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anshil
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 Re: 0.999.   « Reply #25 on: Jul 29th, 2002, 12:08pm »

Well srowen I agree that it doesn't make much sense on paper and can I be it can easly be proven to be wrong as soon I start to assert it to be true. But it does make a sense in imagination. Say x is the smallest rational number bigger than one. You can imagine it as 1.0000......00001, or x is the biggest rational number smaller than 1. That one you can imagine as 0.9999.....9998. And not 0.99999...... which is equal to one, and not smaller.

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S. Owen
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 Re: 0.999.   « Reply #26 on: Jul 29th, 2002, 12:27pm »

on Jul 29th, 2002, 12:08pm, anshil wrote:
 Well srowen I agree that it doesn't make much sense on paper and can I be it can easly be proven to be wrong as soon I start to assert it to be true. But it does make a sense in imagination. Say x is the smallest rational number bigger than one. You can imagine it as 1.0000......00001, or x is the biggest rational number smaller than 1. That one you can imagine as 0.9999.....9998. And not 0.99999...... which is equal to one, and not smaller.

There is no such thing as the "smallest rational number greater than 1". Call it x. What is (1+x)/2 then?
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anshil
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 Re: 0.999.   « Reply #27 on: Jul 29th, 2002, 12:44pm »

Oops, you're right srowen. And I yet discovered another funny fact.

Okay say x is the smalles rationonal number greater than one. So what would (1 + x) / 2 then be? Well I would say it would be still x. So what does the equation yield? (1 + x) / 2 = x ----> x = 1. Ha ha ha!

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 Re: 0.999.   « Reply #28 on: Jul 29th, 2002, 3:09pm »

Very interesting discussion...
Here is what i found after search on google.com on the topic

------------------------------------------------------------
http://mathforum.org/dr.math/faq/faq.0.9999.html

Why does 0.9999... = 1 ?
This answer is adapted from an entry in the sci.math Frequently Asked Questions file, which is Copyright (c) 1994 Hans de Vreught (hdev@cp.tn.tudelft.nl).
The first thing to realize about the system of notation that we use (decimal notation) is that things like the number 357.9 really mean "3*100 + 5*10 + 7*1 + 9/10". So whenever you write a number in decimal notation and it has more than one digit, you're really implying a sum.

So in modern mathematics, the string of symbols 0.9999... = 1 is understood to mean "the infinite sum 9/10 + 9/100 + 9/1000 + ...". This in turn is shorthand for "the limit of the sequence of numbers

9/10,
9/10 + 9/100,
9/10 + 9/100 + 9/1000,
...."

One can show that this limit is 9/10 + 9/100 + 9/1000 ... using Analysis, and a proof really isn't all that hard (we all believe it intuitively anyway); a reference can be found in any of the Analysis texts referenced at the end of this message. Then all we have left to do is show that this sum really does equal 1:

Proof: 0.9999... =     Sum    9/10^n
(n=1 -> Infinity)

=  lim     sum 9/10^n
(m -> Infinity) (n=1 -> m)

=  lim      .9(1-10^-(m+1))/(1-1/10)
(m -> Infinity)

=  lim      .9(1-10^-(m+1))/(9/10)
(m -> Infinity)

= .9/(9/10)

= 1

Not formal enough? In that case you need to go back to the construction of the number system. After you have constructed the reals (Cauchy sequences are well suited for this case, see [Shapiro75]), you can indeed verify that the preceding proof correctly shows

lim_(m --> oo) 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.

Another informal argument is to notice that all periodic numbers such as 0.9999... = 9/9 = 1 are equal to the digits in the period divided by as many nines as there are in the period. Applying the same argument to 0.46464646... gives us = 46/99.

References
R.V. Churchill and J.W. Brown. Complex Variables and Applications. 0.9999... = 1 ed., McGraw-Hill, 1990.

E. Hewitt and K. Stromberg. Real and Abstract Analysis. Springer-Verlag, Berlin, 1965.

W. Rudin. Principles of Mathematical Analysis. McGraw-Hill, 1976.

L. Shapiro. Introduction to Abstract Algebra. McGraw-Hill, 1975.
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drdedos
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 Re: 0.999.   « Reply #29 on: Jul 29th, 2002, 3:36pm »

Kozi Morimoto would not win a race against the Tortoise.

Kozi says "0.9 is pretty close to 1, an approximation of 1.
You add 0.09 to make it 0.99 and its even closer to 1 than 0.9 but still not 1.  Repeat to infinity.  You get ever closer and closer to 1 from the low side, but you never get there?"

Kozi, say you are asked to run across a football field.  Before you get across the football field, you have to get 0.9 of the way across, right?  Now before you get across the remaining 0.1, you have to get across 0.9 of THAT first, right?  In other words, you have to first get to 0.99 of the football field?  Likewise, you have to get 0.999 of the way across the football field before you get all the way across, right?  And so on and so on...  Using the same kind of logic you used, quoted above, you will never make it across the football field.  Sorry to hear that.

BTW, the reference to the Tortoise is from a variant of Zeno's Paradox written by Lewis Carroll.
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Kozo Morimoto
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 Re: 0.999.   « Reply #30 on: Jul 29th, 2002, 4:17pm »

Yeah, that's right!  I was thinking about using the turtle/hare paradox to explain MY side of the arguement, but you beat me to it!  If this isn't the case, does that mean that drdedos actually get to the finish line following the rules?  I don't think so.

I apologise if my earlier statement implied that I was arrogant.  I didn't intend that at all - I was trying to express how my English language skill was too ineloquent to express my thoughts in writing.  The fact that this wasn't apparent also indicates that my English skill is well below par.

Back to .999....

You start with 0.8.  Divide by 10 and add 0.9.  What do you get?  0.98.  Divide that by 10 and add 0.9.  What do you get? 0.998.  If you do this infinite amount of time, don't you end up with 0.999...998? with infinite count of 9s between the 0 and the 8?  Is this number equal to 0.999...?

Along another direction, if 0.999... is equal to 1, doesn't this imply a nonhomogeneous (lumpy) R space?  I'd assume that this effect only happens when 9 changes to 10 in decimal representation ie 1.999... = 2 or 0.4999... = .5 or -7.86999... = -7.87 so it only happens at certain places in R and not others?
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S. Owen
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 Re: 0.999.   « Reply #31 on: Jul 29th, 2002, 8:29pm »

Kozo deserves credit for making everyone think very carefully about their arguments, and for refusing to accept anything less than perfect arguments! This is an excellent discussion.

on Jul 29th, 2002, 4:17pm, Kozo Morimoto wrote:
 Back to .999....   You start with 0.8.  Divide by 10 and add 0.9.  What do you get?  0.98.  Divide that by 10 and add 0.9.  What do you get? 0.998.  If you do this infinite amount of time, don't you end up with 0.999...998? with infinite count of 9s between the 0 and the 8?  Is this number equal to 0.999...?\

I think that we must not confuse "doing something many, many times" and "doing something an infinite number of times."

Again, there is no such thing as "0.999...998" with an infinite number of nines. If this is a terminating decimal, then there must be a finite number of nines preceding the eight. If not, then it makes no sense to write it like it is a terminating decimal. This is an illusion, a subtle abuse of notation.

If you are using "0.999...998" as a sort of shorthand to express the limit of your process, then yes, it has a value, and it is 1, and is equal to "0.999...". But it's important to realize that 1 is the limit of this process, the point to which it is heading, but never reaches in a finite number of steps. To write it as if it is a terminating decimal like "0.999...998" is only deceptive.
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NickH
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 Re: 0.999.   « Reply #32 on: Jul 30th, 2002, 12:56am »

"Along another direction, if 0.999... is equal to 1, doesn't this imply a nonhomogeneous (lumpy) R space?  I'd assume that this effect only happens when 9 changes to 10 in decimal representation ie 1.999... = 2 or 0.4999... = .5 or -7.86999... = -7.87 so it only happens at certain places in R and not others?"

That's a good question, and the answer is no, this does not imply lumpy R.  It's true that any number that can be represented as a terminating decimal has two decimal expansions, as in the examples you have given.  But these are just two different ways of writing the same number.  There is no lumpiness in the underlying reals.  Note also that the same thing happens in other number bases, and sometimes for different numbers.  For example, decimal 1/3 can be written in base 3 as either 0.1 or 0.0222...  However, 1/3 has only the repeating representation 0.333..., in base 10.

Ordinarily it does not matter that there are two ways of writing certain decimal numbers.  However, in Cantor's famous diagonal proof of the uncountability of the reals, we must choose one or the other representation. See: http://c2.com/cgi/wiki?CantorsProof
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Kozo Morimoto
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 Re: 0.999.   « Reply #33 on: Jul 30th, 2002, 1:52am »

Thanks for the counter arguments, its making it a lot more digestible.

So the number I get by divide by 10 and 0.9, eg 0.9998 and ever increasing number of 9s, if there were really really large number of 9s (say large n but not "quite inifinity"), its still less than 0.999.. with n+1 9s?  As soon as they both 'reach' infinite number of 9s, they are all both considered to be 1?

So can you start with any finite number and keep dividing it by 10 and add 0.9 to it, and if you repeat the process to infinity you get 1?

So 'inifity' is a really special situation?  Are there any good newbie article/paper on infinity and its applications/consequences on the web?  I'm very much interested in doing some personal research into it.  I've learned quite a bit on this thread, not so much on solving riddles, but still, learned a lot.
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Willis
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 Re: 0.999.   « Reply #34 on: Jul 30th, 2002, 5:15am »

First, let 0.999... be denoted by x.
Observe that 10x = 9.999...
Observe that 10x - 9 = 0.999... = x,
and so 10x - 9 = x.
Thus, 10x - x = 9, 9x = 9, and x = 1.
But x = 0.999..., and so 0.999... = 1.

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S. Owen
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 Don't think of it as "reaching infinity"   « Reply #35 on: Jul 30th, 2002, 6:07am »

Yeah, infinity is a special situation in the sense that it's not a number, not even a big one, so you can't think of it in the same terms.

Thinking of it as "reaching infinity" eventually is misleading. Theses processes will never reach infinity and never equal 1.

However, 0.999... = 1. This is not saying that some large number of 9's equals 1, but is a statement about where their sum is heading. I think it's misleading to think of it as "where you get after an infinite number of 9s".

When you toss in infinity you are speaking about the limits, the imaginary endpoint that you never quite reach.

on Jul 30th, 2002, 1:52am, Kozo Morimoto wrote:
 Thanks for the counter arguments, its making it a lot more digestible.   So the number I get by divide by 10 and 0.9, eg 0.9998 and ever increasing number of 9s, if there were really really large number of 9s (say large n but not "quite inifinity"), its still less than 0.999.. with n+1 9s?  As soon as they both 'reach' infinite number of 9s, they are all both considered to be 1?   So can you start with any finite number and keep dividing it by 10 and add 0.9 to it, and if you repeat the process to infinity you get 1?   So 'inifity' is a really special situation?  Are there any good newbie article/paper on infinity and its applications/consequences on the web?  I'm very much interested in doing some personal research into it.  I've learned quite a bit on this thread, not so much on solving riddles, but still, learned a lot.

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Kozo Morimoto
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 Re: 0.999.   « Reply #36 on: Jul 31st, 2002, 2:47am »

Hope people don't mind me ranting on...

Talking about inifinity and bounds (ie 0.999...99, as a theoretical entity, is it possible to be holding an end of an infinitely long string in each hand?  Like, imagine like holding a skipping rope, but the middle of the rope gets further and further away until its infinitely far away, like 0.999...998.  Or at infinity, do you actually hold 2 ropes/strings of infinite length instead of 1?
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bartleby
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 Re: 0.999.   « Reply #37 on: Jul 31st, 2002, 6:30am »

Hi Kozo!

Yes, you are on the right track if you listen to srowen.

The whole point to the "Does .999.... really EQUAL 1?" puzzle is really "Do you understand what 'infinity' means?"

Every time you bring up an argument like "What if I had a number like .9999....998", you are demonstrating that you don't quite grasp the concept of infinity... Very few people do, I'm not saying I do either.  You're probably a math Ph.D. candidate if you do.  But you have to think about infinity not as a number, but rather as a concept, or even better as a process.  You can use the concept of infinity to basically say, "Well, if I do something, and I do it again, and again, and again, and it's moving toward (but not quite reaching) a goal, (the concept of a limit), then I can project that if it were to happen an infinite number of times, that goal would be exactly reached."

But infinity isn't a number... you can prove this by saying (with the example of .99999.... = .9 + .09 + .009 + ....) that if infinity is a number, then let's say x = infinity, and if I do this process x times, I am still 1-.1*10-x away, and by the way I could still do it one more time, couldn't I?  So the idea of saying "x equals infinity" is meaningless.

This destroys the concept of ".999....98" because exactly HOW MANY 9's are there in there?  And why can't you have one more?

One thing I was waiting for you to say was to refute the proof:  x = .999... , 10x = 9.999..., 9x = 9, x = 1 by saying "Well, if x = .999....  then 10x should be 9.9999....990, right?"  And as shown above, that doesn't make sense to try and say that!
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Drake
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 Re: 0.999.   « Reply #38 on: Aug 2nd, 2002, 3:02pm »

When I first looked at this, I thought it was dependent on the level of precision as to whether 1 >= .999..., but after thinking about it it makes perfect sense because the precision is infinite as defined by the question.

As a test, try this on a un*x machine:

bc -l
(1/3)+(2/3)

The obvious answer is 1, but it yields .999...

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mook
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 Re: 0.999.   « Reply #39 on: Aug 3rd, 2002, 12:07pm »

i'm no math major, but you've provided some pretty simple proofs which look correct on paper.  However, I think that the interesting discussion is on the semantics of infinity.  If you apply no math, you logically know that .999........ is less than 1, as it will repeat infintely as it approaches 1, never equalling it.
First, let 0.999... be denoted by x.
Observe that 10x = 9.999...
Observe that 10x - 9 = 0.999... = x,
and so 10x - 9 = x.
Thus, 10x - x = 9, 9x = 9, and x = 1.
But x = 0.999..., and so 0.999... = 1.

just to get you thinking....
when you multiply .99999... times 10, how can you possibly ever get as answer.
here's some grade school math for you to ponder:

99
x10
---
00
+99
-----
990

if you try to solve
.9999...
x10
--------

you will NEVER find the answer because you will start by multiplying 0 times 9 an infinte number of times; it may always be zero, but you will keep writing zeros down for infinity.

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Charles miller
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 Re: 0.999.   « Reply #40 on: Aug 4th, 2002, 7:24pm »

I don't understand, even if .999... continues on indefinately, how it could ever = 0. It's length only gets it closer to approximating 1 but can never = one. An infinite number of 9's doesn't change this property. .999...
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S. Owen
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 Re: 0.999.   « Reply #41 on: Aug 4th, 2002, 8:36pm »

Indeed, 0.999...9 does not equal 1 no matter how many 9s you add. It seems logical to then say that adding a infinite number of 9s still does not get you to 1, and so 0.999... < 1.

However it's deceptive to talk about adding "an infinite number of 9s" because you can't.

"0.999... = 1" does not say that you get to 1 after adding an infinite number of 9s. It is only a statement about what this process is getting infinitely close to, but never reaching of course.
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OMG
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 Re: 0.999.   « Reply #42 on: Aug 5th, 2002, 4:11pm »

0.9999 approaches 1, never reaches it, but keep trying...

I say 0.99999999... > 1.000...
because
if at some point in time .9999... = 1.000... intersect
then there must be a point at which they intersected previously, where 0.99999... > 1.0000...
life isn't Cartesian, it is not parallel, they must intersect more than once; think pre-BigBang
math is only a concept, the current universe is reality and only temporary
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S. Owen
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 Re: 0.999.   « Reply #43 on: Aug 5th, 2002, 6:50pm »

Oh man, that sounds just profound enough to be true...
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otter
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 Re: 0.999...   « Reply #44 on: Aug 6th, 2002, 7:39pm »

Wow!  The last time I dealt with anything even remotely resembling this was over 30 years ago.

It seems to me that the entire problem rests with being able to grasp the concept of infinity.  Not "very large numbers", but infinity.  In fact, the statement .999... = 1 is *only* true as the number of nines becomes infinite.  For any finite number of nines, the statement becomes false.

Just as an exercise, though, does the logic change if we ask does .999... = 1.000...?

Otter
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and know the place for the first time.
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Paul Hsieh
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 Re: 0.999.   « Reply #45 on: Aug 6th, 2002, 8:12pm »

Look people --

(1-an) = (1-a) * (1+a+a2+a3+ ... +an-1)

So, 0.999999... = 9/10 * (1.11111...)

Now putting 1/10 in the formula above we see that:

(1-(.1)n) = (1-0.1) * (1+.1+.01+ ... +(.000...01))

Taking limit of n to infinity we get:

(1-(.1)inf) = (0.9) * (1.11111...)

But 0 < (.1)inf < eps, for all eps > 0, therefore (.1)inf = 0.  Which means

1/0.9 = 1.1111...

So, 0.999999... = 9/10 * (1.11111...) = 9/10 * 1/0.9 = 9/9 = 1.

Got it?

---

I think the basic trouble people have is that they see a notation like:

0.9999...

and they don't really have a good concept of what that even means.  Well let me tell you.  It means:

0.9 + 0.09 + 0.009 + 0.0009 + ...

and nothing else.  If you cannot sum that up and get the answer 1.0, then that means you cannot sum an infinite series in general, and that you should not be engaging in discussions of mathematics that go beyond high school.  Zeno's paradox is in the same category.

Let me be clear.  This is just an infinite sum.  Its a very straightforward thing.  Either you can do it, or you can't.  And the answer is 1.  That's all.
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na
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 Re: 0.999.   « Reply #46 on: Aug 8th, 2002, 10:03am »

I think the following is wrong,

(1-an) = (1-a) * (1+a+a2+a3+ ... +an-1)

So, 0.999999... = 9/10 * (1.11111...)

It should be

(1-an) = (1-a) * (1+a+a2+a3+ ... +an) if n is infinite
(hint: how much is inf - 1?) (hint: how much is inf - 1?) (hint: how much is inf - 1?) (hint: how much is inf - 1?) (hint: how much is inf - 1?) (hint: how much is inf - 1?) (hint: how much is inf - 1?) (hint: how much is inf - 1?) (hint: how much is inf - 1?) (hint: how much is inf - 1?) (hint: how much is inf - 1?) (hint: how much is inf - 1?) (hint: how much is inf - 1?) (hint: how much is inf - 1?)

so, 0.999999... =x= 9/10 * (1.11111...)
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S. Owen
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 Re: 0.999.   « Reply #47 on: Aug 8th, 2002, 11:30am »

on Aug 8th, 2002, 10:03am, na wrote:
 (1-an) = (1-a) * (1+a+a2+a3+ ... +an) if n is infinite (hint: how much is inf - 1?) (hint: how much is inf - 1?) (hint: how much is inf - 1?) (hint: how much is inf - 1?) (hint: how much is inf - 1?) (hint: how much is inf - 1?) (hint: how much is inf - 1?) (hint: how much is inf - 1?) (hint: how much is inf - 1?) (hint: how much is inf - 1?) (hint: how much is inf - 1?) (hint: how much is inf - 1?) (hint: how much is inf - 1?) (hint: how much is inf - 1?)   so, 0.999999... =x= 9/10 * (1.11111...)

In taking the limit as n goes to infinity, it doesn't matter whether we put in n-1 for n or not. So there is no problem here involving infinity-1 or anything... we are taking a limit.

But more importantly, the change to:

(1-an) = (1-a) *(1+a+a2+a3+ ... +an)

simply makes this not a true statement (try n=2 for example). Did you mean:

(1-an+1) = (1-a) *(1+a+a2+a3+ ... +an)

Same thing as the original equation, and certainly no different in taking a limit.
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Christopher Delovino
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 Re: 0.999.   « Reply #48 on: Aug 8th, 2002, 3:15pm »

Lots of good answers why .9999.... would equal 1 but so far everything here has been done analytically.  Let's try to graphically prove it.  (I'm assuming that the reader has access to a graphing calculator)

Let's first look at

y = x - 1

This gives us a diagonal line with x = 1 on the x-axis (in coordinate form (1, 0)  ).  Now, I can stop here and say that you can just trace .9999... for x and that will eventually get to 0, but that's no fun.

So let's modify the equation a bit by multiplying 1 using
(x - 1)/(x - 1) giving us

y = ((x - 1)^2)/(x -1)

Now this gives us basically the same graph as the previous equation with the exception that there's a gaping hole at x = 1 now (because a 0 in the denominator is a no-no).

Now if we calculate x = .9, y = -0.1
If we go further, we get (once again in coordinate format)

(.9, -0.1)
(.99, -0.01)
(.999, -0.001)
(.9999, -0.0001)
...

Well, from that we can see that as x gets closer to 1, y gets closer to 0.

If my memory of calculus serves me, if the limit at a point of x approaches a y-value c from the right the left then the value of y is c.  Thus answering Kozo's previous question about 1.00...01 being equal to .9999.

In short, .9999 will never reach 1.  BUT considering that infinity will never come along we just skip the exact value semantics and say that it is 1.  That's basically what infinity represents, "I'm too lazy and no computer has enough processing power to prove this.  We're close enough so let's just say that we're there."
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NoYes
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 Re: 0.999.   « Reply #49 on: Aug 10th, 2002, 5:18pm »

Looks like some people didn't pay full attention in calculus.  I skipped to the end after the first page and a half so I am sry if this was brought up in the half a page I didn't read .

Kozo you are a winner ! He is demonstrating that if you all fallow your ideas of how limits work you will fail.

Near the end of page one anshil said:

"Using the ideas given above, does it mean that 1.000 ... 001 with infinite zeroes between the two 1s make it equal to 1?

No not, it's infinitely small larger than one."

If you believe that you must believe that 0.999... != 1

Because 1-0.9999.... = 0.00000.....1
1.00000....1-1=0.00000....1

Kozo is showing the error of your interpretation of infinity and limit when he uses the .99999....8 = .9999... = 1 = 1.0000...1 example.

As I think Olsen said, we can't reach infinity, well if we can't reach infinity there is no problem using infinity + 1 since both are unreachable, and for that matter infinity - 1 is also unreachable.

1- 1/(infinity) = .9999999...., that seemed to be already agreed upon.  so 1- 1/(infinity-1) = .99999......8, right? Because 1/(infintiy-1) would have to be the number that is greater then 1/infinity by the smallest unit.

1-1/(infinity) just shorthand for the limit as n--> infinity of 1-1/n, for those of you that are being a little more staunch with your notation, but now we are into limits.

If you say infinity doesn't exist or is unreachable then "The limit" doesn't exist!  If you say infinity does exist or is reachable then the limit does exist....but you have some other problems, of the infinity +1 sort .

So again, 0.999999.... != 1 if you say infinity is unreachable, because .9999999 will always be short of 1 by 0.0000...1.  You can say the limit of 0.99999... = 1 because you know that .99999... < 1 and is getting closer the deeper you go into the ... or infinity.

limit != =, in english, limit is not equality, all it means is you are getting close to it, you are never at it, because if you did get to it you would be at infinity and still have infinity + 1 to go.  If you want to argue that you do reach it then I would face you with this problem:

if .99999 = 1 when you are at infinity then at infinity + 1 do you have 1.0000....1?  How about infinity + infinity? do you have 2 then?

Almost forgot to talk about it, but someone brought up the tortoise and the hare (found most famously in GEB...not sure why no one mentioned that...).  To say it will never cross the finish line is to assume you take the same amount of time on each step ( i.e. 0->.5, .5->.75, ect.).  Maybe I have always misread it, but I have never been entertained by that "paradox."  Because of course you will cross the finish line because as the distance being traveled reaches 0 the time to cross it reaches 0.

Said the same thing many different ways, hopefully that will help get the message to many different people reading it in many different ways.

-NoYes
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