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

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In summary: I think that what you meant 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'.
  • #1
Hippasos
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Does this reasoning ever reach infinity?

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

What does it mean?

Thanks!
 
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  • #2


Hippasos said:
Does this reasoning ever reach infinity?

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

What does it mean?

Thanks!
I think it means that n<n+1 as n goes towards infinity.
 
  • #3


So the chain

0<1<2<3...

1. Never ends while reaching infinity?

2. Never ends and does not ever reach infinity?
 
  • #4


Hippasos said:
So the chain

0<1<2<3...

1. Never ends while reaching infinity?

2. Never ends and does not ever reach infinity?

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


" 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


Hippasos said:
" 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?

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

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

Others could talk about this at great length, and there are many discssions here on PF about number systems.
 
  • #7


Hippasos said:
Does this reasoning ever reach infinity?

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

What does it mean?

Thanks!

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


What about transfinite arithmetic? Georg Cantor?
 
  • #9


Hippasos said:
Does this reasoning ever reach infinity?

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

What does it mean?

Thanks!

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


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

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


Hippasos said:
Does this reasoning ever reach infinity?

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

What does it mean?

Thanks!
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:

[tex]\exists n \forall m : n \in \mathbb{N} \land m \in \mathbb{M} \land m < n[/tex]

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


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

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

Specifically:

x < y < z is a mild form of 'abuse of notation' as it's called.

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


Jamma said:
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 ...
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


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...'
Just to be clear -- this comment is not aimed at finite strings of chained inequalities, but instead to the infinite string, right?
 
  • #15


Hurkyl said:
Just to be clear -- this comment is not aimed at finite strings of chained inequalities, but instead to the infinite string, right?
Since I have no idea what you're talking about, probably neither.
 
  • #16


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: [tex]\forall n \in \mathbb{Z}^{+} (n-1 < n)[/tex].
 
  • #17


Jarle said:
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: [tex]\forall n \in \mathbb{Z}^{+} (n-1 < n)[/tex].
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


ZQrn said:
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.

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 [tex]\forall[/tex], 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.
 
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  • #19


ZQrn said:
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...'

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 :frown:)?

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


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


Jamma said:
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 :frown:)?

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:
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 [itex]P \rightarrow Q[/itex], we say precisely [itex]P \leq Q[/itex]. 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.

Jarle said:
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 [tex]\forall[/tex], 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.
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:

7d7cb2b0656827e1c79d4631283990e5.png


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.
 
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  • #22


ZQrn said:
I actually realized it's already there, there is a total order on truth values, it's called logical implication, when we say [itex]P \rightarrow Q[/itex], we say precisely [itex]P \leq Q[/itex].

Just another nitpicky note, but logical implication is not the same as [itex]P \rightarrow Q[/itex], 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 [itex]P \rightarrow Q[/itex] or [itex]Q \rightarrow P[/itex] is always true.
 
  • #23


yossell said:
Just another nitpicky note, but logical implication is not the same as [itex]P \rightarrow Q[/itex], 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 [itex]P \rightarrow Q[/itex] or [itex]Q \rightarrow P[/itex] is always true.
Really? I had no idea.

Then what is logical implication if different from this 'material implication'?
 
  • #24


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

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


yossell said:
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.
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?
 
  • #26


ZQrn said:
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?

No, I didn't mean to imply that. Suppose we work with concrete sentences: Snow is white, Grass is green. Then 'Snow is white -> Grass is green' is true - for both consequent and antecedant are true, but `Snow is white' doesn't logically imply `Grass is green' for, since both sentences are atomic, there is a row in the truth table which makes the premise true and the conclusion false. However, `Snow is white -> Grass is green & Snow is white' does logically imply `Grass is green.' and even somebody who didn't know what snow was white and grass was green can know this - for by looking at all the rows in the 4-place truth table for these sentences, we can see that there is no row where the premise is true and the conclusion false.
 
  • #27


yossell said:
No, I didn't mean to imply that. Suppose we work with concrete sentences: Snow is white, Grass is green. Then 'Snow is white -> Grass is green' is true - for both consequent and antecedant are true, but `Snow is white' doesn't logically imply `Grass is green' for, since both sentences are atomic, there is a row in the truth table which makes the premise true and the conclusion false. However, `Snow is white -> Grass is green & Snow is white' does logically imply `Grass is green.' and even somebody who didn't know what snow was white and grass was green can know this - for by looking at all the rows in the 4-place truth table for these sentences, we can see that there is no row where the premise is true and the conclusion false.
Yes, but in our formal system of 'the real world', surely, everything implies that 'grass is green', because that sentence / formula is invariably true. We could say it's an axiom.

I think you're more talking about the issue of expressions who have bound variables. Say we have a formulae Phi which depends on three variables x,y,z which are all bound in it, and same for Theta with a,b,c. If we then say:

Phi(x,y,z) -> Theta(a,b,c)

Then you mean that for any arbitrary selection of x,y,z,a,b,c the left hand may not be 'greater' than the right, correct? At least, if I understand you correctly, it comes down to that in 'material implication' we deal with constants and in logical implication, our formula depends on variables which may be arbitrarily chosen to still satisfy the inequality.
 
  • #28


No.

I've outlined two distinct concepts. You should be able to see, given my particular explanation, why it is not true that everything logically implies `grass is green.' Now, I don't mind if you say, that's not what you meant by 'logical' implication. But I can't see a way to make progress if you don't understand that there are two concepts at play.

Do you understand the difference between (a) 'A -> B' is true, and (b) 'A -> B' is a truth-functional tautology?
 
  • #29


yossell said:
No.

I've outlined two distinct concepts. You should be able to see, given my particular explanation, why it is not true that everything logically implies `grass is green.' Now, I don't mind if you say, that's not what you meant by 'logical' implication. But I can't see a way to make progress if you don't understand that there are two concepts at play.
I understand that they are different, I'm just pointing out that their definition seem to imply that they are incarnations of the same umbrella where the former deals with constants and the latter with bound variables.

Do you understand the difference between (a) 'A -> B' is true, and (b) 'A -> B' is a truth-functional tautology?
I understand that some people make that distinction 'naively', yes, and what it would conceptually mean. The point is that 'tautologies' strictly speaking do not exist, there is no sentence which is 'universally true'.

(b) is simply a naïve way to say '(a) with respect to some overly established / used axiom scheme of logic / inference'.

Tautologies only apply as long as one's willing to say that some very established rules are true in the platonic sense.
 
  • #30


On truth values, [itex]\rightarrow[/itex] typically isn't an ordering; that's a special feature of two-valued logic.

To the two most common logics I know, the corresponding notion of "truth" is most naturally a Boolean algebra or a Heyting algebra. (Actually, classical logic is just a special case of intuitionistic logic) These algebras are, indeed, partially ordered, with the ordering being
[tex]\big(a \leq b\big) := \big(a = a \wedge b\big)[/tex]​

The [itex]\rightarrow[/itex] is an operator, however, and is adjoint to conjunction:
[tex]a \leq \big(b \rightarrow c) := (a \wedge b) \leq c[/tex]​
 
  • #31


Well, it depends on how you define '->' for multi valued logics.

I mean, say we have a logic where every truth value is in [0,1], surely I could then define P -> Q such that P <= Q.

I'm sure we could still derive some interesting things from it.

But I kind of have to plead ignorance on multi valued logics here, so maybe you have a superb reason on why this can't be done.
 

1. Is this reasoning valid?

Yes, this reasoning is valid. It follows the basic principles of mathematical reasoning and is a commonly used method of representing an infinite sequence.

2. Can this reasoning continue indefinitely?

Yes, this reasoning can continue indefinitely. As long as we keep adding 1 to the previous number, the sequence will continue to increase without limit.

3. Is there a limit to how high the numbers can go?

No, there is no limit to how high the numbers can go in this sequence. As we continue to add 1 to the previous number, the numbers will continue to increase without bound.

4. Can this reasoning be applied to other numbers besides 0, 1, 2, 3, 4, 5?

Yes, this reasoning can be applied to any set of numbers as long as they follow the same pattern of increasing by 1. For example, 0.5<1.5<2.5<3.5<4.5<5.5 or -1<-0.5<0<0.5<1<1.5.

5. How is this reasoning used in real life or in scientific research?

This reasoning is commonly used in mathematics and physics to represent infinite sequences, such as in the concept of limits and in the study of infinite series. It is also used in computer programming and data analysis to represent and manipulate large sets of numbers.

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