You are Here: Home >< Maths

# surjective function watch

1. (Original post by B_9710)
It's most certainly false, like i said take given by . Then f is surjective but 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
2. (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?
3. (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
4. (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.
5. (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?
6. (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 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.
7. (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 ._______.
8. (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.
9. (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?
10. (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
11. (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!
12. (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?
13. (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.
14. (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?
15. (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.
16. (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.

"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.

"if is surjective, the so is "
right?
17. (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.

"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.

"if is surjective, the so is "
right?
The reasoning for g not being surjective is not right - all you have shown is that g is not injective again.
18. (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 . Therefore g(x) is not a surjective function.

is that ok?
19. (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?
Yeah that does it.
20. (Original post by B_9710)
Yeah that does it.
Great thanks so much

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: November 4, 2017
Today on TSR

### Congratulations to Harry and Meghan!

But did you bother to watch?

Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams