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This is from Discrete Mathematics and its Applications

Here is the definition of countable that the book uses enter image description here

and how to determine if two sets have the same cardinality enter image description here

Here is the example that the book gave enter image description here Here is some work I added to the example enter image description here

What I realized from the example was that the function n/2 would map an even positive integer to a positive integer and the function -(n - 1)/2 would map an odd positive integer to a negative/zero integer. I then showed that both functions were one to one and onto. From this, I reasoned that there is a one to one correspondence from the set of positive integers to all integers because there are functions that map from an positive integer to an integer and that one positive integer will map to one unique integer and that every integer has a positive integer that will map to it.
Is this reasoning correct?

I tried to apply this idea to the next example the book provided that I didn't really understand. Here is the book example I don't really understand enter image description hereenter image description here

Is there any equation that can be used to map a positive integer to a positive rational number, just like the equation in the last example to map a positive integer to an integer so I can go through my little proof to understand this better?

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committedandroider
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    I believe that there is such an equation and that it's pretty simple, but you don't need taht in order to know that there is a one-to-one correspondence --- a function. ${}\qquad{}$ – Michael Hardy Jan 21 '15 at 05:53
  • I think the most understandable "equation" is, name a rational, figure out where in the grid it lies by looking at it's numerator and denominator, and then conclude by convincing yourself sooner or later, this diagonal algorithm will reach your number. – Alex R. Jan 21 '15 at 05:54
  • ^ I don't consider the above to have a satisfactory answer, but I think that any answer here would also be an answer there. – apnorton Jan 21 '15 at 05:59
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    Of interest: http://mathworld.wolfram.com/PairingFunction.html – apnorton Jan 21 '15 at 06:01
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    In case that you haven't been paying any attention, ALL your recent questions are being retagged with some/many of the tags being removed. For starters, please stop using the [set-theory] tag, especially in conjunction with the [elementary-set-theory] tag. – Asaf Karagila Jan 21 '15 at 06:02
  • Sorry, just the the first section in this chapter was called set theory and i thought set theory just has to do with sets – committedandroider Jan 21 '15 at 06:08
  • @MichaelHardy How would you be able to tell without the function? For me I have to see that the function has to be one to one and onto to make sense of it – committedandroider Jan 21 '15 at 06:09
  • @AlexR. I agree with that but couldn't you do this with the set of integers as well. If you start counting 0, 1, 2, 3, ... you eventually get to your number, say 554. I felt like I needed to do this because the author does proof with functions – committedandroider Jan 21 '15 at 06:10
  • I don't think this would be a duplicate though. – committedandroider Jan 21 '15 at 06:11
  • @committedandroider : I didn't say you could tell without the function; I said you could tell without an equation. The mathworld link posted in a comment above by anorton gives you the equation, but you can define the function by other means than that. – Michael Hardy Jan 21 '15 at 15:09
  • Yeah I don't think I m going to use that math world equation. Might be a little overkill for this. – committedandroider Jan 21 '15 at 19:51
  • @MichaelHardy What woudl your method be? – committedandroider Jan 21 '15 at 20:15
  • @committedandroider : My method would either be Figure 3 in your posted question or something similar. ${}\qquad{}$ – Michael Hardy Jan 22 '15 at 00:22
  • @MichaelHardy Wouldn't this method be cheating as a bit? In a sense in the graph, you're skipping over all of the duplicates. I think. I think that be evaluating the function f(x) = x^2, counting and skipping over the duplicate cases f(1)-f(-1) for example and then saying that the function is one to one. – committedandroider Jan 28 '15 at 19:43
  • The function shown in the picture is not one-to-one, but one could just omit duplicates and then it's one-to-one. – Michael Hardy Jan 28 '15 at 22:10
  • oh cause you're defining your own function – committedandroider Jan 28 '15 at 23:44

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