Consider the following functions that map N into N:
1N(n) = n
f(n) = 3n
g(n) = n + (-1)^n
h(n) = min{n, 100}
k(n) = max{0, n-5}
(a) Which of these functions are one-to-one?
(b) Which of these functions map N onto N?

- anonymous

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- anonymous

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- anonymous

Hmm, what's the first function supposed to be? I haven't come across that notation before. Do you have a definition for it?

- anonymous

No, that is all that is given. =\

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- anonymous

I'm afraid I wouldn't be able to help with that one then...
For \(f(n)=3n\) (and the others, for that matter), use the definition of one-to-one:
\[f(x_1)=f(x_2)~~\Rightarrow~~x_1=x_2\\
\text{(contrapositive of what you may have been shown: }x_1\not=x_2~~\Rightarrow~~f(x_1)\not=f(x_2).)\]
So assume \(f(n_1)=f(n_2)\) for some \(n_1,n_2\in\mathbb{N}\):
\[\begin{align*}f(n_1)&=f(n_2)\\
3n_1&=3n_2\\
n_1&=n_2\end{align*}\]
So the first function is one-to-one.
To check if it is onto requires some simple reasoning. For any given \(n\), \(f(n)\) will always result with a multiple of 3. So this one isn't onto, since the multiples of 3 are only a subset of \(\mathbb{N}\).

- anonymous

By the way, were you told that \(\mathbb{N}=\left\{0,1,2,...\right\}\) or \(\mathbb{N}=\left\{1,2,...\right\}\) ? I only ask because if it's the latter, then the last function \(k(n)\) doesn't map from \(\mathbb{N}\) to \(\mathbb{N}\).

- anonymous

N = the set of positive integers: 1,2,3 ...

- anonymous

without the 0, right.

- anonymous

Okay.
By the way, now that I think about it, it could be that \(1_\mathbb{N}(n)\) is just an arbitrary way of saying something like \(f(n)\). If that's the case, then you can apply the same exact reasoning for \(3n\).
Assume \(f(n_1)=f(n_2)\) for some \(n_1,n_2\in\mathbb{N}\):
\[\begin{align*}f(n_1)&=f(n_2)\\
n_1&=n_2\end{align*}\]
So \(1_\mathbb{N}(n)\) is one-to-one.
This function is onto, though, since for any \(n\in\mathbb{N}\) you plug in you get the same \(n\). The range of the function will be \(\mathbb{N}\), same as the target set.

- anonymous

so using that reasoning, I don't understand why all of the functions would not be one-to-one

- anonymous

Oh, disregard that thing I mentioned about \(k(n)\). I thought it said min, not max.
And not all are one-to-one.
Anyway, going down the list:
\[g(n)=n+(-1)^n=\begin{cases}n-1&\text{for odd }n\\
n+1&\text{for even }n\end{cases}\]

- anonymous

If \(n_1\) is odd and \(n_2\) is even, then obviously \(g(n_1)\not=g(n_2)\), so you have to assume both are odd or both are even.
\[\begin{align*}g(n_{1})&=g(n_2)\\
n_1\pm1&=n_2\pm1\\
n_1&=n_2\end{align*}\]
So \(g(n) \) is indeed one-to-one.
It's not onto, though. What happens when \(n=1\)?

- anonymous

You get 0?

- anonymous

Right. Not in the natural numbers, so the function isn't even well defined.

- anonymous

The next one's pretty easy.
\[h(n)=\min\left\{n,100\right\}=\begin{cases}n&\text{for }n<100\\100&\text{for }n\ge100\end{cases}\]
What do you think?

- anonymous

h(n1) = h(n2)

- anonymous

That's about as far as I get =\
I don't really understand the concept of one-to-one

- anonymous

Consider the values of \(n\) one case at a time.
Let say \(n_1<100\) and \(n_2<100\). Then you have \(h(n_1)=n_1\) and \(h(n_2)=n_2\), right? So for this case, \(h(n)\) is one-to-one, provided that \(h(n_1)\not=h(n_2)\), as per the definition.
For \(n_1,n_2\ge100\), you have \(h(n_1)=h(n_2)=100\). So it's not one-to-one, since there are two values of \(n\) that give the same output. ("two-to-one" as opposed to "one-to-one")
It's not onto either, since for \(n\ge100\), the output doesn't get higher than 100, thus leaving out all the other natural numbers. Make sense?

- anonymous

Kind of makes sense.

- anonymous

What specifically are you having trouble with?

- anonymous

so for an earlier problem you say g(n1)n1±1n1=g(n2)=n2±1=n2
but in this problem h(n1)=n1 and h(n2)=n2 and n1<100 and n2<100 so wouldn't that mean n1 = n2? so I don't get why h(n1)≠h(n2)

- anonymous

in the first problem n1 = n2

- anonymous

For \(g(n)\), I assume \(g(n_1)=g(n_2)\) and showed that \(n_1=n_2\), meaning that for two equal outputs, you have that the inputs are the same.
If \(n_1,n_2\) are odd, then \(g(n_1)=n_1-1\), likwise for \(n_2\).
Similarly, if they're even, then \(g(n_1)=n_1+1\), and likewise for \(n_2\).
So if \(g(n_1)=g(n_2)\), then \(n_1-1=n_2-1\) or \(n_1+1=n_2+1\), or more succintly, \(n_1\pm1=n_2\pm1\), which gives you \(n_1=n_2\). So \(g\) is one-to-one.

- anonymous

Note that \(g\) was dealt with case-wise. I did the same for \(h\).
\(\min\{n,100\}\) is the smallest number between \(n\) and 100, which is either \(n\) or 100.
Take the first case: If \(n<100\), then \(\min\{n,100\}\) must be \(n\), right?
So we then assume \(h(n_1)=h(n_2)\) \(\color{red}{\text{for some }n_1,n_2<100}\). Then \(h(n_1)=n_1\) and \(h(n_2)=n_2\). This means that \(n_1=n_2\), so the function is one-to -one *for this case only*.

- anonymous

so for the same inputs, the outputs must be the same on both sides of the equation for the function to be one to one?
I don't get what the difference is between n1 and n2 in h(n)

- anonymous

Yes. Given that two outputs are the same, you have to show that the two inputs used to get that output are the same as well.
In other words \(f(a)=f(b)\) is only true if \(a=b\). \(f(a),f(b)\) are the outputs, \(a,b\) are the inputs.

- anonymous

In \(h(n)\), whether there is a difference between \(n_1\) and \(n_2\) depends entirely on what \(h(n_1)\) and \(h(n_2)\) are.
Like I said earlier, if both \(n_1\) and \(n_2\) are less than 100, then their associated outputs are \(h(n_1)=n_1\) and \(h(n_2)=n_2\).
If \(h(n_1)=h(n_2)\), then you necessarily have \(n_1=n_2\), and so \(h\) is one-to-one ONLY if \(n_1\) and \(n_2\) are less than 100.

- anonymous

For the second case, a simple counter-example suffices to show that \(h\) isn't one-to-one.
Take \(n_1=100\) and \(n_2=101\). Then \(h(100)=h(101)=100\). You have two inputs that give the same output. \(h\) is thus not one-to-one.

- anonymous

so if different inputs give the same output the function is not one-to-one. But if the same inputs give the same output the function is one-to-one?

- anonymous

If *only* the same inputs give the same output, the function is one-to-one.

- anonymous

ah ok that makes sense.

- anonymous

So for k(n) = max{0, n-5} would we need to do same thing for each of the cases?

- anonymous

The problem with this function is that it's not well-defined. It doesn't map from \(\mathbb{N}\) to \(\mathbb{N}\). For \(n\le5\), you have \(\max\{0,n-5\}\le0\).
\[k(n)=\max\{0,n-5\}=\begin{cases}0&\text{for }n\le5&&\text{(this is the poorly-defined part)}\\
n-5&\text{for }n>5\end{cases}\]
But yes, you would need to check case-by-case.

- anonymous

so we say n1 <= 5 and n2 <= 5. Then we have k(n1) = n1 and k(n2) = n2?

- anonymous

which means that it is one-to-one for the first case?

- anonymous

No, not quite. \(k(n_1)=k(n_2)=0\), by the definition of \(k\).

- anonymous

oh right

- anonymous

Now supposing we have \(n_1\not=n_2\), we still have that \(k(n_1)=k(n_2)\). Same output for different inputs. So not one-to-one.

- anonymous

so would it map N onto N?

- anonymous

Why do you think so? I want to see your reasoning.

- anonymous

I don't think it would be.

- anonymous

I don't really know how to show it isn't

- anonymous

I don't know, I don't understand how to know if a function maps N onto N either :(

- anonymous

I would say it is. \(n-5\) gives you all natural numbers, provided that \(n>5\), of course.
It's the same reason for why \(1_\mathbb{N}(n)\) was onto.

- anonymous

hmm, ok, so was h(n) onto then?

- anonymous

Alright, well I think I'm starting to understand one-to-one a little better now. I will look over it some more and see if I can understand everything that is happening.
Sorry to take so much time, and thank you so much for all the help and for putting up with my stupidity! :)

- anonymous

Yeah, it's onto.
Not a problem, happy to help!

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