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.
THIS ONE IS FROM RELATION.
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.
look at \(f(f(n))\)
THEN @Zarkon
what do you get when you eveluate \(f(f(n))\)?
WE DON'T GET ANYTHING..IT IS SAID AS --> F IS A FUNCTION FROM N TO N.
yes we do...eveluate it
I DON'T KNOW..MUCH PLEASE ELABORATE.
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 \(f(f(n))=n\)
OKAY.NOW WHAT I NEED TO DO?
what happens when n is odd?
DON'T KNOW.
I THINK IT MUST BE LESS THEN
\[f(f(n))=n\] for all \(n\in\mathbb{N}\cup\{0\}\) you just need to show it
once you have that then ...it shows that \(f\) is its own inverse...thus it has an inverse...and hence it is bijective
HOW WE CAN SHOW IT?
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
OKAY. NOW WE NEED TO PROVE ITS INJECTIVITY NAH.. ?
To prove that f is one to one or injective, you have to show that if f(m)=f(n) then m=n
OKAY. BUT HOW TO DO SO ?
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
MIMIC HO?
Do a similar method if m=even to show that m=n if f[m]=f[n]
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
Did you get it now?
IT SHOWS ITS BIJECTIVITY?
Yes
THANKYOU :)
We showed that is surjective and one to one and that is it.
OKAY.
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