Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

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

What exactly does f:C→C mean?

$$\mathrm{What}\:\mathrm{exactly}\:\mathrm{does}\:{f}:\mathbb{C}\rightarrow\mathbb{C}\:\mathrm{mean}? \\ $$

Answered by RasheedAhmad last updated on 18/Nov/15

A function f whose domain and range both are C (set of complex numbers)

$${A}\:{function}\:\:{f}\:\:{whose}\:{domain}\:{and} \\ $$$${range}\:{both}\:{are}\:\mathbb{C}\:\left({set}\:{of}\:{complex}\right. \\ $$$$\left.{numbers}\right) \\ $$

Commented by Filup last updated on 18/Nov/15

Oh, I see! What other variations are there? Could I have some examples?

$$\mathrm{Oh},\:\mathrm{I}\:\mathrm{see}!\:\mathrm{What}\:\mathrm{other}\:\mathrm{variations}\:\mathrm{are}\:\mathrm{there}? \\ $$$$\mathrm{Could}\:\mathrm{I}\:\mathrm{have}\:\mathrm{some}\:\mathrm{examples}? \\ $$

Commented by RasheedAhmad last updated on 18/Nov/15

Take an example. If g(x)=(1/x) If x∈N i−e domain of g(x) then g(x)∈(0,1] i−e the range of g(x) We can write g:N→(0,1] Mr. prakash jain,123456, Mr. Yozzi and some others can give more satisfying answer.

$${Take}\:{an}\:{example}.\:{If}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{g}\left({x}\right)=\frac{\mathrm{1}}{{x}} \\ $$$${If}\:{x}\in\mathbb{N}\:{i}−{e}\:{domain}\:{of}\:{g}\left({x}\right)\:{then} \\ $$$${g}\left({x}\right)\in\left(\mathrm{0},\mathrm{1}\right]\:{i}−{e}\:{the}\:{range}\:{of}\:{g}\left({x}\right) \\ $$$${We}\:{can}\:{write} \\ $$$${g}:\mathbb{N}\rightarrow\left(\mathrm{0},\mathrm{1}\right] \\ $$$${Mr}.\:{prakash}\:{jain},\mathrm{123456},\:{Mr}.\:{Yozzi} \\ $$$${and}\:{some}\:{others}\:{can}\:{give}\:{more} \\ $$$${satisfying}\:{answer}. \\ $$

Commented by 123456 last updated on 18/Nov/15

g(z)=(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→0,z→∞^∼ (complex infinity) then it can be give by g:C/{0}→C/{0} simple pole at z=0 if we take h(z)= { ((g(z)),(z≠0)),(0,(z=0)) :} (removing problem at z=0) h:C→C in general you can take g:A→B where A⊂C/{0} is a subset of the complex excluding 0 (maximal domain possible) and B⊂C is a subset that depends on A

$${g}\left({z}\right)=\frac{\mathrm{1}}{{z}} \\ $$$$\mathrm{all}\:\mathrm{complex}\:\mathrm{number}\:\mathrm{except}\:\mathrm{0}\:\mathrm{can}\:\mathrm{be}\:\mathrm{pruged} \\ $$$$\mathrm{then}\:\mathrm{the}\:\mathrm{domain}\:\mathrm{is}\:\mathbb{C}/\left\{\mathrm{0}\right\} \\ $$$$\mathrm{the}\:\mathrm{range}\:\mathrm{is}\:\mathrm{all}\:\mathrm{complex}\:\mathrm{except}\:\mathrm{0},\:\mathrm{since} \\ $$$$\mathrm{1}/{z}\rightarrow\mathrm{0},{z}\rightarrow\overset{\sim} {\infty}\:\left(\mathrm{complex}\:\mathrm{infinity}\right) \\ $$$$\mathrm{then}\:\mathrm{it}\:\mathrm{can}\:\mathrm{be}\:\mathrm{give}\:\mathrm{by} \\ $$$${g}:\mathbb{C}/\left\{\mathrm{0}\right\}\rightarrow\mathbb{C}/\left\{\mathrm{0}\right\}\:\mathrm{simple}\:\mathrm{pole}\:\mathrm{at}\:{z}=\mathrm{0} \\ $$$$\mathrm{if}\:\mathrm{we}\:\mathrm{take} \\ $$$${h}\left({z}\right)=\begin{cases}{{g}\left({z}\right)}&{{z}\neq\mathrm{0}}\\{\mathrm{0}}&{{z}=\mathrm{0}}\end{cases}\:\left(\mathrm{removing}\:\mathrm{problem}\:\mathrm{at}\:{z}=\mathrm{0}\right) \\ $$$${h}:\mathbb{C}\rightarrow\mathbb{C} \\ $$$$\mathrm{in}\:\mathrm{general}\:\mathrm{you}\:\mathrm{can}\:\mathrm{take} \\ $$$${g}:\mathrm{A}\rightarrow\mathrm{B}\:\mathrm{where}\:\mathrm{A}\subset\mathbb{C}/\left\{\mathrm{0}\right\}\:\mathrm{is}\:\mathrm{a}\:\mathrm{subset}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{complex}\:\mathrm{excluding}\:\mathrm{0}\:\left(\mathrm{maximal}\:\mathrm{domain}\right. \\ $$$$\left.\mathrm{possible}\right)\:\mathrm{and}\:\mathrm{B}\subset\mathbb{C}\:\mathrm{is}\:\mathrm{a}\:\mathrm{subset}\:\mathrm{that} \\ $$$$\mathrm{depends}\:\mathrm{on}\:\mathrm{A} \\ $$

Commented by Yozzi last updated on 18/Nov/15

An example could be a function f which maps members of the set of real numbers to members of the set of integers. In this case we could define this statement by notation as follows: f: R→Z where f(x)=⌊x⌋ , x∈R. ⌊x⌋ is the greatest integer less than or equal to x. (E.g ⌊π⌋=3 ∵ π=3.14...∣ ⌊e⌋=2 ∵ e=2.718...∣ ⌊0.1⌋=0 ∣ ⌊−236.653⌋=−237∣ ⌊−10⌋=−10) f is a many−to−one function since for all n≤x<n+1 ,where n∈Z, ⌊x⌋=n . This then shows how the output of f are integers given that the inputs are real. The domain of f is R and its codomain is Z . f is bijective since (1) every member of the domain is mapped to only one member of the codmain. So ∄x∈R such that ⌊x⌋ has more than one result (vertical line test proves this) and (2) f is surjective since for every value of numbers in the codomain, ∃x∈R mapped to f(x)∈Z. So, f generates the entire set of integers, which is infinitely countable, since the set of real numbers has a non−finite cardinality. The range of f is Z. Let P={prime numbers} and {Q^� }^+ ={positive irrational numbers}. If f is a function defined by f(x)=(√x) , x∈P then f maps the prime numbers to some members of {Q^� }^+ . So f: P→{Q^� }^+ .

$${An}\:{example}\:{could}\:{be}\:{a}\:{function}\:{f} \\ $$$${which}\:{maps}\:{members}\:{of}\:{the}\:{set}\:{of}\:{real} \\ $$$${numbers}\:{to}\:{members}\:{of}\:{the}\:{set}\:{of} \\ $$$${integers}.\:{In}\:{this}\:{case}\:{we}\:{could}\:{define} \\ $$$${this}\:{statement}\:{by}\:{notation}\:{as}\:{follows}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{f}:\:\mathbb{R}\rightarrow\mathbb{Z} \\ $$$${where}\:{f}\left({x}\right)=\lfloor{x}\rfloor\:,\:{x}\in\mathbb{R}.\:\lfloor{x}\rfloor\:{is}\:{the} \\ $$$${greatest}\:{integer}\:{less}\:{than}\:{or}\:{equal} \\ $$$${to}\:{x}.\:\left({E}.{g}\:\lfloor\pi\rfloor=\mathrm{3}\:\:\because\:\pi=\mathrm{3}.\mathrm{14}...\mid\right. \\ $$$$\lfloor{e}\rfloor=\mathrm{2}\:\:\because\:{e}=\mathrm{2}.\mathrm{718}...\mid\:\lfloor\mathrm{0}.\mathrm{1}\rfloor=\mathrm{0}\:\mid \\ $$$$\left.\lfloor−\mathrm{236}.\mathrm{653}\rfloor=−\mathrm{237}\mid\:\lfloor−\mathrm{10}\rfloor=−\mathrm{10}\right)\: \\ $$$${f}\:{is}\:{a}\:{many}−{to}−{one}\:{function}\:{since} \\ $$$${for}\:{all}\:{n}\leqslant{x}<{n}+\mathrm{1}\:,{where}\:{n}\in\mathbb{Z}, \\ $$$$\lfloor{x}\rfloor={n}\:.\:{This}\:{then}\:{shows}\:{how}\:{the}\:{output} \\ $$$${of}\:{f}\:{are}\:{integers}\:{given}\:{that}\:{the}\:{inputs} \\ $$$${are}\:{real}.\:{The}\:{domain}\:{of}\:{f}\:{is}\:\mathbb{R}\:{and}\:{its} \\ $$$${codomain}\:{is}\:\mathbb{Z}\:.\:{f}\:{is}\:{bijective}\:{since}\: \\ $$$$\left(\mathrm{1}\right)\:{every}\:{member}\:{of}\:{the}\:{domain}\:{is} \\ $$$${mapped}\:{to}\:{only}\:{one}\:{member}\:{of}\:{the}\: \\ $$$${codmain}.\:{So}\:\nexists{x}\in\mathbb{R}\:{such}\:{that}\:\lfloor{x}\rfloor\:{has} \\ $$$${more}\:{than}\:{one}\:{result}\:\left({vertical}\:{line}\:{test}\right. \\ $$$$\left.{proves}\:{this}\right) \\ $$$${and} \\ $$$$\left(\mathrm{2}\right)\:{f}\:{is}\:{surjective}\:{since}\:{for}\:{every}\: \\ $$$${value}\:{of}\:{numbers}\:{in}\:{the}\:{codomain},\:\exists{x}\in\mathbb{R}\:{mapped}\:{to}\:{f}\left({x}\right)\in\mathbb{Z}. \\ $$$${So},\:{f}\:{generates}\:{the}\:{entire}\:{set}\:{of}\:{integers}, \\ $$$${which}\:{is}\:{infinitely}\:{countable},\:{since} \\ $$$${the}\:{set}\:{of}\:{real}\:{numbers}\:{has}\:{a}\:{non}−{finite} \\ $$$${cardinality}.\:{The}\:{range}\:{of}\:{f}\:{is}\:\mathbb{Z}. \\ $$$$ \\ $$$${Let}\:\mathbb{P}=\left\{{prime}\:{numbers}\right\}\:{and} \\ $$$$\left\{\bar {\mathbb{Q}}\right\}^{+} =\left\{{positive}\:{irrational}\:{numbers}\right\}. \\ $$$${If}\:{f}\:{is}\:{a}\:{function}\:{defined}\:{by} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{f}\left({x}\right)=\sqrt{{x}}\:\:,\:{x}\in\mathbb{P} \\ $$$${then}\:{f}\:{maps}\:{the}\:{prime}\:{numbers}\:{to} \\ $$$${some}\:{members}\:{of}\:\left\{\bar {\mathbb{Q}}\right\}^{+} . \\ $$$${So}\:\:\:\:{f}:\:\mathbb{P}\rightarrow\left\{\bar {\mathbb{Q}}\right\}^{+} . \\ $$$$ \\ $$

Commented by Filup last updated on 18/Nov/15

Thank you all so much! I understand now!

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{all}\:\mathrm{so}\:\mathrm{much}!\:\mathrm{I}\:\mathrm{understand}\:\mathrm{now}! \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com