Bijective Functions and Why They're Important | Bijections, Bijective Proof, Functions and Relations

What are bijective functions and why should we care about them? We'll be going over bijections, examples, proofs, and non-examples in today's video math lesson!
A bijection is a function that is both injective and surjective. We mention two very important things about bijections in the lesson. First, suppose f is a function from A to B and f is injective and surjective. Thus, f is bijective. Then, we can define an inverse function from B to A that takes the images of f and sends them back to their respective pre-images. This can only be done with a bijective function, which is very important!
However, there is a lot of possibility for nuance in the definition of inverse functions (we don't give a formal definition in this lesson), and while only a bijective function can have an inverse like we described, other types of functions can have other types of inverses that follow less strict definitions. This lesson is not meant to be an in-depth presentation of inverse functions, so I hope you'll look forward to learning more about them.
Here is a definition for you if you are interested: Let F be a function from A to B. Then a function, G, that maps B to A, is an inverse function of F if and only if g(f(a)) = a for all a in A and f(g(b)) = b for all b in B.
If you need a recap on injective and surjective functions, check out my lessons on them!
Injections: • Injective Functions (a...
Surjections: • Surjective Functions (...
SOLUTION TO PRACTICE PROBLEM:
We want to show that the integers (Z) have the same cardinality as the odd numbers (we'll call the set of odd numbers S). So we want to find a bijection from one to the other; Z to X or X to Z. A good place to start, as we did with the example in the video, is with the definition of an odd number. An odd number is a number equal to 2k + 1 for some integer k. So if we define a function f from Z to X as f(z) = 2z + 1, that should work out! We just need to prove it.
First, we prove f is injective. Take z, w in Z and suppose f(z) = f(w). Then 2z + 1 = 2w + 1. Thus, 2z = 2w and z = w. So f is injective.
Proving surjectivity is even easier. Take x in X. By definition of X, x is odd and therefore there exists an integer k such that 2k + 1 = x. Then, f(k) = 2k + 1 = x. Thus, f is surjective. Therefore, since f is injective and surjective, f is bijective so |Z| = |X|.
I hope you find this video helpful, and be sure to ask any questions down in the comments!
********************************************************************
The outro music is by a favorite musician of mine named Vallow, who, upon my request, kindly gave me permission to use his music in my outros. I usually put my own music in the outros, but I love Vallow's music, and wanted to share it with those of you watching. Please check out all of his wonderful work.
Vallow Bandcamp: vallow.bandcamp.com/
Vallow Spotify: open.spotify.com/artist/0fRtu...
Vallow SoundCloud: / benwatts-3
********************************************************************
+WRATH OF MATH+
◆ Support Wrath of Math on Patreon: / wrathofmathlessons
Follow Wrath of Math on...
● Instagram: / wrathofmathedu
● Facebook: / wrathofmath
● Twitter: / wrathofmathedu
My Music Channel: / seanemusic

[@augustascesnavicius5125]

Got everything I need in the first 2 minutes of the video, thanks man

[@rabsonkayaya1020]

The video has helped me a lot thank you very much sir.

[@winstonacousticstudio445]

clear and thorough!

[@WrathofMath]

Glad to hear it! Thanks for watching!

[@thedeathofbirth0763]

warmlittlediamond If you go to his video on injection, he very eloquently shows the difference between codomain and the image=range by drawing it. He also explains when a function f:A→B is defined in some specific way A and B can be totally arbitrary, it is the definition of the function that makes them matter at all. ie let A={ sarah, steve,joe} and B={ 1,2,floor,cat, fish,dog, chair,3} f:A→B and f is defined as a person and her/his pet then chair and 3 can not be mapped to anything in A but they exist to show that sets A and B simply can be different but it is the definition of your function that gives them importance if you change your definition of the function so that f is defined as the number of pets of each person in B then 1,2,3 can be used , if you change your definition again and say f represents where she/he (in A) is sitting then only chair and floor will be mapped so you can make up many functions using the same sets .

[@aaronstone5142]

thanks this video really saved me

[@kibme5189]

This was awesome thanks, I'm definitely recommending your channel to anyone I come across with lol.

[@WrathofMath]

Thank you!

[@WarmLittleDiamond90]

This was so incredibly helpful. Thanks a lot. I have one question, if you have time? Take an injective function, for example. Not every point in the set B/codomain is mapped to by a point in the set A/domain. So why is the unmapped-to point in the set B to begin with? Why is a point that isn't mapped to by any point in the domain even considered to be part of the codomain? I think there's something fundamental that I'm missing here. I hope my question makes sense! Thanks again!

[@resilient.warfighter]

An example of an injective, but not surjective function would be f:R→R given by f(x) = e^x. All real numbers can go into the function, but the output can never be 0 or any negative real numbers.

[@eru453]

Hello, what if we put on set A, annoter element where f4 isn't define. f4 is still a bijection ?

[@geraldalejandria6886]

This is helpful

[@WrathofMath]

Glad to hear it - thanks for watching!

[@dr.rajanikamble1620]

Its very helpful to understand to bijective function, thank you sir

[@WrathofMath]

So glad to hear it, thanks for watching and let me know if you ever have any questions!

[@pamodpeiris154]

Thanks

[@WrathofMath]

My pleasure, thanks for watching!

[@shaheryarsalafi2967]

Could you share the solution of the example at the end of the video??

[@WrathofMath]

Thanks for watching! The solution is in the description.

[@leeming1317]

I got f1 wrong but the other 3 right