In normal math the integer after 0 is 1, this can be proved by assuming that there exists some integer z between 0 and 1, then showing it leads to z^2 also being an integer, as well as z^3, z^4 .... but z^n < 1 so there must be an infinite number of integers between 0 and 1, or none at all.
Can you instead of taking the second option go with the first? or is there some other property that prevents that?
tl;dr - can there be an infinite number of integers between 0 and 1?
>>7909602
I understand your question but is it not just a matter of definitions?
What IS 1? We define it as the first natural number after 0. 2 is defined as the second one, and so forth and so forth. The integers are just an extension of the naturals to make them a group under addition.
The proof you mentioned in your post is nice but otherwise pretty useless since we are dealing with the definition of 1 here.
>>7909602
But... there aint no infinite number of integer between 0 and 1.
Those are real numbers.
I ondt understand
z^n < 1
What is n here? an integer, a real number or what?
Either way, it is an axiom that there is infinite amount of numbers between 0 and 1.
Check out the supremumaxiom
http://www.math.utk.edu/~freire/teaching/m300f12/supremum.pdf
It's quite an heavy axiom
>>7909614
amount of real numbers*
Just wanted to add that I am an engineer and that we don't go heavily into axioms.
>>7909611
Here I use 1 as the multiplicative identity, and 0 as the additive identity,
So the proof shows that the '1' after 0 is the same '1' which satisfies 1*x = x
>>7909614
>But... there aint no infinite number of integer between 0 and 1.
yes, but wht if there were, would it work
>What is n here?
n is any natural number
baby's first set theory homework
It's constructed this way by axioms.
>>7909602
you could answer this question with a different viewpoint
see '1' as a statement that "there is something".
this isnt even an axiom. you know there is something. look around you. on the other hand it is no 'truth'. the only 'truth' there is that I know of: I know that I know nothing. famous.
now see '2' as the statement that there is concept of "irreversible more" or even 'growth'.
If you say that there is no such thing as a sub-concept (a lateral concept), then there cannot be something between your two concepts of 1 and 2.
so the integer itself is the axiom that there is no sub-concept. but that axiom is not needed if the concept of irreversible more is stated. and if you see 1 as the sum of all sub (sub sub sub) concepts:
0.5 + 0.25 + 0,125 + ... = 1
then you have implied all possible subconcepts already in your first statement, that there is something. And that there is something - we obviously all experience.
no axioms needed. fuck axioms.
>>7909710
Haha. This is what happens when plebs try to do math.
>>7909642
>yes, but wht if there were, would it work
Warning: quality post incoming.
So you want something "like the integers" but with a number between 0 and 1? Hold onto your butts, because I have the arithmetic you want.
You are correct that if the usual rules of multiplying integers apply then we should conclude that having one number strictly between 0 and 1 implies that there should be an infinite number of such numbers. Let [math]\mathbb{Z}[/math] denote the ring of integers under the usual ordering. Then the direct product [math]\mathbb{Z}\times\mathbb{Z}[/math] consists of ordered pairs [math](a,b)[/math] where [math]a,b\in\mathbb{Z}[/math] and addition and multiplication are performed componentwise. This set of elements under these operations form a new ring which we can endow with the dictionary ordering. That is, we say [math](a,b)<(c,d)[/math] when [math]a<c[/math] or [math]a=c[/math] and [math]b<d[/math].
Observe that when [math]a=b[/math] we can identify the element [math](a,b)[/math] with the integer [math]a[/math]. This identification is the diagonal map which can be used to show that the integers [math]\mathbb{Z}[/math] are isomorphic to a proper subring of [math]\mathbb{Z}\times\mathbb{Z}[/math]. Moreover, one can use our dictionary ordering to see that, indeed, an infinite number of "numbers" lie between [math]0\approx(0,0)[/math] and [math]1\approx(1,1)[/math].
Just noticed that [math](0,0)<(1,-2)<(1,1)[/math] but [math](1,-2)^2=(1,4)>(1,1)[/math]. If this upsets you then take the direct product [math]\mathbb{Z}_{\ge0}\times\mathbb{Z}_{\ge0}[/math] instead.
>>7909773
>[math]\mathbb{Z}_{\ge0} \times\mathbb{Z}_{\ge0}[/math]
When will neo-m00t fix the fucking [math]L_{A} T^E\chi[/math]?
>>7909779
Idk, but just remember to space away the brackets or else it may not render.
>>7909773
thanks, exactly what I wanted to know
I was trying to uplift (ab+1)/b or similar to integer status and dint think of making a pair.
>>7909602
>is there some other property that prevents that?
Any nonempty set of positive integers has a least element. If the set of integers between 0 and 1 is nonempty, then it has a least element [math]x[/math]. But [math]x^2[/math] is also between 0 and 1, and less than [math]x[/math].
>>7909982
In [math]\mathbb{Z}_{\ge0} \times\mathbb{Z}_{\ge0}[/math] we have that [math]x\le x^2[/math] for all [math]x[/math], so your argument doesn't hold there. Every nonempty subset of [math]\mathbb{Z}_{\ge0} \times\mathbb{Z}_{\ge0}[/math] has a least element.
>>7910046
Of course, but that breaks other expected properties of the integers. In this case it isn't a ring (no additive inverse).
I'm just explaining how you'd typically go about proving it given some definition or axiomatization of the integers.
>>7909602
>be mathematician
>define integer as whole number
>try to disprove your own arbitrary definition
you guys really like to waste time on pointless shit huh.
>>7910160
But how do you know that [math]0<x<1[/math] implies that [math]x^2<x[/math] using only properties of the integers? I guess that is what I should have responded with in the first place.
>>7910194
Good odds you're trolling, but what OP was trying to ask was whether we could extend the ring of integers to a bigger ordered ring where there are an infinite number of elements between 0 and 1.
The answer is yes, so now we are talking about other technical issues which are related to the original question. Modern algebra is insanely powerful and useful, so understanding these things is pretty far from pointless.
>>7910205
If [math]0 < x < 1[/math] then [math]1 - x[/math] is positive, so [math]x(1 - x) = x - x^2[/math] is also positive.
>>7910221
if tex had a symbol for shrugging i would use it
numbers between 0 and 1 are not integers
https://en.wikipedia.org/wiki/Hyperinteger
Is there a subring of the hyperintegers which has the desired property? It's certainly not the hyperintegers themselves.
>>7909773
A nicer example would be the same sorting and addition rules, but making [math](a,b) \cdot (c,d) = (ac, bc+ad)[/math].
Then you keep [math]x \geq 0 \implies y \geq 0 \implies xy \geq 0[/math].
>>7910407
Awesome, thanks!
You did mean [math]x\ge0\&y\ge0\implies xy\ge0[/math] right?
>>7910431
Yes, [math]x \geq 0 \implies (y \geq 0 \implies xy \geq 0)[/math].
>>7910448
Ah, I read it as [math](x\ge0\implies y\ge0)\implies xy\ge0[/math] for some reason.
The word integer usually means it's just not a fraction.
There's nothing stopping you making up number systems but in conventional mathematics thats how it's literally defined so you can't change it.
>>7910473
I think OP understands this, they are just too new to know all the terminology yet.
>>7910231
great job spotting that
