d ) (Goal is to prove ( a , b ) = ( c , d ).) Then, (2 a , ab ) = (2 c , cd ) ⇒ 2 a = 2 c and ab = cd . For two ordered pairs to be equal, both their 1st and 2nd coordinates must be equal. Thus, a = c and ab − cd = 0. ⇒ ab − ad = 0 since a = c ⇒ a ( b − d ) = 0. 4/15

Example of Bijective Function Let f : ( R + × R + ) → ( R + × R + ) be the function defined by f ( r , s ) = (2 r , rs ). Proof that f is bijective. Proof of Injectivity Let ( a , b ) , ( c , d ) ∈ R + × R + be arbitrary elements of f ’s domain. Assume f ( a , b ) = f ( c , d ) (Goal is to prove ( a , b ) = ( c , d ).) Then, (2 a , ab ) = (2 c , cd ) ⇒ 2 a = 2 c and ab = cd . For two ordered pairs to be equal, both their 1st and 2nd coordinates must be equal. Thus, a = c and ab − cd = 0. ⇒ ab − ad = 0 since a = c ⇒ a ( b − d ) = 0. So, a = 0 or b = d . Now a = 0 is not possible because of the domain. So b = d . Hence f is injective. 4/15

Example of Bijective Function Let f : ( R + × R + ) → ( R + × R + ) be the function defined by f ( r , s ) = (2 r , rs ). Proof that f is bijective. Proof of Surjectivity 5/15

Example of Bijective Function Let f : ( R + × R + ) → ( R + × R + ) be the function defined by f ( r , s ) = (2 r , rs ). Proof that f is bijective. Proof of Surjectivity Let ( u , v ) ∈ R + × R + be an arbitrary element of f ’s co-domain. We want to find at least one element ( r , s ) ∈ R + × R + (in f ’s domain) such that f ( r , s ) = ( u , v ). 5/15

Example of Bijective Function Let f : ( R + × R + ) → ( R + × R + ) be the function defined by f ( r , s ) = (2 r , rs ). Proof that f is bijective. Proof of Surjectivity Let ( u , v ) ∈ R + × R + be an arbitrary element of f ’s co-domain. We want to find at least one element ( r , s ) ∈ R + × R + (in f ’s domain) such that f ( r , s ) = ( u , v ). Well, f ( r , s ) = ( u , v ) ⇔ (2 r , rs ) = ( u , v ) 5/15

Example of Bijective Function Let f : ( R + × R + ) → ( R + × R + ) be the function defined by f ( r , s ) = (2 r , rs ). Proof that f is bijective. Proof of Surjectivity Let ( u , v ) ∈ R + × R + be an arbitrary element of f ’s co-domain. We want to find at least one element ( r , s ) ∈ R + × R + (in f ’s domain) such that f ( r , s ) = ( u , v ). Well, f ( r , s ) = ( u , v ) ⇔ (2 r , rs ) = ( u , v ) ⇔ 2 r = u and rs = v . 5/15

Example of Bijective Function Let f : ( R + × R + ) → ( R + × R + ) be the function defined by f ( r , s ) = (2 r , rs ). Proof that f is bijective. Proof of Surjectivity Let ( u , v ) ∈ R + × R + be an arbitrary element of f ’s co-domain. We want to find at least one element ( r , s ) ∈ R + × R + (in f ’s domain) such that f ( r , s ) = ( u , v ). Well, f ( r , s ) = ( u , v ) ⇔ (2 r , rs ) = ( u , v ) ⇔ 2 r = u and rs = v . ⇔ r = u 2 and s = v r . 5/15

Example of Bijective Function Let f : ( R + × R + ) → ( R + × R + ) be the function defined by f ( r , s ) = (2 r , rs ). Proof that f is bijective. Proof of Surjectivity Let ( u , v ) ∈ R + × R + be an arbitrary element of f ’s co-domain. We want to find at least one element ( r , s ) ∈ R + × R + (in f ’s domain) such that f ( r , s ) = ( u , v ). Well, f ( r , s ) = ( u , v ) ⇔ (2 r , rs ) = ( u , v ) ⇔ 2 r = u and rs = v . ⇔ r = u 2 and s = v r . 2 ) = 2 v Thus, s = u . v ( u u ) ∈ R + × R + . 2 , 2 v Now we need to make sure that ( r , s ) = ( u 5/15

Example of Bijective Function Let f : ( R + × R + ) → ( R + × R + ) be the function defined by f ( r , s ) = (2 r , rs ). Proof that f is bijective. Proof of Surjectivity Let ( u , v ) ∈ R + × R + be an arbitrary element of f ’s co-domain. We want to find at least one element ( r , s ) ∈ R + × R + (in f ’s domain) such that f ( r , s ) = ( u , v ). Well, f ( r , s ) = ( u , v ) ⇔ (2 r , rs ) = ( u , v ) ⇔ 2 r = u and rs = v . ⇔ r = u 2 and s = v r . 2 ) = 2 v Thus, s = u . v ( u u ) ∈ R + × R + . 2 , 2 v Now we need to make sure that ( r , s ) = ( u 2 and since u , v ∈ R + so is 2 u Since u ∈ R + , so is u v . 5/15

Example of Bijective Function Let f : ( R + × R + ) → ( R + × R + ) be the function defined by f ( r , s ) = (2 r , rs ). Proof that f is bijective. Proof of Surjectivity Let ( u , v ) ∈ R + × R + be an arbitrary element of f ’s co-domain. We want to find at least one element ( r , s ) ∈ R + × R + (in f ’s domain) such that f ( r , s ) = ( u , v ). Well, f ( r , s ) = ( u , v ) ⇔ (2 r , rs ) = ( u , v ) ⇔ 2 r = u and rs = v . ⇔ r = u 2 and s = v r . 2 ) = 2 v Thus, s = u . v ( u u ) ∈ R + × R + . 2 , 2 v Now we need to make sure that ( r , s ) = ( u 2 and since u , v ∈ R + so is 2 u Since u ∈ R + , so is u v . 2 , 2 v Thus f ( u u ) = ( r , s ) and f is surjective. 5/15

Inverse Functions Let f be a one-to-one correspondence from the set A to the set B . The inverse function of f , f − 1 is the function that assigns to an element b belonging to B the unique element a in A such that f ( a ) = b . The inverse function of f is denoted by f − 1 . Hence, f − 1 ( b ) = a when f ( a ) = b .. Bijectivity = Injectivity + Surjectivity. 6/15

Example of Inverse Is f : Z 2 → Z 2 defined by f ( x , y ) = ( x − y , − x ) invertible? Injective: Let ( p , q ) , ( r , s ) ∈ Z × Z be arbitrary elements of f ’s domain. 7/15

Example of Inverse Is f : Z 2 → Z 2 defined by f ( x , y ) = ( x − y , − x ) invertible? Injective: Let ( p , q ) , ( r , s ) ∈ Z × Z be arbitrary elements of f ’s domain. Assume f ( p , q ) = f ( r , s ) (Goal is to prove ( p , q ) = ( r , s ).) 7/15

Example of Inverse Is f : Z 2 → Z 2 defined by f ( x , y ) = ( x − y , − x ) invertible? Injective: Let ( p , q ) , ( r , s ) ∈ Z × Z be arbitrary elements of f ’s domain. Assume f ( p , q ) = f ( r , s ) (Goal is to prove ( p , q ) = ( r , s ).) Then, ( p − q , − p ) = ( r − s , − r ) 7/15

Example of Inverse Is f : Z 2 → Z 2 defined by f ( x , y ) = ( x − y , − x ) invertible? Injective: Let ( p , q ) , ( r , s ) ∈ Z × Z be arbitrary elements of f ’s domain. Assume f ( p , q ) = f ( r , s ) (Goal is to prove ( p , q ) = ( r , s ).) Then, ( p − q , − p ) = ( r − s , − r ) ⇒ p = r and q = s 7/15

Example of Inverse Is f : Z 2 → Z 2 defined by f ( x , y ) = ( x − y , − x ) invertible? Injective: Let ( p , q ) , ( r , s ) ∈ Z × Z be arbitrary elements of f ’s domain. Assume f ( p , q ) = f ( r , s ) (Goal is to prove ( p , q ) = ( r , s ).) Then, ( p − q , − p ) = ( r − s , − r ) ⇒ p = r and q = s thus, ( p , q ) = ( r , s ). 7/15