45 replies to this topic

[galexcd]

Removed per request

[ghostrider]

A circle is simply an infinite-gon with all congruent sides. That proves it geometrically.

[husker]

Math is tricky. From what I was taught, a line is just a bunch of points. Now what is a point? It's hard to define. A circle is similar. It's easier if you just take math for what it is and don't think too deeply. But I do encourage thinking, and you bring up a good point. Hang on with me here as I give you my explanation, but remember I'm just in geometry. 179.99 etc. doesn't equal 180, and never will. It will come very close to 180, but it will never reach it. Therefore, a circle can't be a line because it never is 180. Just my ideas, you can think what you want.

[husker]

Quote

Ok I've got the equation: (lets say underlined numbers mean repeating)

x=.9 - Set x to .9 repeating
10x=9.9 - Multiply both sides by 10
10x-x=9.9-x - Subtract X
9x=9 - x is .9 repeating, so subtracting it from 9.9 repeating gives you 9
x=1 - divide by nine
.9=1 - Substitute

.9 times 10 equals 9 not 9.9

[TypoMage]

Why are you guys talking all mathematicly? A circle is the thing that goes on top of stick figures. Or this ( o ) That is a circle I do not really understand all these numbers and stuff?

[jlhaslip]

Quote

A circle is just an infinite number of angles on a line right?

I would say not...

A circle is a collection of contiguous 'points' equal distance from a 'centre' point. I didn't Google this definition, just based it on my understanding of what a circle is/is not. "An infinite number of angles on a straight line"... does not compute, Will Robinson...

[truefusion]

Quote

Well this keeps on going 179.9, 179.99999999, until it hits 179.9-repeating. Well it is mathematically proven that .9-repeating, equals 1.

It is not mathematically proven that .9-repeating equals 1. They tell you to round it off, which would equal 1. The only reason they tell you to round it off is so you won't have to write the number down.

Quote

Ok I've got the equation: (lets say underlined numbers mean repeating)

x=.9 - Set x to .9 repeating
[1]10x=9.9 - Multiply both sides by 10
10x-x=9.9-x - Subtract X
9x=9 - x is .9 repeating, so subtracting it from 9.9 repeating gives you 9
x=1 - divide by nine
.9=1 - Substitute

I think you messed up somewhere in your equation... Like [1], don't you divide here instead of multiplying? Do the opposite?

[husker]

Well to continue my annoying streak , I'd like to point out that it is mathematical impossible for .9 repeating to equal 1. It's like saying in the equation 2^x=y that y could equal 0.

Anyways, it's an interesting idea, but still not going for it.

[RedAlert]

.9~ = 1, and anyone who doesn't think so should go back to school and take advanced math.

Here are some proofs...pick one you can understand:
Let us assume x = .99999~
Now we know that when we multiply something by 10, we
move the decimal one place to the right.
so 10x = 9.9999~ There is no 9 lost by doing this
since there are an infinite number of them.
Now we do simple arithmetic
10x - x = 9.999~ - .9999~
9x = 9 This is allowed because every 9 after the
decimal will cancel with another 9.
x = 1 and x = .999~ so 1 = .9999~

.9999~ = .9 + .09 + .009 + ....
here we represent .9~ as an infinite sum
sum[i:0->inf.](.9*.1^i)
We know how to solve infinite sums.
sum = .9/(1 - .1) = .9/.9 = 1
Since we said the sum was initially .9999~, we can
conclude that .9999~ = 1

1/3 = .33333~
This is true, and can be proven with an infinite sum
as above.
3*1/3 = 3*.33333~
1 = .99999~
We are allowed to multiply by 3 because no part is
going to carry over to the next part. Thus, every part
of the decimal will increase by factor of 3, making it
a 9.

The real numbers are defined as limits of Cauchy
sequences of rational numbers.

*A rational number is a fraction of two integers
*A cauchy sequence is a sequence x(1), x(2), ... such
that for every integer n there exists an integer m
such that |x(j) - x(k)| =< 1/n for all j,k >=m.

Two Cauchy sequences x(1), x(2),... and y(1), y(2),...
are considered equivalent if for every integer n there
exists an integer m such that |x(k) - y(k)| =< 1/n for
all k>=m.

Let x(j) = 1 - (1/10^j)
Let y(j) = 1.

I'll leave it to you to check these are Cauchy
sequences.
These two sequences are equivalent:
Given some integer n, |x(k) - y(k)| = |1-(1/10^k) - 1|
= |1/10^k| =< 1/n if 10^k >= n. So we'll set m =
{smallest integer greater than log(n)}.

Thus the sequences .9, .99, .999, ... and 1, 1, 1...
are equivalent, so they have the same limit, namely,
.999~.

0.9~ = 0.9 + 0.09 + 0.009 + 0.0009 + ...
S = 0.9~
S = 0.9 + 0.09 + 0.009 + 0.0009 + ...
S = 0.9 + (1/10)[0.9 + 0.09 + 0.009 + ...]
S = 0.9 + (1/10)S
(9/10)S = 0.9
S = 1

Therefore, 0.9~ = 1.

If two numbers are not equal, there are an infinite
number of numbers between them. Give me a number
between .9999~ and 1.

All repeating and terminating decimals can be
represented as fractions. If .999~ is not represented
by 1, what fraction does represent it?

[Saint_Michael]

I like to point out that I lost all faith in the circle now with all this math and it's existence nonsense . So to help mess with people's mind's even more check out this wonderful info about the circle.
http://en.wikipedia.org/wiki/Circle

Of course through my researching this wonderful thought there are several other sites that would agree .9999~ = 1

here

http://digg.com/tech_news/.9999999=1

and here

http://polymathematics.typepad.com/polymat..._sorry_it_.html

But like I said you all ruin the beauty of the circle now I to rely on a triangle and a square to get me through life . but to stay on topic just a bit after reading these posts and those 3 websites I would have to agree especially with this part

Quote

Here are some proofs...pick one you can understand:
Let us assume x = .99999~
Now we know that when we multiply something by 10, we
move the decimal one place to the right.
so 10x = 9.9999~ There is no 9 lost by doing this
since there are an infinite number of them.
Now we do simple arithmetic
10x - x = 9.999~ - .9999~
9x = 9 This is allowed because every 9 after the
decimal will cancel with another 9.
x = 1 and x = .999~ so 1 = .9999~

The math teacher followed the same line of thinking, in which I conclude that .99999~ = 1 is both a true and false statement. Meaning that if you round up you will get 1; however, since the number is always repeating itself then it is not a true solid number (can't think of the word for it but you math geeks know what I am referring to). You call this a enigma in itself and odds are you would have to apply occam's razor to make the most sense out of this enigma.

Edited by Saint_Michael, 14 May 2007 - 10:40 AM.