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The question is as follows:

Let $A = B = \{0, 1, 2, 3\}$. Which function $f: A \to B$ is one-to-one?

There are three answers to choose from:

(a) $\space f(x) = x + 1 $

(b) $\space f(x) = x \bmod 3 $

(c) $\space f(x) = 3 - x$

I know its not (b), but aren't both (a) and (c) right? Last I checked all linear functions are one-to-one and each element in the domain is mapped to only one element in the range. I answered (a) and got it wrong. Am I missing something?

graydad
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Diehardwalnut
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4 Answers4

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$3+1=4$ and 4 is not an element of B.

MegaboofMD
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  • So am i getting this right, for this question, A=B={0,1,2,3} doesn't just represent the list of elements to choose from, it also represents the range of elements. If so, does f:A→B mean A is the domain and B is the range? – Diehardwalnut Sep 18 '15 at 03:25
  • Yes it saying that the domain of f is elements of A and the range of f is elements of B. In which case answer a) maps an element of A to an element not in B, thus f is not a function from A to B. – MegaboofMD Sep 18 '15 at 03:32
  • @Diehardwalnut Close. $B$ is the codomain, which means that it must contain the range of the function. For example, $g:A\to B$ defined by $g(x)=1$ is still a valid function, even though the range of $g$ is ${1}$ and not $B={1,2,3,4}$. – Akiva Weinberger Sep 18 '15 at 03:59
  • Yes. Sorry, B is called the codomain. – MegaboofMD Sep 18 '15 at 05:35
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you are indeed missing something. What $f:A \to B$ means is that $f$ is fed values only from $A$ and will spit out values only found in $B$. If this is untrue of $f$ then $f$ is not a function.

If you choose option (a) what is $f(3)$?

graydad
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  • Yes I see now, choosing 3 is correct since 3 is an element of A, however, after 3 is fed to (a) it gives 4 which is not an element of B. – Diehardwalnut Sep 18 '15 at 03:34
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$f:A \rightarrow B$ defined by the formula $f(x) = x + 1$ is not a function. Can you see why?

Raj
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In $(a)$ $f$ is not a function as codomain is not suitably defined.

neelkanth
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