# Does this reasoning ever reach infinity? 0<1<2<3<4<5

1. Jun 17, 2010

### Hippasos

Does this reasoning ever reach infinity?

0<1<2<3<4<5.....

What does it mean?

Thanks!

Last edited: Jun 17, 2010
2. Jun 17, 2010

### ramsey2879

Re: 1<2

I think it means that n<n+1 as n goes towards infinity.

3. Jun 17, 2010

### Hippasos

Re: 1<2

So the chain

0<1<2<3...

1. Never ends while reaching infinity?

2. Never ends and does not ever reach infinity?

4. Jun 17, 2010

### DaveC426913

Re: 1<2

We don't really talk about reaching infinity. Infinity is a concept, not a number; you can not perform most mathematical procedures on it. (For example you cannot claim that infinity/infinity = 1)

We talk about numbers as they approach infinity. I guess that's tantamount to your #2.

5. Jun 17, 2010

### Hippasos

Re: 1<2

" DaveC426913;2765460... Infinity is a concept, not a number ..."

That is interesting - so mathematics is not strictly about numbers but also concepts.

1. Mathematics without infinities i.e. concepts - not possible?

2. Mathematics does/does not require concepts to exist?

3. How do we know if a concept is a mathematical one?

6. Jun 17, 2010

### DaveC426913

Re: 1<2

There are myriad mathematical frameworks out there; arithmetic using the real numbers is merely one tiny set.

http://en.wikipedia.org/wiki/Number

7. Jun 20, 2010

### squelchy451

Re: 1<2

Technically, infinity is not a number
The reasoning that n < n+1 is satisfied for all numbers, I think that's what it's trying to say

8. Jun 20, 2010

### Antiphon

Re: 1<2

What about transfinite arithmetic? Georg Cantor?

9. Jul 28, 2010

### Jamma

Re: 1<2

It means precisely the following:

"n<n+1 for every natural number n (or n=0)"

In fact, you could also see it as an infinite number of statements such as

"5<6", "15<16" or "0<1"

So in a way, a statement is being made which applies to an infinite number of objects, but it says nothing about anything ever reaching infinity (and try to think about what you would actually be saying if it was, you will probably notice that you don't know what you mean precisely by this statement).

10. Jul 28, 2010

### Hurkyl

Staff Emeritus
Re: 1<2

Well, the be precise, the opening post is vague.

e.g. what reasoning is the opening poster referring to? What does he mean by reach? And what exactly is the ellipsis covering up?

In a mathematical document, the intent would usually be clear from the context. I'm drawing a blank when it comes to trying to think of someplace I might see it naturally occurring, however.

11. Jul 29, 2010

### ZQrn

Re: 1<2

Specifically:

x < y < z is a mild form of 'abuse of notation' as it's called. This is because it's neither (x < y) < z nor x < (y < z), specifically, it's x < y /\ y < z. Often transitive relationships are abused in that way, technically you can't do that like you can do x + y + z, which is both (x + y) + z and x + (y + z).

I guess that what you mean here, 'is there a highest number'. More 'mathematically' said:

'Does there exist a natural number n such that for any natural number m. m < n, or that statement formally, as in, completely properly and mathematically written down:

$$\exists n \forall m : n \in \mathbb{N} \land m \in \mathbb{M} \land m < n$$

And this formal sentence just happens to be false. So if I interpreted your question correctly, the answer is 'no'.

To show why: The natural numbers are defined as a set of 'objects' such that every object n has an object in that set called successor(n), the reverse is not true, namely, there is one object which is not a successor of another, that object is called zero conventionally.

So, by axiom, each natural has a successor, and by definition of '<', each natural is lower than its successor. Therefore there doesn't exist a natural which is higher than all other naturals, a natural is never higher than its successor.

'Infinity', as said before is best avoided, it's not an object in most contexts, and typically used in another form of 'abuse of notation', typically I'd recommend and use myself terms such as 'diverges' or 'grows unbounded' in place of 'goes to infinity'.

12. Jul 29, 2010

### Jamma

Re: 1<2

That's what I was trying to suggest, although you worded it better.

So let's agree that x < y < z to mean x < y /\ y < z, x < y < z < a to mean x < y /\ y < z /\ z < a ...

where I am using the ... to mean something similar to the meaning of the ... that the OP was asking about in the first place :rofl:

13. Jul 29, 2010

### ZQrn

Re: 1<2

Well, I'm afraid it's not that simple because of a few reasons.

1: Context free grammar, if we define it like this then mathematical notation is no longer generated by a context free grammar and violates the basic rule that x # y $z is either (x # y)$ z or x # (y \$ z). Which it is can be inferred from the associativity and the precedences of operators.
2: Formalism, where 'meaning' is defined as simple manipulations of symbols, it gets a lot more complex to define the inference if it's not context free.
3: ambiguity, the point is the 'false' is strictly smaller than 'true'. Indeed, we call the mathematical relationships of disjunction and conjunction monotonous because in disjunction the result is always equal or greater than its operants, and in conjunction it's always equal or less than its operants. So a < b < c < d would technically mean (a < b) < (c < d). Essentially implying here in a system of binary logic that the the former subpart is false, and the latter is true. This may be useless in binary logic, but in modal logic or fuzzy logic this has more implications.

I'm not saying it's not possible to make it rigorous, or even to just work with it under the assumption of understanding, I'm just pointing out that technically it's as unmathematical as saying such things as 'the infinitieth digit 0.999...'

14. Jul 29, 2010

### Hurkyl

Staff Emeritus
Re: 1<2

Just to be clear -- this comment is not aimed at finite strings of chained inequalities, but instead to the infinite string, right?

15. Jul 29, 2010

### ZQrn

Re: 1<2

Since I have no idea what you're talking about, probably neither.

16. Jul 29, 2010

### disregardthat

Re: 1<2

0 < 1 /\ 1 < 2 /\ 2 < 3 ... is a reasonable interpretation of 0<1<2<3..., but what does this mean? It is supposed to represent an infinite string of symbols, but that doesn't make much sense. What you want to say can be described by using quantifiers as such: $$\forall n \in \mathbb{Z}^{+} (n-1 < n)$$.

17. Jul 29, 2010

### ZQrn

Re: 1<2

It makes as much sense as things like N := {1,2,3 ...}. And it happens to be true.

The expression 0 < 1 /\ 1 < 2 /\ 2 < 3 ... is simply true, why, because true is a value, and the limit of that expression converges on that value. More formally we could write:

E(0) = 0 < 1
E(n) = N(n-1) /\ n < (n+1)

And then trivially, we can see that: lim (n -> inf) E(n) = true, abbreviated as simply: lim (n -> inf) E(n).

True and false are objects and values, functions defined on those values are things like '/\' or '->' which are often called connectives.

18. Jul 29, 2010

### disregardthat

Re: 1<2

In what context can you make sense of limiting truth values?

EDIT: bad example, I withdraw it

To extend predicates over an infinite domain we will need a logical machinery such as set theory in order to make sense of them. In which case we use quantifiers such as $$\forall$$, and not an infinite string of symbols. N = {1,2,3...} makes sense because we know exactly what it is when we write it, not because it is a representation of an infinitely long sequence of consecutive integers.

Last edited: Jul 29, 2010
19. Jul 30, 2010

### Jamma

Re: 1<2

Did I not make it rigourous (although the end of my post was meant to be more of a quirky joke, I guess I appreciate that it wasn't funny now )?

I guess you're first point is valid, but maths isn't context free, you later used the word "lim" but the "i" in the middle no longer means the imaginary unit "i", neither are you implying that "lim" means "l*i*m". Maths isn't context free when there is no danger in it being so, and rightly so, imagine how long texts would have to be if it was always context free, and how many new symbols we would need. Being this picky isn't as logical as you think, if anything, it is stupid.

And saying that, couldn't you logically say that the "< < < < <" is a giant symbol in which the elements reside? I see no problem in this, the topological propoerty of connectedness isn't a requirement of a mathematical symbol, as the letter i proves.

You could argue that then we cannot tell if each individual "<" should be interpretted as part of the "< < < < <" or on its own, but then you can't tell when the "." of an "i" should be seen as a "." above an iota", but I suppose it's clear from the context isn't it? :rofl:

20. Jul 30, 2010

### Hurkyl

Staff Emeritus
Re: 1<2

Nitpick: "context-free" is a technical term, and it refers to the generation of elements of the language, rather than to the issue of parsing.

21. Aug 2, 2010

### ZQrn

Re: 1<2

See Hurkyl's post, context free is a technical term. Mathematics for the most part is context free and mathematical notation is clearly intended as such.

Also, after some thinking about defining an order on truth values as in false < true, I actually realized it's already there, there is a total order on truth values, it's called logical implication, when we say $P \rightarrow Q$, we say precisely $P \leq Q$. If we posit that false is the lesser one, of course, any direction will work but we then just have to reverse it. Saying that P implies Q is really the same as saying that in the ordering of binary truth, P must be at least as large, or strictly smaller than Q.

Anyway, maybe 'Chomsky hierarchy' is an interesting subject for you to delve into.

I'm not saying that it makes sense, I'm saying that it's possible, nothing stops us from doing so.

We have a binary operation addition, which is commutative, we can thus make a nice special notation for a limit on it, that huge sigma. Similarily, it's often done in the same way with set union and set intersection.

Also, note that set union and set intersection are respectively simply the same as as logical disjunction (or) and logical conjunction (and) in the right context. They are isomorphic.

We may encode all propositions that are 'true' in a certain context simply as members of a set that represents that context, in that way, set union becomes the logical or, and set intersection the logical and. Or conversely, we may encode set membership simply as a boolean functor defined on a domain of discourse, on which elements it is true, it is in the set it encodes, on which elements it returns false, it is not in the set. And in that way again set union becomes logical or, and set intersection becomes logical and. Say that P and R are such functors, then P /\ R gives the set that is the intersection of both really..

So there's really no way to say we can't take a limit over truth values with logical and or logical or as its operation to construct 'partial conjunctions' or
'partial disjunctions' and it's done quite a lot in the sense of this notation, just written down another way:

Isomorphisms are quite common, in fact, under a lot of other isomorphisms, in the right context, conjunction becomes a counterpart to multiplication, and disjunction to addition.

Last edited by a moderator: Aug 2, 2010
22. Aug 2, 2010

### yossell

Re: 1<2

Just another nitpicky note, but logical implication is not the same as $P \rightarrow Q$, which is usually called material implication. Logical implication is not a total order: P does not logically imply Q nor does Q logically imply P, though it is true that one of $P \rightarrow Q$ or $Q \rightarrow P$ is always true.

23. Aug 2, 2010

### ZQrn

Re: 1<2

Then what is logical implication if different from this 'material implication'?

24. Aug 2, 2010

### yossell

Re: 1<2

At the propositional level, A logically implies B if and only if `A materially implies B' is *valid*.

So: (P & P -> Q) logically implies Q.

In general, given two propositional formulas A and B, A logically implies B iff, when you write out the full truth tables for A and B, there is no row of the truth table in which A is true and B is false.

In order to know whether A materially implies B, you have to know something about the actual truth values of A and B - you can't answer the question if you don't know either of the actual truth values. However, you don't have to consider any of the possible truth values.

In order to know whether A logically implies B, you don't have to know the actual truth values of A and B at all - however, you do have to look at the various possible truth values that A and B could have - by doing a truth-table - to answer the question.

25. Aug 2, 2010

### ZQrn

Re: 1<2

Ahh, so logical implication ranges over formulae, id est, expressions which resolve to truth values, while material implication ranges over variables which hold truth values?