![FUNCTIONS
Thus, ho(gof) and (hog)of have the same domain A.
Now f:AB and hog:BD
(hog)of:AD
Part-2 Let aA
[ho(gof)] (a) = h[(gof) (a)]
= h[g(f (a))]
= (hog)[f (a)]
= (hog)of (a)
Hence, from part-1 and part-2 we conclude that
ho(gof)=(hog)of aA
Do you know
(hog)of?
ho(gof) : A D](https://image.slidesharecdn.com/mat1ajrfunm05theorems-240423115231-7a59aeaf/85/FUNCTION-M05-THEOREMS-WITH-PROOFS-FOR-BOARD-LEVEL-4-320.jpg)
![FUNCTIONS
Part 2
Hence, from Part-1 and Part-2 we conclude that foIA=f
Let aA, then (foIA) (a) = f(IA(a))
(ii) To prove that IBof = f
Since f:AB and IB:BB
IB(b) = b, bB
(i.e.,) A
𝐟
A
𝐈𝐁
𝐁
IBof :
= f(a)
Thus f and IBof both have the same domain A
Part 1
We know
[ ∵ IA(a) = a a A ]
A B
By observing
f and IBof they
have the same
domain A](https://image.slidesharecdn.com/mat1ajrfunm05theorems-240423115231-7a59aeaf/85/FUNCTION-M05-THEOREMS-WITH-PROOFS-FOR-BOARD-LEVEL-12-320.jpg)
![FUNCTIONS
Hence from part-1 and part-2 we conclude that IBof = f
Hence proved.
Part2
[ ∵ IB(b)=b, bB and f(a)B]
For any aA, (IBof)(a)=IB(f(a))=f(a)](https://image.slidesharecdn.com/mat1ajrfunm05theorems-240423115231-7a59aeaf/85/FUNCTION-M05-THEOREMS-WITH-PROOFS-FOR-BOARD-LEVEL-13-320.jpg)
FUNCTION M05 THEOREMS WITH PROOFS FOR BOARD LEVEL
- 3. FUNCTIONS Theorem(1): If f:AB, g:BC & h:CD are three functions. Then prove that ho(gof)=(hog)of i.e., composition of functions is associative. Proof Part-1 Given that f:AB and g:BC Now, gof:AC and h:CD Also, g:BC and h:CD gof:AC ho(gof) : A D hog:BD Do you know gof? Do you know ho(gof)? Do you know hog?
- 4. FUNCTIONS Thus, ho(gof) and (hog)of have the same domain A. Now f:AB and hog:BD (hog)of:AD Part-2 Let aA [ho(gof)] (a) = h[(gof) (a)] = h[g(f (a))] = (hog)[f (a)] = (hog)of (a) Hence, from part-1 and part-2 we conclude that ho(gof)=(hog)of aA Do you know (hog)of? ho(gof) : A D
- 5. FUNCTIONS Theorem(2): If f:AB, g:BC are two bijective functions then prove that gof:AC is also a bijective function Proof: Given that f, g are bijections. (i) To prove that gof:AC is one-one Let a1,a2A, then f(a1),f(a2)B Hence, g(f(a1)), g(f(a2))C Let (gof)(a1) = (gof)(a2) g(f(a1)) = g(f(a2)) Hence f, g are both one-one and onto functions Bijection means it is both one-one & on-to To prove gof:AC is bijective, firstly we prove it is one-one Since f:AB is a function Since g:BC is a function Since f and g are functions we know that gof:AC is a function These are images of a1, a2
- 6. FUNCTIONS (ii) To prove that gof:AC is onto Let c be any element in co-domain C Since g:BC is onto Now, bB and f:AB is onto bB aA such that f(a)=b(2) Thus gof:AC is one-one. f(a1) = f(a2) (∵ g is one-one) a1 = a2 (∵ f is one-one) Later we prove gof:AC onto function cC bB such that g(b)=c(1) By the definition of onto function
- 7. FUNCTIONS Hence, gof:AC is both one one and onto (i.e.,) gof is bijective function. As c is arbitrary, every element in C has pre-image in domain A under gof. Thus gof : A C is onto. Hence proved. From eq(1) for cC, (gof)(a)=c for some aA c = g(b) = g(f(a)) c = (gof)(a) From eq(2) b=f(a)
- 8. FUNCTIONS Theorem(3): If f : A B, g : B C are two bijective functions then prove that (gof)-1 = f-1og-1 Proof Given that f : A B, g : B C are bijective Part 1 i.e., A 𝐟 B 𝐠 C are bijections gof: (gof)-1:CA is inverse of (gof) Also f-1:BA, g-1:CB are both bijections. Firstly we prove (gof)-1, f-1og-1 have the same domain AC is a bijection If f,g are bijections, f-1, g-1 are also bijections.
- 9. FUNCTIONS Thus (gof)-1 and f-1 og-1 both have the same domain “C”. i.e., C 𝐠−𝟏 B 𝐟−𝟏 𝐀 are bijections. (f-1og-1):CA Part 2 Let c be any element in C. By data g:BC is a bijection (i.e.,) g:BC is onto. Hence, for cC a unique bB such that g(b) = c b=g-1(c) and (gof)-1:CA By observing (gof)-1, f-1og-1, they have the same domain C By the definition of onto function
- 10. FUNCTIONS Also, f:AB is a bijection (i.e.,) f:AB is onto. Hence, bB a unique aA such that f(a) = b Now c = g(b) a = f-1(b) = g(f(a)) c= (gof)(a) (gof)-1(c) = a(1) Also, (f-1og-1) (c)= f-1(g-1(c)) = f-1(b) = a (f-1og-1) (c) = a (2) From (1) & (2) (gof)-1 (c) = (f-1og-1) (c), cC We know g(b) = c We know g-1(c)=b
- 11. FUNCTIONS Theorem(4): If f:AB is a function and IA & IB are identity functions on A,B respectively then prove that foIA=f=IBof Proof (i) To prove that foIA=f Part 1 We know that IA:AA. foIA : IA(a) = a aA Given that f: A B. Thus A 𝐈𝐀 A 𝐟 𝐁 Thus foIA and f both have the same domain A Firstly we prove foIA, f have the same domain A B By observing foIA, f they have the same domain A
- 12. FUNCTIONS Part 2 Hence, from Part-1 and Part-2 we conclude that foIA=f Let aA, then (foIA) (a) = f(IA(a)) (ii) To prove that IBof = f Since f:AB and IB:BB IB(b) = b, bB (i.e.,) A 𝐟 A 𝐈𝐁 𝐁 IBof : = f(a) Thus f and IBof both have the same domain A Part 1 We know [ ∵ IA(a) = a a A ] A B By observing f and IBof they have the same domain A
- 13. FUNCTIONS Hence from part-1 and part-2 we conclude that IBof = f Hence proved. Part2 [ ∵ IB(b)=b, bB and f(a)B] For any aA, (IBof)(a)=IB(f(a))=f(a)
- 14. FUNCTIONS Theorem(5): If f:AB is a bijective function. IA& IB are identity functions then prove Proof: Part 1 Given f:AB is a bijection then we know, f-1:B A is also a bijection Now, f-1:BA and f:AB (i) To prove that fof-1=IB i) fof-1=IB ii) f-1of=IA (i.e.,) B 𝐟−𝟏 𝐀 𝐟 𝐁 Firstly, we prove fof-1, IB have the same domain fof-1: BB
- 15. FUNCTIONS Part 2 Let bB. Since f:AB is a bijection a unique a A such that f(a)=bf-1(b)=a Thus fof-1 and IB both have the same domain B Also by the definition of identity function, IB(b)=b b B Now, for any bB, Also by the definition of identity function, IB:BB (fof-1)(b) =f(f-1(b)) =f(a)=b=IB(b) Hence, from part-1 and part-2, we conclude that fof-1=IB
- 16. FUNCTIONS (ii) To prove that f-1of=IA Part 1 Since f:AB and f-1:BA (i.e.,) A 𝐟 𝐁 𝐟−𝟏 𝐀 f-1of: AA Also by the definition of identity function IA:AA Thus, f-1of and IA both have the same domain of A
- 17. FUNCTIONS Part 2 Also by the definition of identity function Let a1A since f:AB is a bijection, a unique b1B such that f(a1)=b1 IA(a1)=a1 a1 A Now for any a1A f-1(b1) =a1 (f-1of)(a1) =f-1(f(a1)) =f-1(b1) =a1 =IA(a1) Hence, from part-1 and part-2 we conclude that f-1of = IA
- 18. FUNCTIONS As f: A B is a function so f(a1), f(a2)B Theorem(6): If f: A B and g: B A are two functions such that gof = IA and fog = IB, then g = f-1 Proof: First of all, we prove that f: A B is a bijection. To show that f is one-one Let a1,a2A. Bijection means it is both one-one & onto Firstly, we prove f is one-one Assume f(a1) = f(a2) g(f(a1)) = g(f(a2)) (∵g:BA is a function) (gof)(a1) = (gof)(a2)
- 19. FUNCTIONS IA(a1) = IA(a2) ( ∵ gof = IA) a1 = a2 f: A B is one-one function. To show that f is onto: Let b be any element in B, then as g : BA is a function, aA such that g(b) = a Now b = IB(b) = (fog)(b) = f(g(b)) = f(a) i.e., f(a) = b We know fog = IB
- 20. FUNCTIONS Thus for any bB, then there exists pre image aA under f. Hence f is onto ∴ f: A B is bijection. f: A B is bijection, so f-1 exists. To show that g = f-1 g: B A is a function. f-1: B A is also bijection. g, f-1 have same domain B. Let b be any element in B, then as g : BA is a function, a A such that g(b)=a for some a A
- 21. FUNCTIONS Thus, we get g=f-1 f : AB is bijection, bB aA such that f(a) = b g(b)=f-1(b) a = f-1(b) Hence proved We know g(b)=a
- 22. FUNCTIONS Theorem(7): If f:AB, g:BA are two functions and (i) gof:AC is one-one, then f:AB is to one-one (ii) gof:AC is on-to, then g:BC is to on-to Proof i) Let a1,a2 A Now f(a1)=f(a2) gof(a1)=gof(a2) (∵g:AB is a function) a1=a2 (∵gof is one-one) Thus, f(a1)=f(a2) g(f(a1))=g(f(a2)) a1=a2 i.e,f is one-one function. Hence proved If we take f(a1) = f(a2), we get a1 = a2
- 23. FUNCTIONS Thus g:BC is onto as every element in co-domain C has pre-image in domain B under g. Hence proved (ii) Let c be any element in C. As gof:AC is onto, There exists a A such that Thus pre-image of c is f(a)B under g (gof)(a)=c g(f(a))=c