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?(Original post by B_9710)
It's most certainly false, like i said take given by . Then f is surjective but is not.
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

will'o'wisp2
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 03112017 21:30

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 03112017 21:34
(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
What is your undertsanding of what a surjective function is? 
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 03112017 21:36
(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?
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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 03112017 21:40
(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 
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 03112017 21:50
(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.
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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 03112017 22:10
(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? 
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 03112017 22:16
(Original post by B_9710)
It's surjective because for every there is a such that 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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 03112017 22:17
(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 codomain" does it mean for R>R then that all the real numbers can come out of the function? 
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 03112017 22:17
(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 ._______. 
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 03112017 22:48
(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?
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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 03112017 22:50
(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 
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 03112017 22:54
(Original post by B_9710)
Exactly!
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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 03112017 22:59
(Original post by will'o'wisp2)
good lord i understand why noone 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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 03112017 23:04
(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.
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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 03112017 23:14
(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? 
will'o'wisp2
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 04112017 12:58
(Original post by B_9710)
Yeah that's fine.
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 is injective, the so is "
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 is surjective, the so is "
right? 
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 04112017 13:35
(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 is injective, the so is "
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 is surjective, the so is "
right? 
will'o'wisp2
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 04112017 13:52
(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.
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 . Therefore g(x) is not a surjective function.
is that ok? 
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 04112017 14:09
(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 . Therefore g(x) is not a surjective function.
is that ok? 
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 04112017 14:16
(Original post by B_9710)
Yeah that does it.
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