# From a structure, define 0, 1 and the successor relation

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• May 8th 2012, 11:14 AM
HallsofIvy
Re: From a structure, define 0, 1 and the successor relation
Quote:

Originally Posted by Zhai
Thanks for the replies so far
But "There is no number less than 0" means that there are numbers that are more than 0...

No, it doesn't. Is this a language difficulty? For example, I could say about the set {0} that "there is no number in the set less than 0". That does NOT mean that there are numbers in the set that are more than 0.

Quote:

and that wont exactly define 0...
However, in this case there are, of course, numbers larger than 0. But you seem to think that "there is no number less than 0" means only that "there are numbers that are more than 0" and it does not.

Quote:

And since the structure only contains natural numbers, then I dont think we needed
to worry about numbers less than 0.
But I was thinking about something like this though in First Order: "There is one and only x less than every y"
Then x has to be 0 because its less than every number. Could this be it? Or can someone think of something better?
The problem with that is that you would first have to define "less than every y". You can do that, of course, but are you still in First Order logic?
• May 8th 2012, 11:35 AM
Zhai
Re: From a structure, define 0, 1 and the successor relation
Quote:

Originally Posted by HallsofIvy
No, it doesn't. Is this a language difficulty? For example, I could say about the set {0} that "there is no number in the set less than 0". That does NOT mean that there are numbers in the set that are more than 0.

However, in this case there are, of course, numbers larger than 0. But you seem to think that "there is no number less than 0" means only that "there are numbers that are more than 0" and it does not.

The problem with that is that you would first have to define "less than every y". You can do that, of course, but are you still in First Order logic?

Not language difficulty, but perhaps me not thinking about it as sets.
But its so obvious when you spell it out that way. I should have thought of that.

I dont think I want to go further than First Order if that is what you are thinking.
I was thinking "less than every natural numbers y".
• May 8th 2012, 11:41 AM
Zhai
Re: From a structure, define 0, 1 and the successor relation
As for the last problem: Defining the relation "y is the sucessor of x"
Obviously x < y.
Then I want to say something like this " (x < y) and (y = x +1) "
This way of thinking would work I think...
But I cant use + and 1 since it's not in the language...
So how to formalize it so that y immediately follows x (x immediately comes before y)
without the use of + and 1?

EDIT:
I think I might have a possible solution:
If I say "For all z ( (x < y) and (if x < z then z ≤ y) )" would that work?
This is just an idea I came up with though.
• May 8th 2012, 12:33 PM
emakarov
Re: From a structure, define 0, 1 and the successor relation
Quote:

Originally Posted by Zhai
So how to formalize it so that y immediately follows x (x immediately comes before y)
without the use of + and 1?