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

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  • Опубліковано 23 січ 2025

КОМЕНТАРІ • 56

  • @lilymatchaaa
    @lilymatchaaa 2 роки тому +13

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

  • @user-wr4yl7tx3w
    @user-wr4yl7tx3w Рік тому +1

    best explanation on surjective. clear and concise.

  • @muhammadrifayis8746
    @muhammadrifayis8746 Рік тому +1

    I'm deeply grateful for your wonderful explanation sir.

  • @zomzombos
    @zomzombos 4 роки тому +9

    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?)

  • @samrogers1198
    @samrogers1198 4 роки тому +6

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

    • @WrathofMath
      @WrathofMath  4 роки тому +1

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

  • @Skindust.007..
    @Skindust.007.. Рік тому

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

    • @WrathofMath
      @WrathofMath  Рік тому

      Glad to help, thanks for watching!

  • @pritybose534
    @pritybose534 3 роки тому

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

    • @WrathofMath
      @WrathofMath  3 роки тому +1

      So glad it helped, thanks for watching!

  • @kanyeisschizo4509
    @kanyeisschizo4509 9 місяців тому

    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
      @paulhammond6978 7 місяців тому

      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)

  • @mrstorresmath9473
    @mrstorresmath9473 4 роки тому +2

    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.

  • @shachisharma8908
    @shachisharma8908 2 роки тому +1

    I loved your teaching way...!❤🌟

    • @WrathofMath
      @WrathofMath  2 роки тому +1

      Glad to hear it, thanks for watching!

    • @shachisharma8908
      @shachisharma8908 2 роки тому +1

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

  • @MwauraAlexWokabiMwaura
    @MwauraAlexWokabiMwaura Рік тому

    Where have you proven that last task on your video

  • @shubhivijan6110
    @shubhivijan6110 4 роки тому +1

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

  • @jingyiwang5113
    @jingyiwang5113 Рік тому

    Thank you so much for such amazing explanation! 😀

    • @WrathofMath
      @WrathofMath  Рік тому

      Glad to help - thanks for watching!

  • @tiamiyuyusuff234
    @tiamiyuyusuff234 Рік тому

    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 .

  • @jothiramesh4212
    @jothiramesh4212 4 роки тому

    you sound like kirito, love your math vids!

  • @TheXadass
    @TheXadass 5 років тому +3

    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
      @WrathofMath  5 років тому

      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
      @qingdeng1824 4 роки тому

      Where is the answer for the last question

  • @XxOneilxXx
    @XxOneilxXx 4 роки тому

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

  • @kibme5189
    @kibme5189 2 роки тому

    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
      @WrathofMath  2 роки тому

      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?

  • @rosemarypatronellamatura4689
    @rosemarypatronellamatura4689 4 роки тому

    This is good 👍 and understandable 😌.Thanks so much

    • @WrathofMath
      @WrathofMath  4 роки тому

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

  • @tendaiisiah6453
    @tendaiisiah6453 3 роки тому

    Thank u very much for the lesion

    • @WrathofMath
      @WrathofMath  3 роки тому

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

  • @paulhammond6978
    @paulhammond6978 7 місяців тому

    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))

  • @noubissikely4190
    @noubissikely4190 2 роки тому +1

    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
      @WrathofMath  2 роки тому +1

      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
      @gustafwiklund4740 2 роки тому +2

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

  • @Sakuna2726
    @Sakuna2726 Місяць тому

    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 ?

  • @yenzzziiiee
    @yenzzziiiee 3 роки тому

    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 :>

  • @GEMINDIGO
    @GEMINDIGO 4 роки тому

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

    • @WrathofMath
      @WrathofMath  4 роки тому

      My pleasure, thanks for watching!

  • @vincentzacks9584
    @vincentzacks9584 3 місяці тому

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

    • @WrathofMath
      @WrathofMath  3 місяці тому

      Glad I could help, thanks for watching!

  • @julsbert6098
    @julsbert6098 5 років тому

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

    • @WrathofMath
      @WrathofMath  5 років тому

      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
      @WrathofMath  5 років тому

      Here it is! Thanks again for the question!
      Surjective Ceiling and Floor Functions | Functions and Relations, Surjective Functions: ua-cam.com/video/HNAY6so87Mk/v-deo.html

  • @SODIUM10
    @SODIUM10 10 місяців тому +2

    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
      @paulhammond6978 7 місяців тому

      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})

  • @muhammadsaif4849
    @muhammadsaif4849 Рік тому

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

  • @tanyapandey8436
    @tanyapandey8436 3 роки тому

    🔥💯Thankyou

    • @WrathofMath
      @WrathofMath  3 роки тому

      Of course, thanks for watching! You may be interested in this recent video finding and proving a bijection: ua-cam.com/video/KdClpnYbG0I/v-deo.html

  • @mdakashsk788
    @mdakashsk788 2 роки тому +1

    Not surjective function.

  • @MrMsquare1
    @MrMsquare1 Рік тому

    The answer to the assignment is no

  • @er.v.s.dwivedi5721
    @er.v.s.dwivedi5721 4 роки тому

    Nice video #CBSEPathshala