Does 0.9999(repeating) = 1?

Yes 705 54.95%
No 578 45.05%
Voters: 1283. You may not vote on this poll

Does 0.9999(repeating) = 1?

First Riot Post
Comment below rating threshold, click here to show it.

Mawat

Member

01-23-2013

So, I just level up to five just to post here...

In previous pages, people used limits into certain F( k ) to try to prove that it is 1.However, limit implies the maximum value that the function could have but will NEVER have.

http://na.leagueoflegends.com/board/...6&d=1358972830


Separated, the result of 1/(10^n) with n->infinite tends towards zero but it is never zero.It can be as small as you want but it will never be zero.

However, when we apply limit to this function we don't get the actual function value on the point "infinite"(infinite is not really a point, is a behavior).When we apply limit for whatever value of N, we are actually looking the function in close intervals before and after the point to "limit" the maximum value it couldn't reach.


In some polynomials functions, assuming a definite point, the maximum value of the function on that point(which you get by applying limit with the variable tending towards the point) will be exactly the same as the function calculated on that point (which you get just applying the point in the polynomial).This is not the case because infinite is a behavior and not actually a number, so all we know is that this limit is zero because the function will never reach it, but the actual value will be as small we want but will never be zero.


For some engineering applications, we can actually ignore this minimum interval between 0,999... and 1 but this is not the case.


Comment below rating threshold, click here to show it.

Kodoku

Senior Member

01-23-2013

Quote:
Originally Posted by Mawat View Post
So, I just level up to five just to post here...

In previous pages, people used limits into certain F( k ) to try to prove that it is 1.However, limit implies the maximum value that the function could have but will NEVER have.

http://na.leagueoflegends.com/board/...6&d=1358972830


Separated, the result of 1/(10^n) with n->infinite tends towards zero but it is never zero.It can be as small as you want but it will never be zero.

However, when we apply limit to this function we don't get the actual function value on the point "infinite"(infinite is not really a point, is a behavior).When we apply limit for whatever value of N, we are actually looking the function in close intervals before and after the point to "limit" the maximum value it couldn't reach.


In some polynomials functions, assuming a definite point, the maximum value of the function on that point(which you get by applying limit with the variable tending towards the point) will be exactly the same as the function calculated on that point (which you get just applying the point in the polynomial).This is not the case because infinite is a behavior and not actually a number, so all we know is that this limit is zero because the function will never reach it, but the actual value will be as small we want but will never be zero.


For some engineering applications, we can actually ignore this minimum interval between 0,999... and 1 but this is not the case.
An infinite sum is itself a limit. It's not that we're taking the limit OF the thing we're trying to determine, it's that the thing we're trying to determine IS a limit.

Any infinite summation x1 + x2 + ...
is just a limit. Namely, the infinite summation 0.9 + 0.09 + 0.009 + ... (which is clearly 0.99...) is the limit of the sequence of partial sums (0.9, 0.99, 0.999, ...).

You'll notice that we're taking the limit of the sequence of partial sums, NOT the "limit of the series", whatever that means. That's because an infinite series is just a limit.


Comment below rating threshold, click here to show it.

Worstcase

This user has referred a friend to League of Legends, click for more information

Senior Member

01-23-2013

Quote:
Originally Posted by WhattayaBrian View Post
-snip-
Damn it all. Just goes to show that I shouldn't be trusted with recounting things I learned a long, long time ago. Now I feel bad for giving people incorrect information. Oh well.


Comment below rating threshold, click here to show it.

David Hume

Senior Member

01-23-2013

Quote:
Originally Posted by WhattayaBrian View Post
There are different sizes of infinity, but "Infinity * 2" both a) doesn't make sense mathematically as an operation, and b) if it did, it would be the same size as "Infinity * 1".
Your statement is correct, but just want to point out to everyone else that "Infinity * 2" having the same size (cardinality) as regular Infinity does not imply that they are equal.

Taking "Infinity" to mean the first/"smallest" infinite ordinal number, the cardinality of the natural numbers, expressions like "Infinity + 1" and "Infinity * 2" are perfectly well defined.

In set theory, the standard definition of the ordinal numbers are as follows:
0 is the empty set, { }.
1 is the set containing 0, {0} ( = { {} } )
2 is the set containing 0 and 1 {0, 1} ( = { {}, {{}} }
3 is the set containing 0, 1, and 2 {0, 1, 2}
etc.

In general, we define "n + 1" to be "the set containing n, and also every number inside of n." We also define < to mean that "A < B" if A is an element in the set of B (remember, numbers are just special kinds of sets in this system, and 3 really does contain 2, which is our justification for saying 2 < 3. You can think of the < sign as being equivalent to the ∈ or ⊂ sign if that makes you feel better - they're all equivalent on these particular sets.

So the "first infinity" is ω, the set of all the normal counting numbers, {0, 1, 2, 3, 4, ... }. ω is therefore both the set of all finite numbers and itself a number that is "bigger" than all of those numbers (in the sense that 3 < ω, 5 < ω, 100,000,000 < ω).

Since it's a number, we can +1 it to get the next number. Here goes:
ω + 1
= the set containing everything in ω (0, 1, 2, 3, .... ) and ω itself
= {0, 1, 2, 3, ... , ω}

which is a perfectly well defined number in our system. Since ω is a thing in ω + 1, then
ω < ω + 1
is a true statement - BUT ω and ω + 1 have the same cardinality (which is the bijection / bragging contest thing WhattayaBrian brought up earlier.) So in terms of cardinality, ω and ω + 1 have the same size (i.e., |ω| = |ω + 1| ) but at the same time, in terms of less than/greater than in terms of ordinality, ω < ω + 1.

If you extend "+ 1" into "+ n" in the sane way that gives you normal addition, and define multiplication in terms of this addition, it turns out that ω * 2 is still perfectly well defined number - it comes out to ω + ω, which is {0, 1, 2, 3, ... , ω, ω + 1, ω + 2, ω + 3, ... } which is greater than both ω and ω + 1 (and ω +2, etc) but is the same size as both of them.

so
ω < ω + 1 < ω + 500,000 < ω + ω = ω * 2
but
|ω| = |ω + 1| = |ω + 500,000| = |ω + ω| = |ω * 2| (the |x| means "the cardinality of x")

see http://en.wikipedia.org/wiki/Ordinal_number and http://en.wikipedia.org/wiki/Cardinal_number if your brain can still handle more.


Comment below rating threshold, click here to show it.

Aliath

Senior Member

01-23-2013

Please leave OT.


Comment below rating threshold, click here to show it.

Milranduil

Senior Member

01-23-2013

I cannot fathom that 44% of the people who polled this don't think they're equal.... it breaks my brain.


Comment below rating threshold, click here to show it.

Insentience

Member

01-23-2013

Wow this thread is still going?


Comment below rating threshold, click here to show it.

Milranduil

Senior Member

01-23-2013

Not anymore. Got moved to off-topic. RIP


Comment below rating threshold, click here to show it.

Kodoku

Senior Member

01-23-2013

It was moved to off-topic over 10 pages ago.


Comment below rating threshold, click here to show it.

Kaolla

Senior Member

01-23-2013

Quote:
Originally Posted by David Hume View Post
Your statement is correct, but just want to point out to everyone else that "Infinity * 2" having the same size (cardinality) as regular Infinity does not imply that they are equal.
precisely what i tell people but they don't listen