[PDF] [PDF] 7 Functions Definition 71 A function f : A −→ B consists of: (1) the

[PDF] Homework  Solutions Math 3283W - Math User Home Pages

11 oct 2016 · On other problems the stated solution may be complete These are the only non-surjective functions (are you convinced?), so there are 8 - 2 = 6 (3) Classify each function as injective, surjective, bijective or none of these

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= {−5+4n : n ∈ N ∪ {0}} 3 Consider functions from Z to Z Give an example of ( a) a function that is injective but not surjective;

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That fact implies that /5(x) is injective Now /5(x) > 0 for all x G R and so /5(x) cannot be a surjective function from R to R Indeed, if x > 0, then y = /6(x) = x2> 0 and so g6(y) = Vx2= x for x > 0 If x < 0, then y = /6(x) = -x2< 0 and so g6(y) = - V-y = - Vx2= -ث xث = x for x < 0

[PDF] 2 Properties of Functions 21 Injections, Surjections - FSU Math

The examples illustrate functions that are injective, surjective, and bijective injection It is not a surjection because the range is not equal to the codomain For

[PDF] 7 Functions Definition 71 A function f : A −→ B consists of: (1) the

On the other hand, there is no person who is a million inches high, so f is not surjective either 1 Page 2 The function in (2) is neither injective nor surjective as 

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a function is only injective, you must prove that it is injective and not surjective Similarly, if you is injective Proof that fA,B is not surjective for general A and B:

[PDF] A → B and g : B → C are functions

28 oct 2011 · (a) g is not injective but g ◦ f is injective (b) f is not surjective but g ◦ f is surjective Solution The same example works for both Let A = {1}, B 

[PDF] MATH 2200 SPRING 2018 1 Show that if f : A −→ B and g : B

5 Give an example of the following functions: (a) A function f : Z>0 −→ Z that is injective but not surjective

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The definition of an injection and a surjection gives us a framework for how to go about proving that a function is or is not injective or surjective We give examples  

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7.Functions

Denition 7.1.Afunctionf:A!Bconsists of:

(1)the setA, which is called thedomainoff, (2)the setB, which is called therangeoff, (3)a rule which assigns to every elementa2Aan elementb2B, which we denote byf(a).

It is easy to write down examples of functions:

(1) LetAbe the set of all people and letB= [0;1). Letf(x) be the height of personx, to the nearest inch. (2)f:R!Rgiven byf(x) =x2. (3)f:R![0;1) given byf(x) =x2. (4)f: [0;1)!Rgiven byf(x) =x2. (5)f: [0;1)![0;1) given byf(x) =x2. (6)f:R! f0;1ggiven by f(x) =(

0 ifxis rational

1 ifxis irrational.

(7)f:N! f0;1ggiven by f(x) =(

0 ifxis even

1 ifxis odd.

Denition 7.2.Iff:A!Bandg:B!Care two functions,

then we cancomposefandgto get a new functiongf:A!C.

Ifa2A, then(gf)(a) =g(f(a)).

Iff:R!Ris the functionf(x) = sinxandg:R!Ris the

functiong(x) = 2x, thengf:R!Ris the function (gf)(x) =

2sinx, whilstfg:R!Ris the function (fg)(x) = sin2x.

Denition 7.3.Letf:A!Bbe a function.

We say thatfisinjectiveif wheneverf(a1) =f(a2)for some pair a

1,a22A, then in facta1=a2.

We say thatfissurjectiveif for everyb2B, we may nd an a2Asuch thatf(a) =b. We say thatfis abijectioniffis both injective and surjective. It is interesting to go through the examples above. The function in (1) is neither injective or surjective. Indeed there are lots of people who are ve foot nine, so just because there are two people with the same height, does not mean they are the same person. Sofis not injective. On the other hand, there is no person who is a million inches high, so fis not surjective either. 1 The function in (2) is neither injective nor surjective as well.f(1) =

1 =f(1), but 16=1. There is no real number whose square is1, so

there is no real numberasuch thatf(a) =1. The function in (3) is not injective but it is surjective.f(1) =f(1), and 16=1. But ifb0 then there is always a real numbera0 such thatf(a) =b(namely, the square root ofb). The function in (4) is injective but not surjective. Iff(a1) =f(a2), thena21=a22. As botha10 anda20, this impliesa1=a2. On the other hand, there is still no number whose square is1. The function in (5) is bijective. It is injective, as in (4) and it is surjective as in (3). The function in (6) is not injective but it is surjective.f(0) = 0 = f(1), but 06= 1. On the other hand, ifb2 f0;1gthen eitherb= 0 or b= 1. Butf(0) = 0 andf(p2) = 1. The function in (7) is not injective but it is surjective.f(0) =f(2) =

0, but 06= 2. On the other hand, ifb2 f0;1gthen eitherb= 0 or

b= 1. Butf(0) = 0 andf(1) = 1. (8)f:R3!Rgiven byf(~v) =kvk. (9)f:R3 f0g !R3given byf(~v) =~vkvk. (10)r:R!R3given byf(t) = (cost;sint;t). (11)f:R!(0;1) is given byf(x) =ex. The function in (8) is neither injective nor surjective. There are plenty of unit vectors and there are no vectors of negative length. The function in (9) is neither injective nor surjective. There are plenty of vectors which point in the same direction and the image consists of vectors of unit length. The function in (10) is injective but not surjective. The function in (11) is bijective. Iff:A!Bis a bijective function, thenfhas an inverse functiong:B!A.gf:A!Ais the identity, it sendsatoa. Similarlyfg:B!Bis the identity, it sendsbtob. The inverse of the exponential function is the logarithm g: (0;1)!Rgiven byg(x) = lnx. Denition 7.4.IfAandBare two sets, theproductofAandB, denotedABis the set of ordered pairs

ABf(a;b)ja2A;b2Bg:

Thegraphof a functionf:A!Bis the subset of the product f(a;f(a)ja2Ag AB:

The productRR=R2,RR2=R3, and so on.

2 Iff:R!Ris the functionf(x) =x2, then the graph offis the parabola,y=x2, f(x;x2)jx2Rg R2:

Now suppose thatg:R2!Ris the function

g(x;y) =x2+y2:

The graph ofgis the setz=x2+y2,

f(x;y;x2+y2)j(x;y)2R2g R3: One way to picturegis to slice it using level curves and level sets. Ifg:R2!Ris any function, the level curve ofgat heightcis the set f(x;y)2R2jg(x;y) =cg R2: In the example whenf(x;y) =x2+y2, the level curves are concentric circles of radiuspc, centred at the origin. Ifc= 0 the level curve is the origin and ifc <0 the level curve is empty. Note that graph has a minimum at the origin. Ifg:R3!Ris any function, the level set ofgat heightcis the set f(x;y;z)2R3jg(x;y;z) =cg R3: Ifg(x;y;z) =zx2y2then the level sets are parallel paraboloids. It is interesting to try to visualise various functions. If we consider f(x;y) =x2y2, then this is an upside down paraboloid, which has a maximum at the origin. Note that the level curves are the same as the level curves forf(x;y) =x2+y2(for dierent values ofc, of course). Now consider the functionf(x;y) =x2y2. If we consider xing y=band varyingx, then we get a parabolaz=x2b2. If we consider xingx=aand varyingy, then we get an upside down parabola, z=a2y2. In fact the origin is a saddle point, a point which is neither a maximum nor a minimum. 3quotesdbs_dbs6.pdfusesText_12