Surjective Functions (and a Proof!) | Surjections, Onto Functions, Surjective Proofs

[@lilymatchaaa]

ITS SO HARD BEING DUMB AAAAAAAAAAA IMMA NEED TO ATCH THIS 10X BEFORE GETTING IT

[@user-wr4yl7tx3w]

best explanation on surjective. clear and concise.

[@WrathofMath]

Thank you!

[@zomzombos]

I can only say thank you for this! A concept that is very useful in mathematical olympiad that I struggled to understand it. Just wondering what is the use of surjectivity of a function? (If a function is proven to be surjective, how that information can be helpful?)

[@muhammadrifayis8746]

I'm deeply grateful for your wonderful explanation sir.

[@WrathofMath]

Thanks and welcome!

[@samrogers1198]

Thank you! by far the best video on subjectivity I have seen.

[@WrathofMath]

You're welcome! I am so glad it helped and thanks for watching!

[@Skindust.007..]

from what I got, it is a Surjective function. Thank you so much for the help in the video.🙏🏾

[@WrathofMath]

Glad to help, thanks for watching!

[@shachisharma8908]

I loved your teaching way...!❤🌟

[@WrathofMath]

Glad to hear it, thanks for watching!

[@shachisharma8908]

@@WrathofMath Loads of blessings from India sir..😊

[@SODIUM10]

If anyone doesn't get it, All you have to do is prove that for every possible value of y, there exists a value of x. For example: y = 2x Do this interms of x, x = y/2 Here, whatever you put in y, there is always a value for x. Another example: y=1/x Solve this interms od x, x=1/y Here, this is not a surjective func because for y = 0, the value of x doesn't exists.

[@paulhammond6978]

The f(x) = 1/x example is good, because it's not defined at 0, but is defined everywhere else, so you can make this function surjective simply by excluding one point from the domain and the co-domain. (f would be surjective on R/{0} --> R/{0})

[@shubhivijan6110]

Thats an amazing explanation !!!!! Great work 👍

[@pritybose534]

Thanks.....I was searching so much for the proof finally understood....thanks again..beautifully explained

[@WrathofMath]

So glad it helped, thanks for watching!

[@mrstorresmath9473]

This is the first video I see from your channel. You explain very clearly. Question: How can one prove a function is surjective without the function definition? Example, a function phi from the complex numbers to the complex numbers. I know phi (i) = -i. I want to prove this is surjective. Can't wrap my head around it. Thank you for any help.

[@jingyiwang5113]

Thank you so much for such amazing explanation! 😀

[@WrathofMath]

Glad to help - thanks for watching!

[@kanyeisschizo4509]

2:11 can there be elements in set B that don't get mapped to at all for the function to still be surjective?

[@paulhammond6978]

No. This is precisely what a surjective function is not allowed to do. There must be no elements in the codomain that are not reached by something in the function. (if f: A -> B, im f = B)

[@jothiramesh4212]

you sound like kirito, love your math vids!

[@MwauraAlexWokabiMwaura]

Where have you proven that last task on your video

[@tiamiyuyusuff234]

At the long run i got 1/2[(2y-5)^¹/³ +5] so how can i evaluate to get y as yours? since for any function to be surjective , for all y in Y there exist x in X such that f(x)= y . So sir is this function surjective or not ? I am confused because i don't know how to evaluate my answer to get y as yours .

[@vincentzacks9584]

Much better than my overpaid professor, who can't explain shit 😂

[@WrathofMath]

Glad I could help, thanks for watching!

[@rosemarypatronellamatura4689]

This is good 👍 and understandable 😌.Thanks so much

[@WrathofMath]

Glad to hear it, you're welcome and thanks for watching!

[@tendaiisiah6453]

Thank u very much for the lesion

[@WrathofMath]

My pleasure, thanks for watching and let me know if you ever have any questions!

[@TheXadass]

Thanks a lot for this vid! I appreciate the little extra effort in providing an exercise at the end, really helped me a lot (I think :D)

[@WrathofMath]

You're very welcome and thank you for watching! I'm glad you got something out of the exercise at the end! I've had to scrap lessons half way through editing them when I realize I didn't include an exercise at the end and I meant to. Sometimes it could be stitched on without having to re-record, but usually it would be just a tad too messy, and I hate to have any weird edits distract people from the lessons. Let me know if you ever have any video requests!

[@qingdeng1824]

Where is the answer for the last question

[@kibme5189]

I need help with the exercise, so x= (2y-5)^(1/3). And then f((2y-5)^(1/3))= 2*((2y-5)^(1/3))-5 but how do I make it equal to y like the way you did? I mean how do I know if this is surjective?

[@WrathofMath]

I'm not sure with the details you've provided. Firstly, to know if a function is surjective we need to know the intended codomain of the function, because a function is surjective if the range equals the codomain. Given that you've not mentioned a codomain, we might assume it is the real numbers. Then there is the function, is it supposed to be x= (2y-5)^(1/3), so that x is a function of y? Or is the question about y as an implicit function of x? And what is the f you wrote in your comment? It looks like you have f as f(x) = 2x - 5, is that right?

[@XxOneilxXx]

how does one find how many surjections there are between two finite sets?

[@GEMINDIGO]

Clearly explained all the important elements of a proof.Thanks for that.

[@WrathofMath]

My pleasure, thanks for watching!

[@paulhammond6978]

Don't want to bother with a full proof, but since your f is defined by a cubic expression, it's covering the whole of the reals, and therefore f is surjective. (and there should be a range of values where you would get three possible values of x for each f(x))

[@yenzzziiiee]

Can anyone please tell me why one-to-many relations cannot be considered as functions? how is it different from the surjective function? Also, what is the difference between the Bijective function and the Inverse function? Google is not giving me clear answers and I'm desperate :>

[@tanyapandey8436]

🔥💯Thankyou

[@WrathofMath]

Of course, thanks for watching! You may be interested in this recent video finding and proving a bijection: kzbin.info/www/bejne/gZWmnaOkjsd6Zqs

[@Sakuna2726]

f(x)=1/2(x^3+5)=y X^3=2y-5 x=(2y-5)^1/3€|R Function is surjective Is it kr not ?

[@noubissikely4190]

The answer to the assignment is : not surjective because the root of a number doesn't include negative numbers which are part of real numbers. Please am I right?

[@WrathofMath]

Did I ask about a root function in this video? I just skimmed through it but didn’t see that. But correct, if we have a function f(x) = sqrt(x), which maps the nonnegative reals to the reals, it is not surjective because not every real (all the negatives) are mapped to. If we redefined f to, instead of the reals, have a codomain of the nonnegative reals, then it would be surjective.

[@gustafwiklund4740]

the function is surjektive because we have to utilize the cubic-root not the square-root, this allows the function to have negative values.

[@muhammadsaif4849]

Please write a surjective function f: z×z >R

[@julsbert6098]

Question, can floor functions or ceiling functions be surjective? If yes, how would you prove it? Thanks!

[@WrathofMath]

Thanks for watching and for the question! The short answer is yes, it all depends on the domain and codomain you choose to define for the function. For some choices, it will be surjective, for some choices it will not be. I'll record a lesson on it for you that will explore this in a bit more detail and show some proofs!

[@WrathofMath]

Here it is! Thanks again for the question! Surjective Ceiling and Floor Functions | Functions and Relations, Surjective Functions: kzbin.info/www/bejne/fn-kimmppJ1qg80

[@MrMsquare1]

The answer to the assignment is no

[@er.v.s.dwivedi5721]

Nice video #CBSEPathshala

[@WrathofMath]

Thanks for watching!

[@mdakashsk788]

Not surjective function.