# Is this a valid function?

I am stuck with the question below, Say whether the given function is one to one. $A=\mathbb{Z}$, $B=\mathbb{Z}\times\mathbb{Z}$, $f(a)=(a,a+1)$ I am a bit confused about $f(a)=(a,a+1)$, there are two outputs $(a,a+1)$ for a single input $a$ which is against the definition of a function. Please help me out by expressing your review about the question. Thanks

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Is $Z$ the integers, and is $Z*Z$ the set of pairs of integers? If so, you can use `\mathbb{Z}` for $\mathbb{Z}$, and $\times$ (`\times`) instead of $*$ (the latter could be confused with the free product of groups). – Arturo Magidin Nov 4 '11 at 21:35

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No, there's a single output, which happens to be a pair of two natural numbers. ($\mathbb Z\times \mathbb Z$ is the set of such pairs). There's only one pair coming out of the function. It doesn't matter that it has some internal structure. – Henning Makholm Nov 4 '11 at 21:35
@ArturoMagidin: Yes it is. – Fahad Uddin Nov 4 '11 at 21:40

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Sigh. I meant "happens to be a pair of two integers". Damn the edit window. – Henning Makholm Nov 4 '11 at 22:48

## 2 Answers

If $f\colon A\to B$, then the inputs of $f$ are elements of $A$, and the outputs of $f$ are elements of $B$, whatever the elements of $B$ may be.

If the elements of $B$ are sets with 17 elements each, then the outputs of the function will be sets with 17 elements each. If the elements of $B$ are books, then the output will be books.

Here, the set $B$ is the set whose elements are ordered pairs. So every output of $f$ must be an ordered pair. This is a single item, the pair (just like your address may consist of many words and numbers, but it's still a single address).

By the way: whether the function is one-to-one or not is immaterial here (and it seems to me, immaterial to your confusion...)

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The output is a single pair of two integers. Your definition of $B$ specifies that all outputs of your function are pairs of integers.

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