What-exactly-does-f-C-C-mean-

Question Number 2367 by Filup last updated on 18/Nov/15

What exactly does f : C \to C mean ?

Answered by RasheedAhmad last updated on 18/Nov/15

A f u n c t i o n f w h o s e d o m a i n a n d

r a n g e b o t h a r e C (s e t o f c o m p l e x

n u m b e r s)

Commented by Filup last updated on 18/Nov/15

Oh, I see! What other variations are there ?

Could I have some examples ?

Commented by RasheedAhmad last updated on 18/Nov/15

T a k e a n e x a m p l e . I f

g (x) = \frac{1}{x}

I f x \in N i - e d o m a i n o f g (x) t h e n

g (x) \in (0, 1] i - e t h e r a n g e o f g (x)

W e c a n w r i t e

g : N \to (0, 1]

M r . p r a k a s h j a i n, 123456, M r . Y o z z i

a n d s o m e o t h e r s c a n g i v e m o r e

s a t i s f y i n g a n s w e r .

Commented by 123456 last updated on 18/Nov/15

g (z) = \frac{1}{z}

all complex number except 0 can be pruged

then the domain is C / {0}

the range is all complex except 0, since

1 / z \to 0, z \to \tilde{\infty} (complex infinity)

then it can be give by

g : C / {0} \to C / {0} simple pole at z = 0

if we take

h (z) = {\begin{cases} g (z) & z \neq 0 \\ 0 & z = 0 \end{cases} (removing problem at z = 0)

h : C \to C

in general you can take

g : A \to B where A \subset C / {0} is a subset of

the complex excluding 0 (maximal domain

possible) and B \subset C is a subset that

depends on A

Commented by Yozzi last updated on 18/Nov/15

A n e x a m p l e c o u l d b e a f u n c t i o n f

w h i c h m a p s m e m b e r s o f t h e s e t o f r e a l

n u m b e r s t o m e m b e r s o f t h e s e t o f

i n t e g e r s . I n t h i s c a s e w e c o u l d d e f i n e

t h i s s t a t e m e n t b y n o t a t i o n a s f o l l o w s :

f : R \to Z

w h e r e f (x) = ⌊ x ⌋, x \in R . ⌊ x ⌋ i s t h e

g r e a t e s t i n t e g e r l e s s t h a n o r e q u a l

t o x . (E . g ⌊ π ⌋ = 3 ∵ π = 3.14 \dots ∣

⌊ e ⌋ = 2 ∵ e = 2.718 \dots ∣ ⌊ 0.1 ⌋ = 0 ∣

⌊ - 236.653 ⌋ = - 237 ∣ ⌊ - 10 ⌋ = - 10)

f i s a m a n y - t o - o n e f u n c t i o n s i n c e

f o r a l l n ⩽ x < n + 1, w h e r e n \in Z,

⌊ x ⌋ = n . T h i s t h e n s h o w s h o w t h e o u t p u t

o f f a r e i n t e g e r s g i v e n t h a t t h e i n p u t s

a r e r e a l . T h e d o m a i n o f f i s R a n d i t s

c o d o m a i n i s Z . f i s b i j e c t i v e s i n c e

(1) e v e r y m e m b e r o f t h e d o m a i n i s

m a p p e d t o o n l y o n e m e m b e r o f t h e

c o d m a i n . S o ∄ x \in R s u c h t h a t ⌊ x ⌋ h a s

m o r e t h a n o n e r e s u l t (v e r t i c a l l i n e t e s t

p r o v e s t h i s)

a n d

(2) f i s s u r j e c t i v e s i n c e f o r e v e r y

v a l u e o f n u m b e r s i n t h e c o d o m a i n, \exists x \in R m a p p e d t o f (x) \in Z .

S o, f g e n e r a t e s t h e e n t i r e s e t o f i n t e g e r s,

w h i c h i s i n f i n i t e l y c o u n t a b l e, s i n c e

t h e s e t o f r e a l n u m b e r s h a s a n o n - f i n i t e

c a r d i n a l i t y . T h e r a n g e o f f i s Z .

L e t P = {p r i m e n u m b e r s} a n d

{\bar{Q}}^{+} = {p o s i t i v e i r r a t i o n a l n u m b e r s} .

I f f i s a f u n c t i o n d e f i n e d b y

f (x) = \sqrt{x}, x \in P

t h e n f m a p s t h e p r i m e n u m b e r s t o

s o m e m e m b e r s o f {\bar{Q}}^{+} .

S o f : P \to {\bar{Q}}^{+} .

Commented by Filup last updated on 18/Nov/15

Thank you all so much! I understand now!

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