Define a function f: Z -> Z by f(a) = r, the remainder after dividing a by 5. For example f(2)=2 and f(13)=3
a.) Determine the domain of f.
b.) Determine the range of f.
c.) Is f one-to-one?
d.) Is f onto?
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What are your thoughts about this problem? – littleO Oct 19 '16 at 21:56
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a) The domain is $\mathbb Z$, as seen in the definition of the function.
b) The term "range" is ambiguous. If you are talking about the codomain, then it is $\mathbb Z$ as per the definition of the function. If you are talking about the image, then it is is $\{0,1,2,3,4\}$.
c) No, it is many to one. There are multiple (infinite, actually) values of the domain (preimages) for which each element of the image corresponds to. Thus it is not injective (one to one).
d) No, as the image of the function, which is $\{0,1,2,3,4\}$, is not the same as the codomain of the function, which is $\mathbb Z$. Thus this function is not surjective (onto).
Skeleton Bow
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b) No, it is not $\Bbb Z_5$, the former 5 element subset of $\Bbb Z$ will be the range. d) No, it is not onto, as not all $\Bbb Z$ is the range but only a 5 element subset of it. – Berci Oct 19 '16 at 22:06
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I see, I thought $\mathbb Z_5={0,1,2,3,4}$ which is incorrect. Btw nice picture! – Skeleton Bow Oct 19 '16 at 22:21