Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    14
    ReputationRep:
    isn't y=0.5x a surjective function?
    Offline

    15
    ReputationRep:
    Depends what the co domain is. If the function is say from  \mathbb{R}\rightarrow \mathbb{C} then it is not surjective. A function is surjective if the range or image is equal to the co domain.
    • Thread Starter
    Offline

    14
    ReputationRep:
    (Original post by B_9710)
    Depends what the co domain is. If the function is say from  \mathbb{R}\rightarrow \mathbb{C} then it is not surjective. A function is surjective if the range or image is equal to the co domain.
    well it goes from R-->R
    Offline

    15
    ReputationRep:
    (Original post by will'o'wisp2)
    well it goes from R-->R
    Well in that case yes it is surjective, you should be able to prove it easily from the definition of surjective.
    • Thread Starter
    Offline

    14
    ReputationRep:
    (Original post by B_9710)
    Well in that case yes it is surjective.
    so far i'm looking for a function which is surjective but not surjective when squared

    constants don't work, they're neither injective or surjective

    fraction functions don't work like 1/x

    minus function are pretty much the same as positive ones they just flip...

     \sqrt x???? square it, it's bijective but then i(root x)t fails the vertical line test .__.


    i'm thinking still don't tell me an answer tho ^_^
    Offline

    15
    ReputationRep:
    (Original post by will'o'wisp2)
    so far i'm looking for a function which is surjective but not surjective when squared

    constants don't work, they're neither injective or surjective

    fraction functions don't work like 1/x

    minus function are pretty much the same as positive ones they just flip...

     \sqrt x???? square it, it's bijective but then i(root x)t fails the vertical line test .__.


    i'm thinking still don't tell me an answer tho ^_^
    Remember that functions don't have to be defined for all values of  \mathbb{R} , functions can be defined on any subset of R indeed a finite subset as well.
    • Thread Starter
    Offline

    14
    ReputationRep:
    (Original post by B_9710)
    Remember that functions don't have to be defined for all values of  \mathbb{R} , functions can be defined on any subset of R indeed a finite subset as well.
    but they do tho? like i can't choose 1/x right? cos it doesn't work in the realm of R-->R
    Offline

    15
    ReputationRep:
    (Original post by will'o'wisp2)

     \sqrt x???? square it, it's bijective but then i(root x)t fails the vertical line test .__.

    i'm thinking still don't tell me an answer tho ^_^
    Also  \sqrt x doesn't fail the vertical line test.
    Offline

    15
    ReputationRep:
    (Original post by will'o'wisp2)
    but they do tho? like i can't choose 1/x right? cos it doesn't work in the realm of R-->R
    You could have a function f such that f(x)=1/x for all x in R except 0, and then not have f defined at 0. Or you could define f(0) to be whatever number you want -it's still a function.
    Consider a function  f:\{-1,1\}\rightarrow \{-1,1\}, x\mapsto x , this is a perfectly well defined function and is surjective (bijective in fact).
    • Thread Starter
    Offline

    14
    ReputationRep:
    (Original post by B_9710)
    Also  \sqrt x doesn't fail the vertical line test.
    whoops sorry my bad :/

    but it's still injective onl,,,,,, hol up wait no..... bum, it's still bijective tho cus hold up....... you stick -1 into that and it doesn't work so then it isn't surjective because i can pick a value of x that will go into root x but you can't get a y out so then it's injective?
    • Thread Starter
    Offline

    14
    ReputationRep:
    (Original post by B_9710)
    You could have a function f such that f(x)=1/x for all x in R except 0, and then not have f defined at 0. Or you could define f(0) to be whatever number you want -it's still a function.
    Consider a function  f:\{-1,1\}\rightarrow \{-1,1\}, x\mapsto x , this is a perfectly well defined function and is surjective (bijective in fact).
    agreed but it doesn't fit in with the statement at hand which is

    "if f:\mathbb R \Rightarrow \mathbb R is surjective, the so is g(x) :=[f(x)]^2"

    what i need to find if this is true or false, i think it's false so i have to find 1 example only of which this doesn't work ..
    Offline

    21
    ReputationRep:
    (Original post by will'o'wisp2)
    agreed but it doesn't fit in with the statement at hand which is

    "if f:\mathbb R \Rightarrow \mathbb R is surjective, the so is g(x) :=[f(x)]^2"

    what i need to find if this is true or false, i think it's false so i have to find 1 example only of which this doesn't work ..
    What happens if f(x) is negative?
    • Thread Starter
    Offline

    14
    ReputationRep:
    (Original post by Dalek1099)
    What happens if f(x) is negative?
    still in the set of reals but it's pretty much just flipped

    then again root x seems lke neither now tho cus it feels like if i pick any x value in the set of reals then for root x there's isn't always a y to correspond simply 1 to 1 or just a value(any will do).

    I'm thinking modulus now
    Offline

    15
    ReputationRep:
    (Original post by will'o'wisp2)
    agreed but it doesn't fit in with the statement at hand which is

    "if f:\mathbb R \Rightarrow \mathbb R is surjective, the so is g(x) :=[f(x)]^2"

    what i need to find if this is true or false, i think it's false so i have to find 1 example only of which this doesn't work ..
    I didn't know the function had to be from R to R.
    Why not take the most simple example I can think of, f(x)=x, in that case g(x)=x^2 which is not surjective on R.
    Also  x\mapsto \sqrt x is not a valid function from R to R.
    • Thread Starter
    Offline

    14
    ReputationRep:
    (Original post by B_9710)
    I didn't know the function had to be from R to R.
    Why not take the most simple example I can think of, f(x)=x, in that case g(x)=x^2 which is not surjective on R.
    Also  x\mapsto \sqrt x is not a valid function from R to R.
    that's the one i used for the previous question which had injective instead of surjective

    i'm sure that means i gotta find a function of which surjective is a subset of it and a function which is injective only i'm thinking modulus now but ye root x don't work, it's not a function in R->R

    x is bijective from R->R just pick any value of x and you pick the same value for y
    now x² isn't injective bcus you have -1 and 1 which give out 1 the same value but 2 different inputs so that's just surjective

    that's not wat i need, i was thinking x^3 and then x^6 but that's the same as x and x^2
    Offline

    21
    ReputationRep:
    (Original post by will'o'wisp2)
    still in the set of reals but it's pretty much just flipped

    then again root x seems lke neither now tho cus it feels like if i pick any x value in the set of reals then for root x there's isn't always a y to correspond simply 1 to 1 or just a value(any will do).

    I'm thinking modulus now
    I'm not sure you are getting what surjective means it means that the image covers entirely the set that you choose for the codomain. Therefore, in your problem you need to choose a codomain so that this equals the image of f(x) so that it is surjective then you need to find real numbers like a lets say such that f(x)=a but there exists no x such that that f(x)^2=a and then since the image of f(x) equalled the codomain and so a is in the codomain then f(x)^2 doesn't equal the codomain.
    Offline

    15
    ReputationRep:
    (Original post by will'o'wisp2)
    that's the one i used for the previous question which had injective instead of surjective

    i'm sure that means i gotta find a function of which surjective is a subset of it and a function which is injective only i'm thinking modulus now but ye root x don't work, it's not a function in R->R

    x is bijective from R->R just pick any value of x and you pick the same value for y
    now x² isn't injective bcus you have -1 and 1 which give out 1 the same value but 2 different inputs so that's just surjective

    that's not wat i need, i was thinking x^3 and then x^6 but that's the same as x and x^2
    What is your question exactly - I'm not sure what you are trying to do.
    • Thread Starter
    Offline

    14
    ReputationRep:
    (Original post by Dalek1099)
    I'm not sure you are getting what surjective means it means that the image covers entirely the set that you choose for the codomain. Therefore, in your problem you need to choose a codomain so that this equals the image of f(x) so that it is injective then you need to find real numbers like a lets say such that f(x)=a but there exists no x such that that f(x)^2=a and then since the image of f(x) equalled the codomain and so a is in the codomain then f(x)^2 doesn't equal the codomain.
    i don't think i do either :/ what i understood it to be as that for any x value i can pick out, there' s a corresponding y value which gives me that x value i just picked

    i just don't then undertstand what it means for "its image to be equal to its co-domain" does it mean for R->R then that all the real numbers can come out of the function?
    • Thread Starter
    Offline

    14
    ReputationRep:
    (Original post by B_9710)
    What is your question exactly - I'm not sure what you are trying to do.
    The question is, is

    "if f:\mathbb R \Rightarrow \mathbb R is surjective, the so is g(x) :=[f(x)]^2"

    true or false?
    Offline

    15
    ReputationRep:
    (Original post by will'o'wisp2)
    The question is, is

    "if f:\mathbb R \Rightarrow \mathbb R is surjective, the so is g(x) :=[f(x)]^2"

    true or false?
    It's most certainly false, like i said take  f:\mathbb{R}\rightarrow \mathbb{R} given by  f(x)=x . Then f is surjective but  f^2 is not.
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Did TEF Bronze Award affect your UCAS choices?
    Useful resources

    Make your revision easier

    Maths

    Maths Forum posting guidelines

    Not sure where to post? Read the updated guidelines here

    Equations

    How to use LaTex

    Writing equations the easy way

    Student revising

    Study habits of A* students

    Top tips from students who have already aced their exams

    Study Planner

    Create your own Study Planner

    Never miss a deadline again

    Polling station sign

    Thinking about a maths degree?

    Chat with other maths applicants

    Can you help? Study help unanswered threads

    Groups associated with this forum:

    View associated groups
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Quick reply
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.