1. Show that f is a bijective

Transcription

1 3 Functions 4 SECTION C Inverse Functions B the end o this section ou will be able to show that a given unction is bijective ind the inverse unction You need to know our work on injections and surjections rom section B thoroughl to understand this section. It is a much more diicult section than the previous two (3A and 3B). C Bijective Functions Guess what the term bijective unction means? A unction which is both injective (one to one) and surjective (onto). A bijective unction is also called a bijection. Hence a bijection is a unction rom a set A (domain) to a set B (codomain) which is both injective and surjective. Thereore a bijective unction is both one to one and onto. A B Fig 6 Domain Fig 6 shows a bijective unction. Codomain Example 7 Let : unction. Solution be deined b x x. Show that is a bijective

2 3 Functions 5 What do we need to prove? We have to prove that the given unction x x is one to one and onto. First we show that the unction is one to one that is is injective. We show How? Well x x and gives x x equating these we have x x which gives x Hence the given unction is one to one. Next we show is onto (surjective). How? Let x where is in the codomain and then ind an x in the domain. x x Transposing Since (because it is in the codomain) which means that is a real number, thereore x is also a real number so we have ound an x in the domain,. Hence is onto (surjective). Since the given unction x x is both, one to one and onto, we conclude that it is a bijective unction. One o the diiculties in this section is it is eas to get conused with all the notation used to show a given unction is bijective. For a unction to be injective (one to one) we consider elements x and in the domain: This is an INJECTION x x Fig 7 To show surjection (onto) we consider to be in the codomain and then ind an x in the domain which corresponds to this via the unction.

3 3 Functions 6 x This is a SURJECTION Fig 8 From the last section ou might have noticed that an unction can be made surjective (onto) b changing the codomain o that unction. Similarl we can change the domain to make the unction injective (one to one). In the next example we use the method o completing the square to ind an appropriate codomain or the unction to be surjective. I ou have orgotten this method o completing the square or ou have not covered it, then see Appendix B. Example 8 Let : be deined b so that is surjective (onto)? x x 4x 5. What is the largest codomain Solution We appl the technique o completing the square on Hence we have x Because 4 5 x x 4x 5. x 4x 5 x 4 5 Because x x 4x 4 x x x 4 5 x What do we know about the square part, x? It is alwas positive or zero because squaring an real number does not give a negative number. Hence x 0 Thereore x x 0. Hence x. Thereore the largest codomain is going to be the set o real numbers which are greater than or equal to. How do we write this? B x x.

4 3 Functions 7 Next we prove that the unction given b the ormula in Example 8 is bijective or the appropriate domain and codomain. Example 9 Let A x x and B x x deined b Solution. Let : A B be x x 4x 5. Prove that is a bijective unction. We have to prove that the given unction x x 4x 5 is one to one and onto. First we show that the unction is one to one that is is injective. We need to show How? gives x x x 4x x x 4x 4 0 Rearranging x x 4 x 0 Remember x x x x x 4 0 Factorizing x 0 or x 4 0 x or x 4 Solving or x Consider the solution x 4. Remember is a member o the domain, A x x, which means it is a real number greater than or equal to. I then x 4 and we have x. I then x 4 which means x cannot be in the domain, A x x. Wh not? Because A x x we dismiss the solution x 4. and so onl contains real numbers. Hence Thereore we onl have the solution x. What can we conclude rom this? The given unction is one to one. What else do we need to show or to be bijective? The given unction x x 4x 5 is onto. How? The procedure outlined in the last section that is let x be in the codomain and then ind an x in the domain which corresponds to this. How? B solving x or x. We have

5 3 Functions 8 x x x x How do we ind x rom x B transposing x 4 5 From Example 8? x Subtracting x Taking Square Root x Adding Remember x means x or x. Choose x. Wh? Because the domain is A x x and x, real numbers, and we need to locate an x in the domain or to be onto. Is x in the domain? Yes because is in the codomain and thereore so x is a real number greater than or equal to. We have x Fig 9 A B Hence x A x x which means we have ound an x in the domain. We conclude that is onto. Since is one to one (injective) and onto (surjective) thereore is bijective. C Inverse Functions What does the term inverse unction mean? Inverse in everda lie means opposite or reverse. What is the inverse o A. Ting shoe laces? B. Multipling b 5? C. Adding 3 and then multipling b 43? The inverses are

6 3 Functions 9 A. Unting shoe laces. B. Dividing b 5. C. Dividing b 43 and then subtracting 3. Inverse unctions simpl unlock what ou have done. Note the inverse o part C. You undo the last operation irst. For example let : A B be a unction such that mean? x. What does this It means that the unction takes x to. What do ou think the inverse unction does? The inverse unction must take back to x. x Inverse Function Fig 0 For example i 00 then the inverse unction sa g must assign 00 to, that is g 00. Consider another example i France Paris then the inverse unction sa g must take Paris back to France, that is g Paris France. I Blue Green then the inverse unction sa g must take Green back to blue, g Green Blue. The deinition o the inverse unction is: Deinition (3.4) Let : A B be a unction. I or an arbitrar x in A we have x B then the unction g : B A given b g x is called the inverse unction o. The inverse unction is normall denoted b. We have in

7 3 Functions 30 x x Fig For brevit we sa the unction and codomain o Since : A B? thereore : B A so the domain is B and codomain is A o is the inverse o. What is the domain because the inverse unction reverses. An important proposition in unctions is that the unction : A B has an inverse i and onl i the unction is bijective. We prove this proposition later in this section. This result means that bijective unctions and inverse unctions are equivalent. To see wh consider the ollowing examples. Let be deined b x x : 0 4. Then Hence is not injective (one to one) but is surjective (onto). But where does the inverse unction, 4? Is it or? There is no unique value o initial value. A g B, take 4? What is the value o the inverse unction at 4. We do not necessaril go back to our 4 Fig We need the unction to be injective (one to one) or it to have an inverse. Otherwise which value would 4 go back to? Now consider a unction which is injective but not surjective. Let : be deined b x x

8 3 Functions 3 Then is injective but not surjective (onto). What is 3 There is no x in the domain, This means that 3 numbers., such that x x 3 equal to? is not deined as we are ONLY allowed to have real? 3 In act non o the negative numbers are assigned b the unction. Hence we need to be surjective (onto) or to have an inverse unction. Otherwise 3 Fig 3 cannot go back to an value in the domain. Next we prove this important result that the unction has an inverse i and onl i it is bijective. This is a diicult proo to ollow. It is eas to get conused with all the similar statements. It seems that we are presenting a circular argument to prove our result but we are not. Read the proo careull. Proposition (3.5) Let : A B be a unction. Then has an inverse is bijective. Proo. Remember a proo means we have to prove the result both was, that is i has an inverse then is bijective and also i is bijective then has an inverse.. Assume has an inverse denoted b g sa. What do we need to prove or this part? Need to prove is injective and surjective. How do we show is injective? Required to prove gives Let x z. x x

9 3 Functions 3 x z g Fig 4 Then b the deinition o the inverse unction Hence injective. g z x and g z x g z. Thereore we have shown that the given unction is What else do we need to show? The unction is surjective. How? B the procedure outlined in the last section, that is assume x member o the codomain, B, and then show that there is an x in the domain, A, which corresponds to this. Let x then b the deinition o the inverse unction g x we have located an x in the domain such that unction is surjective. Since is injective and surjective thereore is bijective. is a x. Hence thereore the given. Now let s go the other wa, that is assume is bijective and deduce that has an inverse unction. Let the unction be bijective. Required to prove that there is a unction g : B A which reverses what the unction does. Deine g : B A b g x where x is the element in A such that x. We need to prove that such a g is a unction. For g : B A to be a unction we need ever element in B assigned to onl one element in A because this is the deinition o a unction. (Everthing in B has onl one destination). Let B then there is an x in A such that x. Wh? Because is onto (surjective) so ever element in B is assigned to.

10 3 Functions 33 x g Codomain o x Fig 5 Thereore everthing in B gets assigned b g to A. Need to prove that g assigns elements in B to onl one element in A. We have z x. Suppose g and thereore Since is injective then b z x z then b our deinition o g we have (3.) A unction is injective x x we have x z. Hence g assigns everthing in B to a unique element in A. Thereore g is a unction and is the inverse unction o. Tr reading through the proo again and see i ou can do parts o the proo without looking. Hence a unction has an inverse i and onl i it is bijective. In Example 7 we showed the ollowing unction is bijective: : deined b x x. How do we ind the inverse o this unction? The procedure to ind the inverse unction is the same technique as to show a given unction is surjective. Let Example 0 Let : x x then solve this equation or x. be deined b x x. Determine the inverse unction. Solution Let x and solve this equation or x. x What is x A equal to? x B

11 3 Functions 34 x is a unction o x thereore we have Remember x Not Equal x Replacing the b x x in the above because the inverse unction is a unction o x. We have actuall ound the inverse unction x in Example 7 when we showed the unction was surjective. Example Let A x x and B x x b x x 4x 5. Find the inverse unction x Solution. Let : A B be deined We have alread shown that the given unction,. x x 4x 5, is bijective in Example 9. Remember the inverse unction is given b letting x and then solving this equation or x which is what we used to show the given unction is surjective (onto). From Example 9 we have b solving x or x. What is the inverse unction, x x x, equal to? is a unction o x thereore we have Replacing b x x x in the above Example Let A x x and B x x 3 b 3x x x Find the inverse unction Solution x. and : A B be deined We can show the given unction is bijective and in the process ind the inverse unction. How do we show the given unction is bijective?

12 3 Functions 35 First show it is injective (one to one) b letting x x : We have 3x x x and and then deduce 3. Equating these and simpliing 3x 3 x 3x 3 x Multipling through b x 3x 3x 3x 3 Expanding Brackets 3x 3 Simpliing x Dividing through b 3 Hence is injective. What else do we need to show? Required to prove that is surjective. How? B considering to be in the codomain and inding an x in the domain such that Now x. We need to solve x or x. 3x x x 3 x Multipling b x x 3 x Opening Brackets x 3 x Collecting like terms x 3 Factorizing x 3 x is in the domain provided 3. Since is in the codomain, 3 B x x 3, thereore 3. Hence we have ound an x in the domain such that x. Thereore is surjective. Hence the given unction is bijective. What is the inverse unction x equal to? x Replacing the b x x x 3 in the above Remember this is same as showing that the given unction is surjective. SUMMARY A unction is a bijective unction i it is both injective (one to one) and surjective (onto).

13 3 Functions 36 Let : A B be a unction. I x is given b x. then the inverse unction A unction has an inverse i and onl i is bijective. : B A To ind the inverse unction we onl have to show the unction is bijective and in the process we ind the inverse unction b letting this equation or x. x and then solve