# Real Analysis

1. Oct 13, 2015

### Matt B.

How would I go about showing that the set {5n | n = 1,2,...} is equivalent to ℤ.

I previously attempted to define a function f as f(n) = 5n/2 , when n is even AND (-1)[(5n-1)/(2)] , when n is odd. I then went on to show that the chosen function is 1-1 and onto, but my professor said this was inaccurate.

Any help would be appreciated! Thank you.

2. Oct 13, 2015

### Staff: Mentor

What are the numbers in your set? It should be almost trivial to set up a 1-1 mapping between the two sets, much simpler than the one you show below.

BTW, homework questions should be posted in the Homework & Coursework sections, not here in the technical math sections.

3. Oct 13, 2015

### jbriggs444

What is (-1)[(5n-1)/(2)] when n=1 ?

4. Oct 13, 2015

### Matt B.

-2??

5. Oct 13, 2015

### jbriggs444

OK. And what did you want it to be?

6. Oct 13, 2015

### Matt B.

Do I not need my defined function to be equivalent to all values in ℤ or just any values of ℤ?

7. Oct 13, 2015

### jbriggs444

Mark44 may have been leading you down a better path in post #2.

Your function is supposed to map values from one set to values in another set. It is supposed to be one to one and onto. We have established that your function takes 1 to -2.

Is 1 a member of the set that you are trying to map to ℤ?

8. Oct 13, 2015

### Matt B.

I'm confused. So am I simply evaluating 5n at different values for n to deduce my original set that I am trying to map to ℤ? Then that would just be {5,10,15,20,25...} so then my function would just be f(n) = 5n ?

9. Oct 13, 2015

### WWGD

Think of a number in your set as a multiple of a number in $\mathbb Z$ and show for very number in $\mathbb Z$ there is one such multiple.

WWGD4444.

I think the advice of :
Mark44
JBriggs444
WWGD4444

Should do it ;).

10. Oct 13, 2015

### Staff: Mentor

Yes, that's the simple one I was hinting at. Clearly this is a one-to-one function, so each number in the set {n} gets mapped to a number in the set {5n}, and vice versa.

Edit: I wasn't thinking very clearly on this reply...

Last edited: Oct 13, 2015
11. Oct 13, 2015

### jbriggs444

That gets a bijection between {5, 10, 15, ... } and {1, 2, 3, ... }. But I understood the original post to be asking for a bijection with ℤ = { ..., -3, -2, -1, 0, 1, 2, 3, ... }

12. Oct 13, 2015

### WWGD

Ultimately, you can find a bijection between {$5, 10, 15,...$} and $\mathbb N$ and then compose with a not-so-hard bijection between $\mathbb N, \mathbb Z$.

13. Oct 13, 2015

### Staff: Mentor

For some reason, I was thinking (without thinking very hard) that it was the positive integers. I was really thinking $\mathbb{N}$ rather than $\mathbb{Z}$.

14. Oct 13, 2015

### VKnopp

What do you mean equivalent? I beg to differ. $\mathbb{Z}$ is much more dense then your set. I think what you mean is the cardinal is the same. You can prove this via 1 to 1 correspondence.

15. Oct 13, 2015

### Staff: Mentor

"Set equivalence" is appropriate here.
See https://proofwiki.org/wiki/Definition:Set_Equivalence
The set in the OP should have the same cardinality as the set that is to be found.

16. Oct 13, 2015

### Matt B.

So is it not possible to find a 1-1, onto function that relates the set {5, 10, 15,...} onto the set of all integers?

17. Oct 13, 2015

### Staff: Mentor

We didn't say that. See if you can find a way to map some of the numbers in {5, 10, 15, ...} to {0, 1, 2, 3, ...} and the rest of the numbers in {5, 10, 15, ...} to {..., -3, -2, -1}. Or you can follow WWGD's advice in post #12.

18. Oct 13, 2015

### Matt B.

Here was my original attempt at the solution. Since I'm not following the discussion, can somebody point out what I am misinterpreting?

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19. Oct 13, 2015

### Matt B.

Sorry, it's me again. Since I am trying to map {5, 10, 15, 20, ...} onto {...,-3,-2,-1,0,1,2,3} could I use the function f such that f(n) = n/5 when n is even and (-1)(n/5) when n is odd??

20. Oct 14, 2015

### Staff: Mentor

Your set, let's call it S, is {5, 10, 15, 20, ...}
Your function maps the members of S as follows:
5 --> -1
10 --> 2
15 --> -3
20 --> 4
and so on.
No member of S gets paired with 0, 1, -2, 3, -4, and so on, so your function is not onto $\mathbb{Z}$, hence not one-to-one, so this doesn't work.

Your function just scales the input value by multiplying it. To pick up all of the missing numbers, the function also has to translate or shift, the input values. Can you think of a way for 5 --> -1, 15 --> -2, 25 --> -3, and so on, and for 10 --> 0, 20 --> 1, 30 --> 2, and so on?

21. Oct 14, 2015

### aikismos

@Matt B.

Here are some clues using a different problem:

Let's say $S := \{ 7n : n \in ℕ\}$

In the elements $(e)$ in $(7, 14, 21, 28, 35, 42, ...)$ you were quick to see that by taking the even values and dividing you could reduce the sequence to the naturals...

7 14 21 28 35 42 ... (divide by 7)
1 2 3 4 5 6 ... (take only even values)
2 4 6 ... (take only those which are even)
1 2 3 ... (you've arrived at the naturals!)

So, in this case $14|e \rightarrow \frac{e}{14} = ℕ$. Of course, part of this is shifting from the $ℕ$ to $S$, which is why the divisibility of $e$ by 14 is required (NB $7(2n) \rightarrow 14n$)

Like Mark44 said, you need to find -W and the remainder of $S$. From your original post, it's obvious you are reaching out and finding two important handles on the problem. Yes, the the negation is necessary to reverse the sequence, and it has to do with the odds. But note that the remainder of $S$ is not $-ℕ$ but $-W$. Think about the difference in the codomains of $\{2n + 1\}$ and $\{2n - 1\}$. What do you notice? You had the right idea, but put it together wrong...

7 14 21 28 35 42 ... (divide by 7)
1 2 3 4 5 6 ... (take only odd values)
1 3 5 ... (now what?)
???
0 -1 -2 -3 ... (this is what you want, right?)

Once you're done, you have to rewrite in terms of $e$ and $S$ by combining the set definition with your notation for naturals!
All you have to do is put it together with what you already know. If you really understand it, then you shouldn't have any problems transferring this thinking to your homework problem.