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    (Original post by B_9710)
    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.
    but i thought x^2 was surjective? because you can pick 2 values such as 1 and -1 which give the same answer so it can't be injective?

    so i've abandoned what i thought was the definition of surjective and turned to the horizontal line test which tells me if the function is injective or not, combined with the vertical line test then i know that certain fucntions and whatnot but i dont' understand then how x^3 is surjective too
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    (Original post by will'o'wisp2)
    but i thought x^2 was surjective? because you can pick 2 values such as 1 and -1 which give the same answer so it can't be injective?

    so i've abandoned what i thought was the definition of surjective and turned to the horizontal line test which tells me if the function is injective or not, combined with the vertical line test then i know that certain fucntions and whatnot but i dont' understand then how x^3 is surjective too
    Yeah the function f(x)=x^2 is not injective nor surjective.
    What is your undertsanding of what a surjective function is?
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    (Original post by B_9710)
    Yeah the function f(x)=x^2 is not injective nor surjective.
    What is your undertsanding of what a surjective function is?
    wat but i thought it was within the realm of R->R

    For any x^2 value that is chosen, there is a corresponding y value in the domain which gives you the x^2 value


    but dats not right, cus then i start classing something like 2x+1 as surjective cus you can pick out any y that gives you the 2x+1 value i choose
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    (Original post by will'o'wisp2)
    wat but i thought it was within the realm of R->R

    For any x^2 value that is chosen, there is a corresponding y value in the domain which gives you the x^2 value


    but dats not right, cus then i start classing something like 2x+1 as surjective cus you can pick out any y that gives you the 2x+1 value i choose
    All because there is more than one x value that gives you the y value doesnt mean the function is not surjective - it just means it is not injective.
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    (Original post by B_9710)
    All because there is more than one x value that gives you the y value doesnt mean the function is not surjective - it just means it is not injective.
    ye so turns out then than any x^3 -nx where n just a natural number then that makes one of those lumpy graph with a nice clear hump and underhump

    so for me that's surjective only according to my weak ass definition

    so how is x^3 surjective then? surely it's only injective because there's 1 x value for each y value right?
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    (Original post by will'o'wisp2)
    ye so turns out then than any x^3 -nx where n just a natural number then that makes one of those lumpy graph with a nice clear hump and underhump

    so for me that's surjective only according to my weak ass definition

    so how is x^3 surjective then? surely it's only injective because there's 1 x value for each y value right?
    It's surjective because for every  q \in \mathbb{R} there is a  p\in \mathbb{R} such that  f(p)=q namely p is the cube root of q. That's what surjective means that every value in the co domain is mapped to by some element in the domain.
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    (Original post by B_9710)
    It's surjective because for every  q \in \mathbb{R} there is a  p\in \mathbb{R} such that  f(p)=q namely p is the cube root of q. That's what surjective means that every value in the co domain is mapped to by some element in the domain.
    -________________________________ _____-

    but then x^2 is surjective tho cus if i say 5 then you can pick root 5 which squares to make x ._______.
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    (Original post by will'o'wisp2)
    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?
    Yes if you are taking R to be the codomain then all real numbers must come out of the function and thus all you need to do is find a real number that doesn't come out of (f(x))^2.
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    (Original post by will'o'wisp2)
    -________________________________ _____-

    but then x^2 is surjective tho cus if i say 5 then you can pick root 5 which squares to make x ._______.
    But what if I say -1, what real number are you gonna square to get -1?
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    (Original post by Dalek1099)
    Yes if you are taking R to be the codomain then all real numbers must come out of the function and thus all you need to do is find a real number that doesn't come out of (f(x))^2.
    (Original post by B_9710)
    But what if I say -1, what real number are you gonna square to get -1?
    Oh right i understand sudenly now

    so all i need to do is pick a value for which the function of which i suppose is surjective can't produce got it

    finally afetr and hour of my idiocy
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    (Original post by will'o'wisp2)
    Oh right i understand sudenly now

    so all i need to do is pick a value for which the function of which i suppose is surjective can't produce got it

    finally afetr and hour of my idiocy
    Exactly!
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    (Original post by B_9710)
    Exactly!
    good lord i understand why no-one wants to help me .---------------.

    so that means i did the first one wrong then

    so for the question i posed even if i swap the word surjective with injective i can still use x and x^2 as examples?
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    (Original post by will'o'wisp2)
    good lord i understand why no-one wants to help me .---------------.

    so that means i did the first one wrong then

    so for the question i posed even if i swap the word surjective with injective i can still use x and x^2 as examples?
    The function f(x)=x^2 is not injective so the claim that if f is injective then f^2 is injective is false - so if that's what you mean then yes.
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    (Original post by B_9710)
    The function f(x)=x^2 is not injective so the claim that if f is injective then f^2 is injective is false - so if that's what you mean then yes.
    this is the last post before i pick this up again tomorrow

    i've put

    i claim that g(x) is not injectiive. To substantiate let f(x)=x, then g(x)=x². Choose x_1=1 and x_2=-1. Then g(x_1)=1 and g(x_2)=1 but 1=/=-1 therefore g(x) is not an injective function.

    Does that still work?
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    (Original post by will'o'wisp2)
    this is the last post before i pick this up again tomorrow

    i've put

    i claim that g(x) is not injectiive. To substantiate let f(x)=x, then g(x)=x². Choose x_1=1 and x_2=-1. Then g(x_1)=1 and g(x_2)=1 but 1=/=-1 therefore g(x) is not an injective function.

    Does that still work?
    Yeah that's fine.
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    (Original post by B_9710)
    Yeah that's fine.
    so then this

    i claim that g(x) is not injectiive. To substantiate let f(x)=x, then g(x)=x². Choose x_1=1 and x_2=-1. Then g(x_1)=1 and g(x_2)=1 but 1=/=-1 therefore g(x) is not an injective function.
    will answer the question

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


    can be copied and modified slightly and this

    i claim that g(x) is not surjectiive. To substantiate let f(x)=x, then g(x)=x². Choose x_1=1 and x_2=-1. Then g(x_1)=1 and g(x_2)=1 but 1=/=-1 therefore g(x) is not an surjective function.

    will answer the question

    "if f:\mathbb R \Rightarrow \mathbb R is surjective, the so is g(x) :=[f(x)]^2"
    right?
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    (Original post by will'o'wisp2)
    so then this

    i claim that g(x) is not injectiive. To substantiate let f(x)=x, then g(x)=x². Choose x_1=1 and x_2=-1. Then g(x_1)=1 and g(x_2)=1 but 1=/=-1 therefore g(x) is not an injective function.
    will answer the question

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


    can be copied and modified slightly and this

    i claim that g(x) is not surjectiive. To substantiate let f(x)=x, then g(x)=x². Choose x_1=1 and x_2=-1. Then g(x_1)=1 and g(x_2)=1 but 1=/=-1 therefore g(x) is not an surjective function.

    will answer the question

    "if f:\mathbb R \Rightarrow \mathbb R is surjective, the so is g(x) :=[f(x)]^2"
    right?
    The reasoning for g not being surjective is not right - all you have shown is that g is not injective again.
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    (Original post by B_9710)
    The reasoning for g not being surjective is not right - all you have shown is that g is not injective again.
    ok, so then let's say

    i claim that g(x) is not surjectiive. To substantiate let f(x)=x, then g(x)=x². Let g(x)=-1, then -1=x² but \nexists x \in \mathbb R : x^2 =-1. Therefore g(x) is not a surjective function.

    is that ok?
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    (Original post by will'o'wisp2)
    ok, so then let's say

    i claim that g(x) is not surjectiive. To substantiate let f(x)=x, then g(x)=x². Let g(x)=-1, then -1=x² but \nexists x \in \mathbb R : x^2 =-1. Therefore g(x) is not a surjective function.

    is that ok?
    Yeah that does it.
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    (Original post by B_9710)
    Yeah that does it.
    Great thanks so much
 
 
 
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