mathsfun Posted November 5, 2004 Share Posted November 5, 2004 can somebody pls help me figure this problem out?? f : Z SUB 7 --> Z SUB 7, f([n] SUB 7) = [2n] SUB 7 is this one below correct?? f : R --> R, f(x) = 2x + 1 f'(x) = 2 > 0 and f(x) = 2x + 1 passes the horizontal line test so f is 1 to 1 but how do I figure out if it is onto?? is this one correct below?? f : Z -->Z, f(n) = 2n + 1 f'(n)= 2 > 0 and f(n) = 2n + 1 passes the horizontal line test so f is 1 to 1 but how do I figure out if it is onto?? thank you Link to comment Share on other sites More sharing options...
MandrakeRoot Posted November 5, 2004 Share Posted November 5, 2004 f is clearly injective since the image of any two distinct points is distinct (a simple property of the sum that you can without a doubt assume) f is surjective since every element has an original in the case where you consider the function f from R into R ! (try finding which one that would be) When taking the function from Z into Z this is not true since 2 does not have an original ! (In other words 2 is not in the range of your function !) I've tried to explain it rather simple, i hope that worked out ? Mandrake Link to comment Share on other sites More sharing options...
matt grime Posted November 5, 2004 Share Posted November 5, 2004 I am not sure that I would say this approach is remotely sound. It is philosophically dubious, and mathematically unnecessary, and since at no point do you use the fact you are reducing mod 7 you should worry. the map n to 2n mod 10 isn't bijective for instance. The simple and better way is: suppose f(n)=f(m) then 2n=2m mod 7 ie 7 divides 2(n-m), as 2 and 7 are coprime, it must be that 7 divides n-m, ie n=m mod 7. Thus proving f is injective. f is surjective is now either trivial, by some larger result, or needs proving. Here is the larger result: if f is a function between two finite sets and f is injective, then f is surjective. proof: if f is not surjective, then there is some element not mapped to, hence, omitting this element, we can define a map that is injective from a set to one strictly smaller. by the pigeon hole principal this cannot be injective # so f was surjective. Or, we could say: Since 2 and 7 are coprime, there are integers a nd b such that 2a+7b=1. Let [n] be in Z_7, then 2an+7bn=1, ie 2an=n mod 7. n ws arbitrary hence the map is surjective. Link to comment Share on other sites More sharing options...
mathsfun Posted November 6, 2004 Author Share Posted November 6, 2004 thank u, appreciate the input also I am not worried bc all this math stuff for fun I ask questions bc I do really want to understand & it can difficult at times when u r learning on your own Link to comment Share on other sites More sharing options...
MandrakeRoot Posted November 8, 2004 Share Posted November 8, 2004 Asking questions is always good in order to understand things. Mandrake Link to comment Share on other sites More sharing options...
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