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BigBalls
Profile Blog Joined May 2003
United States5354 Posts
July 26 2005 03:08 GMT
#61
kryzch thats a completely different question.

The question we are posing would be (lim [n-> oo] 100n) / (lim [n->oo] n), which isnt 1, but instead indeterminate. the answer to your question is clearly 100.
if you guys could use google and post direct links to the maphacks here it would be greatly appreciated. - Nazgul
Krzych
Profile Joined July 2003
Poland693 Posts
July 26 2005 03:20 GMT
#62
On July 26 2005 10:21 BigBalls wrote:
(...)
Now, 2*Z+ is a SUBSET of Z+. This means that every element in 2Z+ is an element of Z+. It is a proper subset, meaning there are elements in Z+ that are not in 2Z+. Both of these sets are the same size.


Hey, BigBalls, I know you're good at math, but I guess you have screwed something. Can you actually say, that Z+ and 2Z+ are the same size (which means they have the same amount of elements?) and at the same time say that 2Z+ is a subset of Z+ (which means that there are no elements that can be found in 2Z+ and cannot be found in Z+) and that there are some elements that belong to Z+, but not to 2Z+ ???
If there was a finite number of elements that are in Z+ and not in 2Z+ there would be no problem, because both of them have infinite number of elements. But there is an infinite number of odd numbers, so this just looks wrong.
I have never seen proof for what you have said, and my math isn't good enough to prove it right or wrong. Plus it can be infinity which fucks with my brain and doesn't let me to understand this.
Krzych
Profile Joined July 2003
Poland693 Posts
July 26 2005 03:23 GMT
#63
On July 26 2005 12:08 BigBalls wrote:
kryzch thats a completely different question.

The question we are posing would be (lim [n-> oo] 100n) / (lim [n->oo] n), which isnt 1, but instead indeterminate. the answer to your question is clearly 100.


Yup, you're right. But I just wanted to say, that infinity can be comprehended by humans. At least in it's not that vicious incarnation :-)
rSWisdom[9]
Profile Joined August 2004
United States117 Posts
July 26 2005 03:25 GMT
#64
so this runner passes an infinite amount infinitely small distances. infinitely small distances are distances of 0, so we get infinity / 0. use L'hopital's rule and it technically could work out, but not necessarily! we need more information! :O
BigBalls
Profile Blog Joined May 2003
United States5354 Posts
July 26 2005 03:30 GMT
#65
On July 26 2005 12:20 Krzych wrote:
Show nested quote +
On July 26 2005 10:21 BigBalls wrote:
(...)
Now, 2*Z+ is a SUBSET of Z+. This means that every element in 2Z+ is an element of Z+. It is a proper subset, meaning there are elements in Z+ that are not in 2Z+. Both of these sets are the same size.


Hey, BigBalls, I know you're good at math, but I guess you have screwed something. Can you actually say, that Z+ and 2Z+ are the same size (which means they have the same amount of elements?) and at the same time say that 2Z+ is a subset of Z+ (which means that there are no elements that can be found in 2Z+ and cannot be found in Z+) and that there are some elements that belong to Z+, but not to 2Z+ ???
If there was a finite number of elements that are in Z+ and not in 2Z+ there would be no problem, because both of them have infinite number of elements. But there is an infinite number of odd numbers, so this just looks wrong.
I have never seen proof for what you have said, and my math isn't good enough to prove it right or wrong. Plus it can be infinity which fucks with my brain and doesn't let me to understand this.


Define a map from 2Z+ to Z+ by x -> x/2. This map is clearly onto, every element in Z+ is mapped to by an element from 2Z+. It is also 1-1. How do we prove 1-1?

Suppose there is an element in Z+ that is mapped to by more than one element. Thus, x in 2Z+ and y in 2Z+ both map to z. Thus, x/2 = z = y/2, which means x=y. thus, the mapping is 1-1.

Since the map is both 1-1 and onto, the sets are the same size, although infinite.
if you guys could use google and post direct links to the maphacks here it would be greatly appreciated. - Nazgul
BigBalls
Profile Blog Joined May 2003
United States5354 Posts
July 26 2005 03:30 GMT
#66
i guess its wrong to even mention the word size when infinity is around, so saying there is a bijection between sets of different sizes is of enough merit, would you disagree?
if you guys could use google and post direct links to the maphacks here it would be greatly appreciated. - Nazgul
jtan
Profile Blog Joined April 2003
Sweden5891 Posts
Last Edited: 2005-07-26 03:43:56
July 26 2005 03:42 GMT
#67
On July 25 2005 18:33 Bill307 wrote:
Show nested quote +
On July 25 2005 04:38 jtan wrote:
An infinite amount of intervals that are getting infinitly small equals a finite lenght


Wrong. 1 + 1/2 + 1/3 + 1/4 + 1/5 + ... diverges to infinity, evening though the intervals are getting infinitely small.

There are a number of ways to determine whether a series converges to a finite number or diverges to infinity. That's calculus 2 .

Btw BigBalls is right. And probably some other people as well, but his answer was particularly terse .

haha yes I know, and your right, but it's a pretty simple explanation for the problem in question.

Also, btw, Zenos final conclusion was "motion is an illusion" which sounds cool:D
http://mathforum.org/isaac/problems/zeno1.html
Enter a Uh
Krzych
Profile Joined July 2003
Poland693 Posts
July 26 2005 03:42 GMT
#68
On July 26 2005 12:30 BigBalls wrote:
i guess its wrong to even mention the word size when infinity is around, so saying there is a bijection between sets of different sizes is of enough merit, would you disagree?


Now, that you have cleared that for me I can't disagree :D
Taveren
Profile Joined May 2004
United States241 Posts
July 26 2005 03:54 GMT
#69
slightly off topic.

im glad im not the only math nerd alive...anyone do math competitions ^^ maybe i've met u guys before
Tav[X]
BigBalls
Profile Blog Joined May 2003
United States5354 Posts
July 26 2005 03:56 GMT
#70
yeah.

i got a 30 on the putnam this year, which was good enough for 320th out of 3700 or so.

The putnam is the college math competition, very very tough, median score is a 0, probably 80% score under a 10.

I used to do some in high school, got a 115 on the AMC cause i made like 5 stupid mistakes, which ended up costing me a chance at the USAMO cause i only got a 60 on the AIME (the second competition which was much more difficult). oh well, they were fun tests to take, wish i would have concentrated more on the first test so I could had a shot at the USAMO.
if you guys could use google and post direct links to the maphacks here it would be greatly appreciated. - Nazgul
BigBalls
Profile Blog Joined May 2003
United States5354 Posts
July 26 2005 03:57 GMT
#71
and i find it kinda funny how i say using the word size is bad then use it again in the sentence, ha
if you guys could use google and post direct links to the maphacks here it would be greatly appreciated. - Nazgul
Taveren
Profile Joined May 2004
United States241 Posts
July 26 2005 04:00 GMT
#72
i overslept for putnam b/c i was too drunk the night before lol.

usamo wasn't too bad..i got 1question right...putting me into top 25 percent..i personally think putman practice look so much easier than usamo problems.

god...i havent done serious math since MOSP 3 years ago. keke.
Tav[X]
BigBalls
Profile Blog Joined May 2003
United States5354 Posts
July 26 2005 04:06 GMT
#73
Well, I think the putnam problems are more challenging, BUT, americans are generally awful at geometry, and that's all the USAMO is, so it's probably a bit more intimidating, and I guess harder overall because of the lack of algebra, calculus and discrete math problems.
if you guys could use google and post direct links to the maphacks here it would be greatly appreciated. - Nazgul
Taveren
Profile Joined May 2004
United States241 Posts
July 26 2005 04:08 GMT
#74
i'll make sure not to drink before this year's putnam...my strength is geometry and inequalities.....any of those on last year's putnam?
Tav[X]
BigBalls
Profile Blog Joined May 2003
United States5354 Posts
July 26 2005 04:09 GMT
#75
yeah, those are always the questions i avoid.

im an algebra/discrete/logic guy, not much of a geometry person
if you guys could use google and post direct links to the maphacks here it would be greatly appreciated. - Nazgul
Taveren
Profile Joined May 2004
United States241 Posts
July 26 2005 04:12 GMT
#76
i think inequalities are the easiest problems ever...just takes some time....it always reduces to AMGM or cauchy.
Tav[X]
BigBalls
Profile Blog Joined May 2003
United States5354 Posts
July 26 2005 04:30 GMT
#77
http://www.unl.edu/amc/a-activities/a7-problems/putnam/

Those are the problems from every year. Last year I got A1, A3, and B2, and should have gotten A2, but I forgot the stupid area of triangle = 1/2 ab sin C formula lol
if you guys could use google and post direct links to the maphacks here it would be greatly appreciated. - Nazgul
imRadu
Profile Blog Joined September 2002
1798 Posts
July 26 2005 04:34 GMT
#78
http://today.reuters.com/news/newsarticle.aspx?type=oddlyEnoughNews&storyid=2005-07-25T130436Z_01_L23243672_RTRIDST_0_ODD-ITALY-VAMPIRES-DC.XML
Its really good to see that some people dont let education get in the way of their ignorance
Tontow
Profile Joined September 2004
United States73 Posts
Last Edited: 2005-07-26 08:24:07
July 26 2005 08:22 GMT
#79
On July 26 2005 10:26 Tontow wrote:
Its lim(m --> ∞) sum(n = 1)^m (9)/(10^n) = 1 ......


Infinite 1 dollar is equal to infinite 100 dollars. And, Zeno is talking about an infinitely small portion of infinity, but
Show nested quote +

-However, It is also mathematically feasible to prove Zeno correct. Assuming that we are dividing “time” an infinite amount of times, then somewhere along the line there is bound to be a repeating decimal.

--------------------------------------------------------------------------------

3. Zeno points out, given that we are assuming that space is continuous,
--------------------------------------------------------------------------------


Now, here is where it gets tricky. The following formula stats that any that any decimal that repeats -- (and, thus, is infinite) – will infinitely approach and equal the next highest whole number, thus rounding it up. (Note: the proof was confirmed by my college math teacher.)
http://www.blizzard.com/press/040401.shtml
lim(m --> ∞) sum(n = 1)^m (9)/(10^n) = 1
0.9999... = 1
Thus x = 0.9999...
10x = 9.9999...
10x - x = 9.9999... - 0.9999...
9x = 9
x = 1.


I think the answer is to be shown as:

-2 -1 0 1 2
<--------------------------------------------------------------------------->
(0 < (X / infinity) <_ 1) / infinity
(0 is smaller than (X divided by infinity), (X divided by infinity) is smaller than or equal to 1) divided by infinity.

On July 26 2005 12:08 BigBalls wrote:
kryzch thats a completely different question.

The question we are posing would be (lim [n-> oo] 100n) / (lim [n->oo] n), which isnt 1, but instead indeterminate. the answer to your question is clearly 100.





Could you do a proof of that?
I’m not shore 100 is correct Because the way you wrote it:
-The limit of N is infinity. (N = infinity)
- 100N = Infinity
-N/N = Infinity




And remember that you stated that:

On July 26 2005 12:30 BigBalls wrote:
Show nested quote +
On July 26 2005 12:20 Krzych wrote:
On July 26 2005 10:21 BigBalls wrote:
(...)
Now, 2*Z+ is a SUBSET of Z+. This means that every element in 2Z+ is an element of Z+. It is a proper subset, meaning there are elements in Z+ that are not in 2Z+. Both of these sets are the same size.


Hey, BigBalls, I know you're good at math, but I guess you have screwed something. Can you actually say, that Z+ and 2Z+ are the same size (which means they have the same amount of elements?) and at the same time say that 2Z+ is a subset of Z+ (which means that there are no elements that can be found in 2Z+ and cannot be found in Z+) and that there are some elements that belong to Z+, but not to 2Z+ ???
If there was a finite number of elements that are in Z+ and not in 2Z+ there would be no problem, because both of them have infinite number of elements. But there is an infinite number of odd numbers, so this just looks wrong.
I have never seen proof for what you have said, and my math isn't good enough to prove it right or wrong. Plus it can be infinity which fucks with my brain and doesn't let me to understand this.


Define a map from 2Z+ to Z+ by x -> x/2. This map is clearly onto, every element in Z+ is mapped to by an element from 2Z+. It is also 1-1. How do we prove 1-1?

Suppose there is an element in Z+ that is mapped to by more than one element. Thus, x in 2Z+ and y in 2Z+ both map to z. Thus, x/2 = z = y/2, which means x=y. thus, the mapping is 1-1.

Since the map is both 1-1 and onto, the sets are the same size, although infinite.










Here is how I arrived at the answer of (0 < (X / infinity) <_ 1) / infinity :


I take into account the theorem (It is important to keep this in mind):

lim(m --> ∞) sum(n = 1)^m (9)/(10^n) = 1
0.9999... = 1
Thus x = 0.9999...
10x = 9.9999...
10x - x = 9.9999... - 0.9999...
9x = 9
x = 1.


X/infinity
X can’t equal 100 because it is impossible to get a repeating decimal that is greater than 1.
X/infinity will eventually have a repeating decimal. And thanks to the theorem, ”lim(m --> ∞) sum(n = 1)^m (9)/(10^n) = 1”, we will eventually run into a paradox.
-It is impossible for X/infinity to equal 0 and so I use “ 0 < “.
-It is impossible for X to equal anything greater than 1 since we are constantly dividing. However, X can equal 1 thanks to the theorem ”lim(m --> ∞) sum(n = 1)^m (9)/(10^n) = 1” and so I use “<_ 1”

And so I end up with:


(0 is smaller than (X divided by infinity), (X divided by infinity) is smaller than or equal to 1) divided by infinity.



To summarize and simplify for everyone:
(1). Start with any given number.
(2). Continue to divide that number until you end up with a repeating decimal; I can guarantee that the repeating decimal will not be greater than 1.
(3). Given the theorem ”lim(m --> ∞) sum(n = 1)^m (9)/(10^n) = 1”. That repeating decimal is = 1
(4). The given number we have now is 1. And your back at step (1).

Thus: (0 < (X / infinity) <_ 1) / infinity
BigBalls
Profile Blog Joined May 2003
United States5354 Posts
July 26 2005 08:33 GMT
#80
what are you even arguing against me???

his problem was lim(n -> infinity) (100n/n), which is 100
if you guys could use google and post direct links to the maphacks here it would be greatly appreciated. - Nazgul
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