S sita Oct 2017 39 0 Rumba Nov 28, 2017 #1 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 a moderator: Nov 28, 2017

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?

romsek Math Team Sep 2015 2,777 1,552 USA Nov 28, 2017 #2 domain is $\mathbb{Z}$ codomain is $\mathbb{R}$ range is $\{0\}\cup \{n^2: ~ n\in \mathbb{N}\}$

S sita Oct 2017 39 0 Rumba Nov 28, 2017 #3 romsek said: domain is $\mathbb{Z}$ codomain is $\mathbb{R}$ range is $\{0\}\cup \{n^2: ~ n\in \mathbb{N}\}$ Click to expand... So is my image the same as your range?

romsek said: domain is $\mathbb{Z}$ codomain is $\mathbb{R}$ range is $\{0\}\cup \{n^2: ~ n\in \mathbb{N}\}$ Click to expand... So is my image the same as your range?

romsek Math Team Sep 2015 2,777 1,552 USA Nov 28, 2017 #4 the range is the image of the entire domain

S sita Oct 2017 39 0 Rumba Nov 28, 2017 #5 romsek said: the range is the image of the entire domain Click to expand... 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.

romsek said: the range is the image of the entire domain Click to expand... 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.

romsek Math Team Sep 2015 2,777 1,552 USA Nov 28, 2017 #6 sita said: Delete Click to expand... 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.

sita said: Delete Click to expand... 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.