I'm trying to come up with example functions that are $N \rightarrow N$ for each category:

  1. One-to-one but not onto.
  2. Onto but not one-to-one.
  3. Nether one-to-one nor onto.
  4. Both one-to-one and onto.

Here's what I've got:

  1. $f(n) = 2n$, because it's a linear function.
  2. $f(n) = n-1$, This only works when n is $\geq$ 2, but I'm thinking that's good enough?
  3. $f(n) = 0$, maybe this is a stretch but because y is always outside of $N$ I'm sure this is neither one-to-one or onto.
  4. $f(n) = n$, here x=y which matches both definitions.

So, I'm posting because I'm unsure of what I've got. Especially with 1 and 2. Here are the definitions I've got:

Onto: For every $y \in Y$ at least one $x \in X$ such that $f(x) = y$

One-to-one: For any $y \in Y$ there is at MOST one $x$ such that $f(x) = y$

This means that in order for a function to be onto there can be an x with two y's but it's not required. If it doesn't, doesn't that automatically make it also one-to-one?

  • $\begingroup$ Do you include $0$ in the natural numbers? $\endgroup$ – graydad Oct 1 '14 at 19:57
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    $\begingroup$ For onto but not one to one, an idea close to yours will work. Let $f(1)=f(2)=1$, and let $f(n)=n-1$ for $n\ge 3$. (I am assuming your $\mathbb{N}$ does not include $0$.) $\endgroup$ – André Nicolas Oct 1 '14 at 20:01
  • $\begingroup$ @graydad No, I don't consider 0 to be natural. Good question. :) $\endgroup$ – Ethan Oct 1 '14 at 20:14
  • $\begingroup$ @AndréNicolas So once I specify that f(1) = 1 that allows me to get a repeat answer/y (as you said f(1) = f(2)) to make it unto. That's helpful, thanks. :D $\endgroup$ – Ethan Oct 1 '14 at 20:14
  • $\begingroup$ If you want a fancier example, let $f(2n)=n$ and let $f(2n-1)=n$. So the values of the function are $1,1,2,2,3,3,\dots$, seriously not one to one. $\endgroup$ – André Nicolas Oct 1 '14 at 20:18

It seems that you are having some misconceptions here.

Foreword: I will assume that $0 \in \mathbb{N}$, although nothing really changes if do differently.

  1. This is one-to-one and not onto, but this has nothing to do with it being linear. Linear functions can be one-to-one or not and onto or no. Moreover it is delicate to speak about linear functions when you are working with $\mathbb{N}$ usually linear functions require an underlying field, such as $\mathbb{R}$. Your function is one-to-one simply because different natural number have different doubles and is not onto because $1$ is never reached as the double of another natural number.

  2. This is technically not a function, because $0$ does not go anywhere ($-1$ is not a natural number). This can easily fixed, for example, sending $0$ to $0$ (or, by the way, to any natural number). Then your function is defined as $$ f(n) = \begin{cases} 0 & \text{if $n=0$} , \\ n-1 & \text{otherwise} . \end{cases} $$ This function is onto (each natural number is reached), but not one-to-one (there are two numbers that are sent to $0$: both $0$ itself and $1$).

  3. If you, as I do, consider $0 \in \mathbb{N}$, then this is a perfectly valid function and, as you see, is neither one-to-one nor onto. If you do not take $0 \in \mathbb{N}$ it is enough to define $f(n) = 1 \quad \forall n \in \mathbb{N}$ and you are done anyway.

  4. The identity function is, of course, both onto and one-to-one. This is a perfectly valid example.

Now, as you can see a function can independently be one-to-one or not and onto or not. It is not true that, in general, onto implies one-to-one and neither it is, on the contrary, that one-to-one implies onto.

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  • $\begingroup$ Very well explained. What is and isn't "reached" is helpful vocabulary and you certainly put this all into perspective. Thank you. $\endgroup$ – Ethan Oct 1 '14 at 20:18
  • $\begingroup$ By the way, when you are learning elementary maths it may be helpful to draw simple diagrams to visualize things. Like those in en.wikipedia.org/wiki/Injective_function. I found them extremely helpful at the beginning, also for more complex situations. $\endgroup$ – Giovanni Mascellani Oct 1 '14 at 20:27

One and four look good to me. For two you haven't quite defined a function because you didn't define $f(1)$. You could make that a piecewise function with $$f(1)=1 \quad \text{and} \quad f(n) = (n-1) \space \text{if} \space n>1$$ This will make $f$ onto but not one-to-one as $f(1)=f(2).$

Three also doesn't make sense to me, as have $g: \mathbb{N} \to \Bbb{N}$ where $g(n)=0$. The way you defined $g$ is by making it map outside of $\Bbb{N}$? That doesn't work, but you could fix that by just having $g(n)=1$. Then it is true that $g$ is not one-to-one and not onto.

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  • $\begingroup$ f(1) = f(2) ... that blew my mind a little, and it explains a lot about how this works. xD And I see what you mean by g(n)=1. At first I didn't want to put that because that means every n maps to 1, but I see now that because every other natural number is not mapped at all, it is therefore not unto. :) Thanks! $\endgroup$ – Ethan Oct 1 '14 at 20:11

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