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Is a number preceding infinity, finite? |
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| Oct22-11, 10:02 PM | #1 |
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Is a number preceding infinity, finite?
Hi,
I'm not sure if this is the right section, but I'm talking about numbers :). The questions is as written in the title: Is a number preceding infinity, finite? |
| Oct22-11, 10:16 PM | #2 |
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Long version: When you use the word "precede" it means there is a unique element which is less than it, but still larger than all others contenders. So on the integers, 3 precedes 4, because 3 < 4 but 3 > 1 and 2 . 10001 precedes 10002. And so on. But on the rational numbers, nothing precedes, say, 2. Why? Suppose there was, call it x. Then [itex]x < \frac{2+x}{2} < 2[/itex]. So x can't precede it. Similarly, take the smallest infinity [itex]\aleph_0[/itex]. If n is a natural number that precedes it, when what about n+1? What you can do, it look at larger infinities. So [itex]\aleph_0[/itex] precedes [itex]\aleph_1[/itex] which precedes [itex]\aleph_2[/itex] and so on. |
| Oct22-11, 10:50 PM | #3 |
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I see it is as anything preceding infinity is finite. The reason is because inifinity is defined as a never ending value. Nothing can be greater than a never ending value, but I know plenty of values that are not never ending, and hence must be less than infinity. Of course infinity has some kind of strange philosophical aspect attached to it, and it sounds as though its separated from every other number. |
| Oct23-11, 01:11 AM | #4 |
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Is a number preceding infinity, finite?If x=0, ie, we argue that that x=0 is the predecessor to 2, then the inequality gives 0<1<2, contradicting the definition of "preceding". No matter what value of x we take, we can find a rational between the two. To qualify for the former, there must be nothing in between the two values. Edit: On second thoughts, scratch this. I'm thinking too much into it. Let's just use "less than" and be done with it. We have two FAQs on this: Cardinal and ordinal numbers and How does cardinality work? |
| Oct23-11, 07:30 AM | #5 |
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Recognitions:
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| Oct23-11, 07:49 AM | #6 |
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I highly recommend you read the FAQ I linked. |
| Oct23-11, 07:52 AM | #7 |
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| Oct23-11, 08:00 AM | #8 |
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Recognitions:
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| Oct23-11, 08:02 AM | #9 |
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Maybe I don't understand your post. Explain. |
| Oct23-11, 08:54 AM | #10 |
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I'm not a mathematician, and I don't understand why you say "There needs to be some ordering of the extended real numbers before you can discuss what number precedes another number" in terms of THIS discussion, but since I DON'T understand it, maybe I'm missing something. |
| Oct23-11, 09:13 AM | #11 |
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Technically infinity is not a number unless you are talking about the ordinals. But in this post the numbers seem to be the reals plus infinity. But then why not minus infinity as well? And is minus infinity less than plus infinity? Are the finite numbers greater than minus infinity? Is minus infinity equal to plus infinity? There is no arithmetic for the numbers plus infinity. |
| Oct23-11, 10:12 AM | #12 |
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The problem with the OP was that he didn't specify which infinity he was talking about. If he wanted to talk about the extended reals so be it. But he must learn there are other possibilities. Now, if we are to restrict the discussion to the extended reals, the answer is still no. The extended reals are the real numbers adjoined with positive and negative infinity. We define [itex]-\infty < x < \infty[/itex] for all [itex]x \in \mathbb{R}[/itex]. So negative infinity is an infinite that is less than positive infinity. |
| Oct23-11, 10:17 AM | #13 |
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Regarding the definition of preceding:
1. Come before (something) in time. 2. Come before in order or position. It doesn't have to be one position before. However, it should be safe to say that a number preceding another (where the numbers are ordered from lowest value to greatest value) has a lesser value. Hence, is the same as saying that it is less than. Let me be a little bit more specific about my question. Given a set of real numbers, the largest possible number would be the (base - 1) recurring - using decimal, this would be 9 recurring. Because 9 is recurring indefinitely, surely one can state that it is infinite? I suppose using this definition the same could be said about 1 recurring, but one would argue that this value is still less than 9 recurring. I suppose this answers my question...1 recurring is infinite and is less than 9 recurring, hence a number less than infinity is not necessarily finite... Does anyone agree with what I just said? Hmm...perhaps I should be talking about natural numbers to avoid the negative numbers which do not apply in my scenario. What do the books of mathematics state about infinity in a set of natural numbers? |
| Oct23-11, 10:17 AM | #14 |
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As I see it [itex]\aleph_0[/itex] and [itex]\aleph_1[/itex] are the same value. |
| Oct23-11, 10:40 AM | #15 |
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Blog Entries: 2
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| Oct23-11, 10:41 AM | #16 |
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To King: in mathematics the term "successor" really does mean "one position after", for example, in Peano Axioms the successor function takes a number and returns the next number. "Predecessor" would mean something something similar, and in fact, it does in my field of work. So please, don't use the term.
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| Oct23-11, 10:44 AM | #17 |
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Aleph null and Aleph one are NOT the same value. I suggust you read up on them. |
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