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Does this reasoning ever reach infinity? 0<1<2<3<4<5 |
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| Jun17-10, 11:04 AM | #1 |
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Does this reasoning ever reach infinity? 0<1<2<3<4<5
Does this reasoning ever reach infinity?
0<1<2<3<4<5..... What does it mean? Thanks! |
| Jun17-10, 01:38 PM | #3 |
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So the chain
0<1<2<3... 1. Never ends while reaching infinity? 2. Never ends and does not ever reach infinity? |
| Jun17-10, 01:59 PM | #4 |
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Does this reasoning ever reach infinity? 0<1<2<3<4<5We talk about numbers as they approach infinity. I guess that's tantamount to your #2. |
| Jun17-10, 02:20 PM | #5 |
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" 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? |
| Jun17-10, 03:00 PM | #6 |
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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. |
| Jun20-10, 06:23 PM | #7 |
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The reasoning that n < n+1 is satisfied for all numbers, I think that's what it's trying to say |
| Jun20-10, 09:26 PM | #8 |
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What about transfinite arithmetic? Georg Cantor?
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| Jul28-10, 04:37 PM | #9 |
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"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). |
| Jul28-10, 04:57 PM | #10 |
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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. |
| Jul29-10, 02:33 AM | #11 |
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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'. |
| Jul29-10, 09:43 AM | #12 |
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where I am using the ... to mean something similar to the meaning of the ... that the OP was asking about in the first place
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| Jul29-10, 02:03 PM | #13 |
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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...' |
| Jul29-10, 04:11 PM | #14 |
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| Jul29-10, 05:23 PM | #15 |
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| Jul29-10, 08:22 PM | #16 |
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Recognitions:
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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].
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| Jul29-10, 08:29 PM | #17 |
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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. |
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