Question

Need some help with finding a couple bijective, injective functions

Answer 1

I attached the question

Answer 2

@Hero

Answer 3

injective means no two domain elements go to the same range element right?

Answer 4

so there are probably meany ways to do it, think fractions

Answer 5

I'm thinking we have to map something like 1, 1/2, 1/4, 1/8, ... to something

Answer 6

like map it onto 1/2, 1/4, 1/8, 1/16 ...

Answer 7

but i'm not sure how to do that

Answer 8

i would just go with \(n\to \frac{1}{n}\)

Answer 9

course that doesn't really work, because you have a problem at 0

Answer 10

so just send \(0\to 0\)

Answer 11

How so?

Answer 12

you can't just say \(\mathbb{Z}\to [-1.1]\) via \(z\mapsto\frac{1}{z}\) because that would be undefined for \(0\)

Answer 13

so in addition so \(z\mapsto\frac{1}{z}\) add \(0\mapsto 0\)

Answer 14

it is clearly injective right? no two elements of \(\mathbb{Z}\) get mapped to the same number in \([-1,1]\)

Answer 15

Sorry, i'm just a little lost @satellite73

Answer 16

ok lets go slow

Answer 17

injective means no two domain elements go to the same range element
yes?

Answer 18

yes

Answer 19

your domain is \(\mathbb{Z}=...-3,-2,-1,0,1,2,3...\)

Answer 20

if you fiip each of them you get some number in \([-1,1]\) like \[-\frac{1}{3},-\frac{1}{2},-1,1,\frac{1}{2},\frac{1}{3},...\]

Answer 21

clear?

Answer 22

Oh okay

Answer 23

the only problem with this is, you can't flip \(0\) because \(\frac{1}{0}\) is undefined

Answer 24

alright

Answer 25

so you have to make a special case for 0, say \(0\mapsto 0\)

Answer 26

nothing else gets sent to 0, so this is still an injection

Answer 27

Okay that makes sense

Answer 28

good !

Answer 29

Is there a way to find a function without using special cases?

Answer 30

if there is ii will leave that up to you, i can't see one

Answer 31

0 has to go somewheres

Answer 32

Okay

Answer 33

you shouldn't be scared of that, it is your function, you can define it anyway you like

Answer 34

you could even say \(0\mapsto \frac{2}{3}\)nothing else goes there

Answer 35

Oh i see, so we need to make a special case for 0, but it still stays an injective function

Answer 36

oui

Answer 37

:) so how about a bijective funciton for f : [0, 1] → [0, 1)?

Answer 38

what do i look like, a genius?

Answer 39

little bit :p

Answer 40

clearly you want to say \(x\mapsto x\) right?

Answer 41

except you have a problem at 1

Answer 42

yup

Answer 43

What is the problem at 1?

Answer 44

if \(x\mapsto x\) then \([01]\to [0,1]\) not to \([0,1)\)

Answer 45

Oh

Answer 46

so how to take care of that?
if you don't like special cases for the first one, you are really not going to like this one

Answer 47

why don't we consider mapping the sequence 1, 1/2, 1/4, 1/8 ... onto 1/2, 1/4, 1/8, 1/16...?

Answer 48

i think that should work

Answer 49

or just \(\frac{1}{n}\) to \(\frac{1}{n+1}\) if \(x=\frac{1}{n}\) for some integer \(n\)

Answer 50

and of course for the rest, \(x\mapsto x\)

Answer 51

How am I supposed to write out my answers for these types of questions? I feel like just writing something like x -> x wouldn't really make sense without writing that what part it's for

Answer 52

just say it in words

Answer 53

say "if \(x=\frac{1}{n}\) for integer \(n\) then \(x\mapsto \frac{1}{n}\), otherwise \(x\mapsto x\)

Answer 54

oops i meant \(x\to \frac{1}{n+1}\) but you get the idea

Answer 55

oh okay

Answer 56

and how am I supposed to write out a special case?

Answer 57

i guess you also have to add \(0\to 0\)

Answer 58

just add that at the end?

Answer 59

since \(0=\frac{0}{n}\) for all n

Answer 60

oh maybe not actually

Answer 61

since \(\frac{0}{n+1}=0\) anyways

Answer 62

you can say it in english probably best
more words if you like, but in math usually terse is best
say the map that sends \(\frac{1}{n}\to \frac{1}{n+1}\) is a map from the sequence \(\{\frac{1}{n}\}\) to \(\{\frac{1}{n+1}\}\) and if \(x\neq \frac{1}{n}\) for integer \(n\) then \(x\mapsto x\)

Answer 63

Alrighty

Answer 64

and for the first part we still have to put in the special case for 0?

Answer 65

yes if you use the one i wrote

Answer 66

you can't just say \(n\mapsto \frac{1}{n}\) without making a special case for 0

Answer 67

Makes sense :) Thanks @satellite73