Who intuits that 10 is closer to infinity than 5?

Being highly intuitive, I intuited on this for several couplings of minutes, and I came to construe that 5 is in fact no more adjoining to infinity than 1, infinity having no ultimate terminus.

0 Replies

g__day said:

Quote:

Whilst everyone understands 10 > 5, would anyone here view that 10 is therefore closer to infinity?

I would, given that my concept of infinity is something like "an amount so vast that it's limits are indeterminant". And words like "vast" make me think of big, bigger and biggest and then even more than biggest which makes me remember that ten is more than five, so yeah, I'd say it's closer to infinity.

Let me guess though- I'm wrong?

malek said:

Quote:

Being highly intuitive, I intuited on this for several couplings of minutes, and I came to construe that 5 is in fact no more adjoining to infinity than 1, infinity having no ultimate terminus.

But if you think of a number line with infinity stretching out into the unmeasurable distance at one end, ten would be closer to that end than five, right? Or am I thinking too linearly (as well as literally). I have a tendency to do that...

0 Replies

"unmeasurable distance at one end"

But there is no end, therefore the comparative distance can never be measured.

Therefore, seeing that definitive evidence can never be obtained, it reduces the whole exercise to a guessing game.

If the thread title used the words " is probably" as opposed to "is", I may have answered differently.

0 Replies

I would intuit that zero is closer to infinity than either 10 or 5.

0 Replies

I would intuitly agree with you.

0 Replies

I still intuit that ten is closer to infinity than five though because even though the distance into infinity can't be measured, the distance between five and ten can and ten comes after five and before infinity.

I feel more confident in my answer now that you've said it's a guessing game- actually the question was "how do you intuit"- so it's not wrong to say that I intuit differently than anyone else...it's just true.

I don't remember learning or even thinking about this in highschool though- I vaguely remember something about imaginary numbers, but I couldn't tell you what that was about anymore.

0 Replies

Positive or negative infinity? In the ring of integers, or the fields of the rational, real, & complex?

As for proof, I'd first try induction and the definition of a well ordered group.

Rap

0 Replies

Five or ten what? Sheep.

0 Replies

Spendius wrote:

Quote:

Five or ten what? Sheep.

So are you telling us it's sheep or asking?

Personally, I don't think it matters- it could be five or ten units of anything to illustrate, although something like sheep would lend itself less to the theoretical aspect of the question- as given time or space you'd eventually come to the end of your flock and I'd hate to think of an infinite number of sheep-that'd be kind of messy and boring and land-depleting-wouldn't it?

I'd like to hear raprap's intuitive take on both positive and negative infinity. I'm in the mood to learn something.

0 Replies

Original question has no answer - as someone already observed here, the type of infinity must be specified before any proof can be presented concerning the 2 positive integers, 5 and 10.

Looking up Georg Cantor might help.

0 Replies

High Seas wrote:

Looking up Georg Cantor might help.

Natural numbers are countably infinite, consequently a low cardinality. The dust that remains when you take the naturals from the rationals, and the rationals from the reals, and the reals from the complex magnify the numbers of elements and make the answer uncountable.

However, in my mind a logic exists. You could partially prove the theorem using two sets, one set (A) containing all numbers (natural or otherwise) from 5 to positive infinity (inclusive?) and the other (B) all numbers from 10 to positive infinity. The first set (A) then contains numbers that are not in the second one (B), whilst the intersection of the two sets (A intersection B=B) is the same as the second set (B). Consequently, there is no one to one relationship from the elements in A to B and the sets do not have the same cardinality (number of elements).

So set A is larger than B. QED(?)

Rap

0 Replies

Well I raised this just to show how folk think about an infinity, say a positive integer infinity. The first thing to observe is - is it treated like any other special number (e.g. pi or e) , or is it viewed as not a number?

Imagine the initial question was re-stated to ask ;

Is Infinity + 10 > Infinity + 5?

Would this trigger any increased clarity in how to deal with the question?

If you say the above re-worked equation must hold true, then the first equation is also true. If you say the above equation is false then the first equation is also false.

The key thing is to consider what determines the way you think. Get it right and the answer falls out.

0 Replies

g__day wrote:

Is Infinity + 10 > Infinity + 5?

Not necessarily. This treats infinity as an undefined number. IMHO this this is akin to which is larger--5 divided by zero or 10 divided by zero? Whereas a legitimate question would be to ask which grows faster the limit of 5/e or 10/e as e goes to zero.

Rap

0 Replies

"infinity" is only a partial representation for a number. Without knowing where the infinities originate from, you cannot determine if one infinite value is greater or less than another. It is entirely possible that the variable representing "infinity + 5" is greater than "infinity + 10".

The question elicits the reader to make the false assumption that both infinities are initially equal...but it is also obvious that "infinity - 1000 = infinity"...so you can't make that assumption.

So the way you have stated the problem, "infinity + 10" ?= "infinity + 5" is indeterminate....as is "(infinity - 10) ?= (infinity - 5)" for the same reason (your initial question)

However, if we state that:

limit Mad

->infinity of (X+10) = A
limit: x->infinity of (X+5) = B

Although A and B both equal infinity, having known where these infinities came from we can say that A>B and A-B = 5.

Likewise, you could say that

limit: x->infinity of (X^2) = C
limit: x->infinity of (X) = D

And although C and D both equal infinity, you can say that C >= D

0 Replies

*leaves thread in silence, never to return...*

0 Replies

Thanks for everyone's feedback!

The key insight mathematicans look for is someone to say you're mixing numbers, and operations on them, with concepts that aren' t numbers.

Whilst you can use limits to represent some small groups of infinities, as a general rule:

INFINITIES AREN'T NUMBERS AND CAN'T BE USED AS SUCH!

Infinities and numbers don't mix. So infinity + numbers leaves you in the space of an infinity. No number is closer to infinity than any other because no number is close to infinity at all, full stop - by definition all are an infinite distance away.

So in some specialised fields of mathematics you can size infinities according to specific rules, but these rules and finite numbers don't mix.

You where all kinda on the right track, the infinity on the LHS and RHS of the equation may well have been the same, but a infinity + 5 and an infinity + 10 can't be dealt with as numbers in a field subject to rules of a field, and that was where the trick is. It's so subtle most people miss it.

The ideas of operations like + - x / > < which are common in real, integer and complex number fields can't be generally applied to infinities, and most often are poorly construed when numbers are interacting with them.

For these branches of maths you have to effectively drag the numbers and the operations from the finite into an infinite domain and apply only the rules that constrain that unique domain - not the rules of everyday maths in Real or Complex number fields.

Well done all!

0 Replies

raprap wrote:

However, in my mind a logic exists. You could partially prove the theorem using two sets, one set (A) containing all numbers (natural or otherwise) from 5 to positive infinity (inclusive?) and the other (B) all numbers from 10 to positive infinity. The first set (A) then contains numbers that are not in the second one (B), whilst the intersection of the two sets (A intersection B=B) is the same as the second set (B). Consequently, there is no one to one relationship from the elements in A to B and the sets do not have the same cardinality (number of elements).

So set A is larger than B. QED(?)

Rap

No. There is a one-to-one mapping of set B to set A:
b -> b-5
The sets have the same cardinality. Your proof doesn't apply to infinite sets.

0 Replies

markr wrote:

raprap wrote:
However, in my mind a logic exists. You could partially prove the theorem using two sets, one set (A) containing all numbers (natural or otherwise) from 5 to positive infinity (inclusive?) and the other (B) all numbers from 10 to positive infinity. The first set (A) then contains numbers that are not in the second one (B), whilst the intersection of the two sets (A intersection B=B) is the same as the second set (B). Consequently, there is no one to one relationship from the elements in A to B and the sets do not have the same cardinality (number of elements).

So set A is larger than B. QED(?)

Rap

No. There is a one-to-one mapping of set B to set A:
b -> b-5
The sets have the same cardinality. Your proof doesn't apply to infinite sets.

B asymptotically tends to {B - 5}, is that what you say, or do you mean A? Or don't you mean asymptotic at all? Pls clarify.

0 Replies

An element, b, in set B is mapped to b-5 in set A.

0 Replies

Who intuits that 10 is closer to infinity than 5?

Related Topics

Quick Links

My Account

able2know