# Is a number preceding infinity, finite?

by King
Tags: finite, infinity, number, preceding
 P: 43 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?
P: 838
 Quote by King 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?
tl,dr version: no "number" (whatever that means) precedes infinity (which infinity?).

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 $x < \frac{2+x}{2} < 2$. So x can't precede it.

Similarly, take the smallest infinity $\aleph_0$. If n is a natural number that precedes it, when what about n+1?

What you can do, it look at larger infinities. So $\aleph_0$ precedes $\aleph_1$ which precedes $\aleph_2$ and so on.
P: 43
 Quote by pwsnafu But on the rational numbers, nothing precedes, say, 2. Why? Suppose there was, call it x. Then $x < \frac{2+x}{2} < 2$. So x can't precede it.
Why is this? If x=0 the inequality makes sense.

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.

P: 838
Is a number preceding infinity, finite?

 Quote by King Why is this? If x=0 the inequality makes sense.
The inequality is true for any rational number less than two. That's the point.
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.
 I see it is as anything preceding infinity is finite.
"preceding" is not the same thing as "is less than". You are thinking of the latter.
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.

 The reason is because inifinity is defined as a never ending value.
That is not the definition of infinity. Importantly, you are using the word "infinity" in the singular. There are many infinities, some larger than others. $\aleph_0 < \aleph_1 < \aleph_2 < \ldots$

We have two FAQs on this: Cardinal and ordinal numbers and How does cardinality work?
 Nothing can be greater than a never ending value,
Note that every real number has a never ending decimal representation. If we used your definition then every real number is infinite.
 but I know plenty of values that are not never ending, and hence must be less than infinity.
Just because a property is true for some numbers does not mean it is true for every number.
P: 1,718
 Quote by King 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?
what other possibility would there be?
P: 838
 Quote by lavinia what other possibility would there be?
Infinite. For example the infinity $\aleph_0$ is strictly less than the infinity $\aleph_1$, but none the less $\aleph_0$ is still an infinity.

PF Gold
P: 6,489
 Quote by lavinia what other possibility would there be?
So you believe pwsnafu is wrong, yes? why is that? What math do you use to say you are right and he is wrong?
P: 1,718
 Quote by pwsnafu Infinite. For example the infinity $\aleph_0$ is strictly less than the infinity $\aleph_1$, but none the less $\aleph_0$ is still an infinity. I highly recommend you read the FAQ I linked.
the posts here discuss the real numbers not the ordinals.
P: 1,718
 Quote by phinds So you believe pwsnafu is wrong, yes? why is that? What math do you use to say you are right and he is wrong?
I didn't say anybody was wrong. There needs to be some ordering of the extended real numbers before you can discuss what number precedes another number. this post seems to be discussing te extended reals rather than ordinals.

Maybe I don't understand your post. Explain.
PF Gold
P: 6,489
 Quote by lavinia I didn't say anybody was wrong. There needs to be some ordering of the extended real numbers before you can discuss what number precedes another number. this post seems to be discussing te extended reals rather than ordinals. Maybe I don't understand your post. Explain.
The problem w/ answering the OPs question is, as explained by pwsnafu, the term "number preceding infinity". The point, as I see it, is that infinity minus 10 is STILL infinity, so I claim that "a number preceding infinity" is infinite, as does pwsnafu. You, on the other hand, seem to be saying that it is finite. I don't see how you can say that, and I don't see any confusion about my contending that you have said pwsnafu and I are wrong. We contend the answer is "infinite" and you contend that the answer is "finite". We can't both be right.

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.
P: 1,718
 Quote by phinds The problem w/ answering the OPs question is, as explained by pwsnafu, the term "number preceding infinity". The point, as I see it, is that infinity minus 10 is STILL infinity, so I claim that "a number preceding infinity" is infinite, as does pwsnafu. You, on the other hand, seem to be saying that it is finite. I don't see how you can say that, and I don't see any confusion about my contending that you have said pwsnafu and I are wrong. We contend the answer is "infinite" and you contend that the answer is "finite". We can't both be right. 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.
Infinity minus 10 has no meaning. When one says that a number precedes another number it must preceded it in some ordering.

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.
P: 838
 Quote by lavinia There is no arithmetic for the numbers plus infinity.
Firstly, we have many ways of treating infinity as a number: ordinals, extended reals, real projectives, surreals, and the Riemann sphere to name a few. All of these define operations with it, so it is an arithmetic. And by the axiom of choice, there always is a well ordering. Whether we can write that ordering down is a another matter.

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 $-\infty < x < \infty$ for all $x \in \mathbb{R}$. So negative infinity is an infinite that is less than positive infinity.
 P: 43 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?
P: 894
 Quote by pwsnafu Infinite. For example the infinity $\aleph_0$ is strictly less than the infinity $\aleph_1$, but none the less $\aleph_0$ is still an infinity. I highly recommend you read the FAQ I linked.
The links you cited don't give any information on infinity. I agree that infinity $\aleph_0$ and $\aleph_1$ are both infinity, but it seems to me to be wrong to say that one infinity is less than another since infinity is not a number and it makes no sense to talk about infinity plus or minus a numerical value. Where do you get that one positive infinity is less than another positive infinity?
As I see it $\aleph_0$ and $\aleph_1$ are the same value.
P: 894
 Quote by King 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?
If you are talking of never ending decimal representatations with a fixed decimal point, you are talking of a finite limit which is not an infinite value, but a finite value.
P: 838
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.

 Does anyone agree with what I just said?
That is not the definition of infinity people on this thread have been using. But if that is the definition you want to use, you are correct. Every real number has an infinity long decimal expansion. But that is not a useful definition

 Quote by ramsey2879 The links you cited don't give any information on infinity. I agree that infinity $\aleph_0$ and $\aleph_1$ are both infinity, but it seems to me to be wrong to say that one infinity is less than another since infinity is not a number and it makes no sense to talk about infinity plus or minus a numerical value. Where do you get that one positive infinity is less than another positive infinity?
The definition of $\aleph_1$ is the cardinality of the set of ordinal numbers. This is uncountable hence is strictly greater than $\aleph_0$. Applying the axiom of choice we can conclude there is nothing in between. You can read it on Wikipedia.
PF Gold
P: 6,489
 Quote by ramsey2879 As I see it $\aleph_0$ and $\aleph_1$ are the same value.
Well, the thing is, math operations don't really CARE what you think and they do not follow your "thought".

Aleph null and Aleph one are NOT the same value. I suggust you read up on them.
P: 894
 Quote by pwsnafu 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. That is not the definition of infinity people on this thread have been using. But if that is the definition you want to use, you are correct. Every real number has an infinity long decimal expansion. But that is not a useful definition The definition of $\aleph_1$ is the cardinality of the set of ordinal numbers. This is uncountable hence is strictly greater than $\aleph_0$. Applying the axiom of choice we can conclude there is nothing in between. You can read it on Wikipedia.
Fine. Now we both know what we are talking about. If the cardinality of the countable natural numbers is $\aleph_0$ then $\aleph_0$ plus or minus 10 is also countable; you just count 10 more or 10 less. Therefore, $\aleph_0$ + 10 is the same value as $\aleph_0$. That is what I was trying to say.

 Related Discussions Calculus & Beyond Homework 9 Calculus & Beyond Homework 8 General Math 2 Special & General Relativity 0