It looks like you're using an Ad Blocker.

Please white-list or disable AboveTopSecret.com in your ad-blocking tool.

Thank you.

 

Some features of ATS will be disabled while you continue to use an ad-blocker.

 

Thinking straight and intro to math.

page: 1
5

log in

join
share:

posted on Jul, 15 2014 @ 09:30 AM
link   
Hi, all.

There are a lot of topics here dedicated to science and I was wondering if we could start a new thread dedicated to mathematics and make it accessible to curious mathphobes. Things like elementary logic, proof techniques, basic foundations, popular topics like infinity and so on.

To start off, here's a simple proof of the fact that there are different sizes of infinities:

Claim: For every set S, |S| < |P(S)|.

Let x be in S.
Since f maps into P(S), f(x) in P(S).
Thus f(x) is a subset of S.
So, x in f(x) or x is not in f(x).

Let C be a set of all x in S such that x not in f(x).
Since C is a subset of S, C in P(S).

Assume that f maps S onto P(S).
Then there's an x in S such that f(x) = C.
Either x in C or x not in C.

Suppose x in C.
By definition x not in f(x).
So x not in C.
Contradiction.

Suppose x not in C.
By definition x in f(x).
So x in C.
Contradiction.

Thus f does not map S onto P(S).
So, P(S) does not have the same size as: |S| != |P(S)|.
Since |S| != |P(S)| and |S| =< |P(S)|(this needs proving), |S| < |P(S)|.

Let N be the set of natural numbers.

N = [1, 2, 3, 4, ..., 1000000000000000000000000, ..., 1424345566767768866534245454676888890894345345242, ...].

Then, by the theorem above |N| < |P(N)| < |P(P(N))| < ...



posted on Jul, 15 2014 @ 09:36 AM
link   
a reply to: groupsandrings

Well, I am not into math, but I am sure others around here are.

Welcome to ATS!



posted on Jul, 15 2014 @ 09:46 AM
link   
a reply to: groupsandrings

I'm not sure if you realize, but this is a forum for introducing people, not curricular materials. Try the science forum. The ATS search function can help you get there, and I know there's plenty of people who would love to discuss infinity with you.

Also, welcome!



posted on Jul, 15 2014 @ 09:46 AM
link   
As long as no one tries to say that .99999... = 1 I'm happy. I do like math though. Welcome to ATS though.
edit on 15-7-2014 by Krazysh0t because: (no reason given)



posted on Jul, 15 2014 @ 09:47 AM
link   

originally posted by: Krazysh0t
As long as no one tries to say that .99999... = 1 I'm happy. I do like math though. Welcome to ATS though.


That is effectively what math does, though. There is no such thing as "perfect accuracy" in math, because it's always just short of being right on the dot. The moment math overlaps into real world applications, you are introducing imperfect factors which are very difficult to be precise with. Numbers are perfect, physics is not.
edit on 15-7-2014 by AfterInfinity because: (no reason given)



posted on Jul, 15 2014 @ 09:52 AM
link   
a reply to: AfterInfinity

I understand rounding. I'm talking about the crazies who think that .999999999... = 1. The difference I'm getting at is the operator being used in the expression. .999999... ≈ 1 is the correct expression. There is nothing wrong with rounding, but at the same time you have to recognize that you aren't getting the exact result either.



posted on Jul, 15 2014 @ 09:53 AM
link   

originally posted by: Krazysh0t
a reply to: AfterInfinity

I understand rounding. I'm talking about the crazies who think that .999999999... = 1. The difference I'm getting at is the operator being used in the expression. .999999... ≈ 1 is the correct expression. There is nothing wrong with rounding, but at the same time you have to recognize that you aren't getting the exact result either.


Exactly.



posted on Jul, 15 2014 @ 10:00 AM
link   
a reply to: groupsandrings

I've got to say your introduction is the first one to ever make my brain hurt.

Nevertheless, Welcome to ATS. Your mathematical adeptness might be able to solve quite a few lingering problems that have been floating around.

But first you must pass a test- you must find the difference between a Heteronic string and Sugra. Only then can you enter this illustrious an infamous order.

I jest, I'm only trying to find common ground. Enjoy your time here.



posted on Jul, 15 2014 @ 10:05 AM
link   
Let n = .9999...

Then 10n = 9.9999...

So,

10n = 9.9999..
-n = .9999...
___________
9n = 9.000

Hence, n = 1.



posted on Jul, 15 2014 @ 10:11 AM
link   
Welcome to ats ..
Only thing other than english that makes my head hurt is maths .. will leave you all to it think time for another drink ..



posted on Jul, 15 2014 @ 10:13 AM
link   

originally posted by: groupsandrings
Let n = .9999...

Then 10n = 9.9999...

So,

10n = 9.9999..
-n = .9999...
___________
9n = 9.000

Hence, n = 1.


Actually, if n = .9999, then 9n is 8.9991

EDIT: I found this on Wikipedia:


Some students interpret "0.999..." (or similar notation) as a large but finite string of 9s, possibly with a variable, unspecified length. If they accept an infinite string of nines, they may still expect a last 9 "at infinity"


Seems to me that numbers are less perfect than I thought. This makes zero sense to me. If I have 9.9999 cents, I do NOT have a dime.
edit on 15-7-2014 by AfterInfinity because: (no reason given)



posted on Jul, 15 2014 @ 10:20 AM
link   
a reply to: AfterInfinity

No problem.

I started this thread in the hopes that someone who likes math and spends a lot of time on ATS might see this and get an idea to start a nice thread in an appropriate forum.



posted on Jul, 15 2014 @ 10:29 AM
link   
thats all well and good but you miss the point ...

i.e. hi, my name is .... how are you?



posted on Jul, 15 2014 @ 10:30 AM
link   
a reply to: AfterInfinity

Consider 0.9.

Multiply it by 10.

What you gonna get? 9.

Consider 0.99.

Multiply it by 10 and you get 9.9

Consider, then exactly 0.9999999.

Multiply by 10 and you get 9.999999.

Next step is multiplying 0.9999... by 10 which yields 9.999...




top topics



 
5

log in

join