Question

let f:NU{0}-> NU{0} be defined by f(n) = n+1,if is even.
=n-1,if n is odd.
show f is a bijection.

Answer 1

THIS ONE IS FROM RELATION.

Answer 2

bijective function or one-to-one correspondence) is a function giving an exact pairing of the elements of two sets. Every element of one set is paired with exactly one element of the other set, and every element of the other set is paired with exactly one element of the first set. (There are no unpaired elements.)

PLEASE HELP.

Answer 3

look at \(f(f(n))\)

Answer 4

THEN @Zarkon

Answer 5

what do you get when you eveluate \(f(f(n))\)?

Answer 6

WE DON'T GET ANYTHING..IT IS SAID AS -->
F IS A FUNCTION FROM N TO N.

Answer 7

yes we do...eveluate it

Answer 8

I DON'T KNOW..MUCH PLEASE ELABORATE.

Answer 9

I'll do one part

suppoe \(n\ge0\) and is even

then \(f(f(n))=f(n+1)\)

now \(n+1\) is odd thus

\(f(n+1)=n+1-1=n\)

thus \(f(f(n))=n\)

Answer 10

OKAY.NOW WHAT I NEED TO DO?

Answer 11

what happens when n is odd?

Answer 12

DON'T KNOW.

Answer 13

I THINK IT MUST BE LESS THEN

Answer 14

\[f(f(n))=n\]

for all \(n\in\mathbb{N}\cup\{0\}\)


you just need to show it

Answer 15

once you have that then ...it shows that \(f\) is its own inverse...thus it has an inverse...and hence it is bijective

Answer 16

HOW WE CAN SHOW IT?

Answer 17

To show that a function is a bijection, you hav to show that
it is surjective and injective.

Let us prove that if is surjective. Let m be an integer.
If m is odd, then m+1 is even anf f[m+1]=m+1-1=m
If m is even them m-1 is odd and f[m-1]=m-1+1=m
So f is srujective

Answer 18

OKAY. NOW WE NEED TO PROVE ITS INJECTIVITY NAH.. ?

Answer 19

To prove that f is one to one or injective, you have to show that
if f(m)=f(n) then m=n

Answer 20

OKAY. BUT HOW TO DO SO ?

Answer 21

Suppose that f[m] = f[n]
If m is odd, then f[m] = m+1 is even and hence f[n] is even
so n is odd, f[n]=n+1
m+1 = n+1 then m=n
Mimic this for m even

Answer 22

MIMIC HO?

Answer 23

Do a similar method if m=even to show that m=n if f[m]=f[n]

Answer 24

If m is even, then f[m] = m-1 is odd and hence f[n] is odd so n is evne, f[n]=n-1

m-1 = n-1 then m=n

Answer 25

Did you get it now?

Answer 26

IT SHOWS ITS BIJECTIVITY?

Answer 27

Yes

Answer 28

THANKYOU :)

Answer 29

We showed that is surjective and one to one and that is it.

Answer 30

OKAY.