0=-1 (Calculus Puzzle)

[Ether]

Originally Posted by jblay

1=.99999...

The ellipsis means infinitely repeating decimal so

.99999... means limit(sum(9/10^i,i,1,n),n,inf) = 1

there is no contradiction.

[Peck]

Has this turned into a thread of trying to find a math trick that Ether doesn't know how to work it? If so, i want a chance to disprove it before he dose .

[Brian Ha]

Peck, sorry to tell you but i doubt ull win haha

Although ether, can you prove or disprove Riemann Sums? Now that would be cool.

[Ether]

Originally Posted by Brian Ha

Although ether, can you prove or disprove Riemann Sums?

fas est dividere cuius rei demonstrationem mirabilem sane detexi. Hanc marginis exiguitas non caperet

[Peck]

Originally Posted by Ether

fas est dividere cuius rei demonstrationem mirabilem sane detexi. Hanc marginis exiguitas non caperet

ingles por favor sinor

[PAR_WIG1350]

Originally Posted by Peck

ingles por favor sinor

1) *Inglés or favor señor.

2) according to google translate: It is a wonderful demonstration of what share of corruption. Shortness of this margin does not take

but that seems like it was lost a bit in the translation

It is actually taken from Fermat who wrote about a theorem (a^x + b^x = c^x has no integral solution set a,b,c for any integer value for x greater than 2) in the margins of a book in 1637. It is actually the description of the proof for this theorem which he came up with and it translates more accurately to

I have discovered a truly marvelous proof of this, which this margin is too narrow to contain.

unfortunately his proof seems to have disappeared off the face of the earth, assuming he even wrote it down.
Over the years people proved that it held true for specific exponents. It wasn't until 1995 that it all came together in an extremely complex proof that was probably not what Fermat had in mind, but a general proof of the theorem non the less.

[EricH]

Originally Posted by PAR_WIG1350

1) *Inglés or favor señor.

2) according to google translate: It is a wonderful demonstration of what share of corruption. Shortness of this margin does not take

but that seems like it was lost a bit in the translation

Sounds like the classic "I have discovered a truly marvelous proof of this, which this margin is too narrow to contain" from Fermat. Which, on investigating the Wikipedia entry on Fermat's last theorem, is exactly what it is.

Well played, Ether.

[Ether]

Originally Posted by EricH

Well played, Ether.

Thank you

[Peck]

Originally Posted by PAR_WIG1350

1) *Inglés or favor señor.

my computer is annoying whenever i try to insert alt symbols. also, i was using spanish, not latin and por was correct.

[RoboDesigners]

Has this turned into a thread of trying to find a math trick that Ether doesn't know how to work it? If so, i want a chance to disprove it before he dose .

I have an article here one of my math professors gave me that has a dozen "proofs" that 1=2. (She gave me this after I showed her my "proof," which happened to be in the article... I guess I didn't come up with it first...)

One that's not so hard to disprove is as follows:

0 = (1 - 1) + (1 - 1) + (1 - 1)...

Rearrange parenthesis:
0 = 1 + (-1 + 1) + (-1 + 1) + (-1 + 1)...
0 = 1 + 0 + 0 + 0....
0 = 1

(Ether's going to get this oh-so-fast...)

//Andrew

[Dusk Star]

Originally Posted by RoboDesigners

I have an article here one of my math professors gave me that has a dozen "proofs" that 1=2. (She gave me this after I showed her my "proof," which happened to be in the article... I guess I didn't come up with it first...)

One that's not so hard to disprove is as follows:

0 = (1 - 1) + (1 - 1) + (1 - 1)...

Rearrange parenthesis:
0 = 1 + (-1 + 1) + (-1 + 1) + (-1 + 1)...
0 = 1 + 0 + 0 + 0....
0 = 1

(Ether's going to get this oh-so-fast...)

//Andrew

Not if the other Andrew gets there first!

after you rearrange parenthesis, you should have the one at the end that was freed up, therefore
0 = 1 + (-1 + 1) + (-1 + 1) + (-1 + 1)... ...-1
0 = 1 + 0 + 0 + 0... ...-1
0 = 0

[Nate Laverdure]

Originally Posted by Dusk Star

after you rearrange parenthesis, you should have the one at the end that was freed up...

But can't the one at the end just be moved into the next (1 - 1) term?

[Ether]

Originally Posted by Dusk Star

Not if the other Andrew gets there first!

after you rearrange parenthesis, you should have the one at the end that was freed up,

There is no 1 "at the end". It's an infinite sum.

[Dusk Star]

Originally Posted by Ether

There is no 1 "at the end". It's an infinite sum.

Ah, ok. Not entirely how I was trying to communicate it, but I guess being an infinite sum invalidates that anyways.

[Ether]

Originally Posted by RoboDesigners

0 = (1 - 1) + (1 - 1) + (1 - 1)...

Rearrange parenthesis:
0 = 1 + (-1 + 1) + (-1 + 1) + (-1 + 1)...
0 = 1 + 0 + 0 + 0....
0 = 1

(Ether's going to get this oh-so-fast...)

Sorry, I was offline all day.

The problem with the above "proof" is that the associative law of arithmetic is not universally valid for infinite sums. So you aren't allowed to re-arrange (or remove) the parentheses.