# 1=0.

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• 08-13-2008
Blackroot
Code:

`#define 1 0`
Haha... HAHAHAHAHA! Take that LOGIC! /dance.
• 08-14-2008
misterMatt
What about the whole .999... = 1 'proof'?

http://en.wikipedia.org/wiki/0.999...
• 08-14-2008
vart
Quote:

sqrt(-1) = i
sqrt(-1) = +i
and
sqrt(-1) = -i
as well
• 08-14-2008
cpjust
Quote:

Originally Posted by vart
sqrt(-1) = +i
and
sqrt(-1) = -i
as well

Oh that's right! So in that case:
Code:

`sqrt( 1 ) = sqrt( -1 ) * sqrt( -1 ) = (+i) * (-i) = 1`
Problem fixed. :)
• 08-14-2008
Stonehambey
Quote:

Originally Posted by misterMatt
What about the whole .999... = 1 'proof'?

http://en.wikipedia.org/wiki/0.999...

There is nothing strange about that. 0.999... is indeed equivalent to one. Some people have a hard time accepting it though.

As for the sqrt one. Remember that "square root" is not a function, since it yields to possibilities (maybe if you do a masters degree in maths you might come across a type of function which gives two answers, but for most people, it is sufficient to define a function as a mapping blah blah which has ONE element in the co-domain ;)). So we in fact start having trouble on the second line where it says

sqrt(1) = 1

what it should say is

sqrt(1) = 1 <-- AND -1 AS WELL!!

All the square roots do a good job at confusing you ;)

Has anyone seen the differential one? It's one of my favourites

x^2 = x + x + x + x +... (x times)

(d/dx)(x^2) = (d/dx)(x + x +...)

2x = 1 + 1 + 1 +... (x time)

2x = x

2 = 1
• 08-14-2008
indigo0086
cute. I'm glad I'm not in bed with mathematics.
• 08-14-2008
vart
you forgot to d/dx of the (x times) part
• 08-14-2008
Stonehambey
Quote:

Originally Posted by vart
you forgot to d/dx of the (x times) part

Sorry, I just meant the (x times) bracket to mean that x squared is x added to itself x times, it's not actually part of the maths :)
• 08-14-2008
@nthony
Quote:

Originally Posted by Stonehambey
There is nothing strange about that. 0.999... is indeed equivalent to one. Some people have a hard time accepting it though.

I disagree, it's rightfully hard as no one can say what infinity is and what happens at it, so using it to equate anything with a defined unity may seem quite strange.

I propose the problem be restated as a limit instead (since limits admit that we cannot define what happens at infinity, but merely what appears to be happening as we approach it) (apologize for on-the-spot math...):

lim[n->inf] Σ[i=1..n]9/10^i =
9 * lim[n->inf] Σ[i=1..n](1/10)^i =
9 * lim[n->inf] ( (1/10) ^ (n+1) - (1/10) ) / ((1/10) - 1) =
9 * lim[n->inf] ( (1/10) ^ (n+1) - 0.1 ) / -0.9 =
9 * ( 0 - 0.1 ) / -0.9 = 1

Therefore it's safe to say that the limit of 0.9999... as the number of 9s grow without bound is 1 (but it is impossible to define its value at infinity as this itself is an undefined value).
• 08-14-2008
vart
Quote:

it's not actually part of the maths
yes it is - you have
Code:

`sum_0^x 1`
so the upper bound of sum depends on x, and as such should be taken in the account when appliyng differential on x
• 08-15-2008
Stonehambey
Quote:

Originally Posted by @nthony
I disagree, it's rightfully hard as no one can say what infinity is and what happens at it, so using it to equate anything with a defined unity may seem quite strange.

I propose the problem be restated as a limit instead (since limits admit that we cannot define what happens at infinity, but merely what appears to be happening as we approach it) (apologize for on-the-spot math...):

lim[n->inf] Σ[i=1..n]9/10^i =
9 * lim[n->inf] Σ[i=1..n](1/10)^i =
9 * lim[n->inf] ( (1/10) ^ (n+1) - (1/10) ) / ((1/10) - 1) =
9 * lim[n->inf] ( (1/10) ^ (n+1) - 0.1 ) / -0.9 =
9 * ( 0 - 0.1 ) / -0.9 = 1

Therefore it's safe to say that the limit of 0.9999... as the number of 9s grow without bound is 1 (but it is impossible to define its value at infinity as this itself is an undefined value).

lol, I told you people have a hard time accepting it ;)

x = 0.9999...

10x = 9.9999...

9x = 10x - x = 9.999... - 0.999... = 9

9x = 9

x = 1

Quote:

yes it is - you have
Code:

sum_0^x 1

so the upper bound of sum depends on x, and as such should be taken in the account when appliyng differential on x
It was just a comment, honestly. The error is in the very first line, if anyone wants a clue (or you could just look it up on google or something :P )
• 08-15-2008
Kernel Sanders
Quote:

Originally Posted by Stonehambey
lol, I told you people have a hard time accepting it ;)

x = 0.9999...

10x = 9.9999...

9x = 10x - x = 9.999... - 0.999... = 9

9x = 9

x = 1

I never liked that one, because 9.999... has one more '9' than 0.999... has. It's confusing because we're dealing with infinity here, but you're still manufacturing an extra decimal place. I'm with @nthony that the limit of 0.999... = 1, but that does not mean that .999... = 1

**EDIT**
From the wiki link
Quote:

There are many proofs that 0.999… = 1, of varying degrees of mathematical rigour. A short sketch of one rigorous proof can be simply stated as follows. Consider that two real numbers are identical if and only if their difference is equal to zero. Most people would agree that the difference between 0.999… and 1, if it exists at all, must be very small. By considering the convergence of the sequence above, we can show that the magnitude of this difference must be smaller than any positive quantity, and it can be shown (see Archimedean property for details) that the only real number with this property is 0. Since the difference is 0 it follows that the numbers 1 and 0.999… are identical. The same argument also explains why 0.333… = 1⁄3, 0.111… = 1⁄9, etc.
That does it for me better than anything ever has, but I still don't like it (I never liked dealing with infinity). The difference is 1/inf, which is defined as zero, but to me it's just another limit thing. The limit of 1/x is 0, but it still never gets there

Quote:

It was just a comment, honestly. The error is in the very first line, if anyone wants a clue (or you could just look it up on google or something :P )
Is it because that only holds true for positive numbers?
• 08-15-2008
Stonehambey
Yeah, it's ok for integers, but that's it I think. Saying that x squared is x added to itself x times isn't universally true for all real numbers.

vart, sorry I see what you're saying now, I initially thought you were implying that I should somehow apply d/dx to a comment ( I took what you said a bit too literally)

As a passing comment. People usually have no problems whatsoever when they see stuff like 1/9 = 0.111... It's widely accepted that point one recurring is the decimal equivalent of one ninth. However multiply both sides by 9 and you get 0.999... = 1 and something doesn't quite 'click'.

0.999... recurring is just an infinite sum WHICH IS DEFINED as 1, so it's correct to say that it equals one. Computers don't have a concept of infinity, but mathematics does, trying to picture 0.999... in your head or on a comp screen is trouble because you can't "imagine" what an infinite stream of numbers looks like. That is where the problems lies I think, intuitively 0.999... isn't equal to one because it will always have a little bit left over. But that doesn't make any sense, because this "little bit" (which is the result of the automatic truncation of the number your brain does when you try to picture it) is ALWAYS added on.

I realise I'm not going to convince some people, this thread and argument could go on forever (oh, the irony!), so I think I'll retire at this point :)
• 08-15-2008
Kernel Sanders
Quote:

Originally Posted by Stonehambey
Yeah, it's ok for integers, but that's it I think. Saying that x squared is x added to itself x times isn't universally true for all real numbers.

vart, sorry I see what you're saying now, I initially thought you were implying that I should somehow apply d/dx to a comment ( I took what you said a bit too literally)

As a passing comment. People usually have no problems whatsoever when they see stuff like 1/9 = 0.111... It's widely accepted that point one recurring is the decimal equivalent of one ninth. However multiply both sides by 9 and you get 0.999... = 1 and something doesn't quite 'click'.

Well, 1/9 isn't actually equal to .111... because 1/9 is impossible to represent using base 10 if you get right down to it

Quote:

0.999... recurring is just an infinite sum WHICH IS DEFINED as 1, so it's correct to say that it equals one. Computers don't have a concept of infinity, but mathematics does, trying to picture 0.999... in your head or on a comp screen is trouble because you can't "imagine" what an infinite stream of numbers looks like. That is where the problems lies I think, intuitively 0.999... isn't equal to one because it will always have a little bit left over. But that doesn't make any sense, because this "little bit" (which is the result of the automatic truncation of the number your brain does when you try to picture it) is ALWAYS added on.

I realise I'm not going to convince some people, this thread and argument could go on forever (oh, the irony!), so I think I'll retire at this point :)
Yeah - I can't say I'm convinced, but that doesn't mean I don't believe it's true. A lot of people who are a hell of a lot better at math than I have spent a hell of a lot more time reasoning about it and come to the conclusion that they are equal. I never was very comfortable with infinity and I guess this is just an artifact of that
• 08-15-2008
cpjust
In that case, what is 0.999...8 equal to? i.e. an infinite number of 9's with an 8 at the end. That probably doesn't make much sense mathematically and maybe even logically, but then again, if 0.999... = 1, then that means 1.000...1 = 1 also right? and if you subtract 1 from that it leaves 0.000...1. If you then subtract that from 0.999... you get 0.999...8.
Just because it can be "proven" mathematically, doesn't necessarily mean it's true. It's just true as far as we can currently prove. ;)
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