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 November 28th, 2017, 03:10 PM #1 Newbie   Joined: Oct 2017 From: Rumba Posts: 20 Thanks: 0 Question about my image dealing with domain Z -> R where f(x) = x^2 Domain is Z Codomain is R Image i'm not entirely sure about, I got {0,1,4,9...} Am i correct to say this? Last edited by greg1313; November 28th, 2017 at 04:35 PM.
 November 28th, 2017, 03:17 PM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,089 Thanks: 1085 domain is $\mathbb{Z}$ codomain is $\mathbb{R}$ range is $\{0\}\cup \{n^2: ~ n\in \mathbb{N}\}$
November 28th, 2017, 03:27 PM   #3
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Quote:
 Originally Posted by romsek domain is $\mathbb{Z}$ codomain is $\mathbb{R}$ range is $\{0\}\cup \{n^2: ~ n\in \mathbb{N}\}$
So is my image the same as your range?

 November 28th, 2017, 03:32 PM #4 Senior Member     Joined: Sep 2015 From: USA Posts: 2,089 Thanks: 1085 the range is the image of the entire domain
November 28th, 2017, 03:41 PM   #5
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Quote:
 Originally Posted by romsek the range is the image of the entire domain
If z is domain, that includes ...-3,-2,-1,0,1,2,3....

How can the image be the entire domain?

Dont understand what you mean.

November 28th, 2017, 04:30 PM   #6
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Quote:
 Originally Posted by sita Delete
Image OF the entire domain.

put the whole domain through the function, everything that comes out is the range. This may or may not be the co-domain. In this case it's just a subset of the co-domain.

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