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Why 1 = 2


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30 replies to this topic

#11 Garrek99

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Posted 28 June 2007 - 01:56 AM

I am not arguing about whether or not the limit of 0.99999999999 is 1. That is obvious due to the definition of the word "limit". And, if you are going to use limits in the proof of .999999 = 1 then its the same as just saying that the limit of .99999 = 1.
Because, inherently we accept that (for ie) any number to the power of negative infinity equals to 0. That's probably the first thing a teacher will tell you when he/she introduces limits and such. But we both know that 0 and x^-inf are two different things conceptually.
Just as duality exists in the way scientists accept light, wave or particle, there is duality in this, no more no less. I stand firm on the belief that .9999999 = 1 practically but not conceptually. And the trick with the equation is that you can't always treat irrational numbers like natural numbers.

I looked up the wikepedia link and here is what I copied from it:
"A 19th-century reaction against such liberal summation methods resulted in the definition that still dominates today: the sum of a series is defined to be the limit of the sequence of its partial sums."
So, it is simply agreed that to refer to a function like that one can express it in terms of its limit. And the limit of 0.9999.... is 1 but they are not the same.

Although, after reading the wikepedia page on 0.99999 I retract everything that I have said. It's just mind numbing how deep that rabbit hole goes. People far more able than me have messed with it and other similar, complicated dudu. There is a lot to learn out there. I am not disenchanted though, just humbled.
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#12 Garrek99

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Posted 29 June 2007 - 09:11 PM

Update! I've been inspired. I realized something important yesterday. Here is a layman’s proof on why 1 and 0.9999999... are distinct:

What is an infinitesimal? ( 1-0.99999...= ) It is the difference between every "consecutive" number whether it is rational, real, irrational or any as long as it has its place on the number line. If numbers are the limits of the subset they represent beneath them than its just from the number system that we use that 1 is neatly capped at 1 because otherwise 1 is as unlimited as 0.9999999.... If after-all every number describes and infinitesimal point on the number line then their separations are also infinitesimal and therefore 1 and 0.9999999... are as distinct as any other number on the number line.
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#13 Incognitum

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Posted 04 July 2007 - 11:11 AM

Update! I've been inspired. I realized something important yesterday. Here is a layman’s proof on why 1 and 0.9999999... are distinct:

What is an infinitesimal? ( 1-0.99999...= ) It is the difference between every "consecutive" number whether it is rational, real, irrational or any as long as it has its place on the number line. If numbers are the limits of the subset they represent beneath them than its just from the number system that we use that 1 is neatly capped at 1 because otherwise 1 is as unlimited as 0.9999999.... If after-all every number describes and infinitesimal point on the number line then their separations are also infinitesimal and therefore 1 and 0.9999999... are as distinct as any other number on the number line.




For layman's purposes you are right of course; numbers are arbitrary, and by definition nominal. It's only when dealing with them in a 'laboratory' setting as it where, that these kinds of distinctions are put under greater scrutiny.
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#14 Garrek99

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Posted 08 July 2007 - 11:19 PM

I feel like I need to add something to an earlier reply. I keep thinking about this and as most things that aren't studied to their core by me (which is everything) I come to an inconclusive conclusion. hehe

Here is a problem: Say 1 and 0.9999999... are separated by an infinitesimal and can somehow be viewed as "consecutive", where or what is the next consecutive down? It doesn't exist on our number line. We can't just right 0.99999... ...8 cause its an infinite series, and 0.9999999... - infinitesimal makes no sense either. It seems the "number line" isn't really a number line. It's like a line with some points marked on the surface and in between these marked points there are bottomless canyons that only God knows what goes on there.
I've enjoyed our discussion on this Incog.
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#15 oldcabman

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Posted 12 August 2007 - 06:23 AM

Here are the given equations:

x = 2
x(x-1) = 2(x-1)
x2-x = 2x-2
x2-2x = x-2
x(x-2) = x-2
x = 1

Take the equation on the second line and solve it:
x(x-1) - 2(x-1) = 0
x**2 -3x + 2 = 0
(x-1)(x-2)=0

(Note: the term x**2 is supposed to be "x-squared")

Therefore, x=1 AND x=2 are valid solutions for the original equation. This is Algebra I. Nothing wrong at all. Equations defined with a 2nd power term often have two different numerical solutions.
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#16 zAlbee

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Posted 13 August 2007 - 02:10 AM

oldcab,

There is something wrong. This proof shows that, from x=2, you can reach x=1. After that you can say

x = 2
x = 1
1 = 2
1 = 1+1
1 = 2+2
1 = 4
etc. etc. and you can prove that every number is equal to every other number, which is of course false. :)

As everyone pointed out, the mistake is dividing by x-2 without checking. When writing a real proof, you might write:

x(x-2) = x-2
Divide both sides by x-2, which is OK if x-2 != 0
x = 1

And then you would realize right away that the step is invalid, because x=2 from before, and x-2 is 0.
Depending on your proof, there may not have been a definition x=2 beforehand, then a step like this is OK. But you still should write the condition, or at least verify any solutions found to make sure a divide by zero did not occur. Even if the solution is not affected, you may be docked points for not doing the check.

Therefore, x=1 AND x=2 are valid solutions for the original equation.


More explicitly, this means that x=1 OR x=2, given the premise x(x-1) - 2(x-1) = 0. There is absolutely nothing wrong with what you wrote -- when you start from the second line. What if you started from the first line, x=2? Then x=1 is not a valid solution anymore. In fact, these are only possible solutions, and there is nothing to say that both of them must be valid solutions at all, or that any situations exist where either is valid (say, you add the premise x>10).

But the original proof starts from the first line, and all 6 equations must be true at the same time, because no conditions were stated. So the conclusion is x=1 AND x=2 AND (all the other equations are true), given the premise x=1. And this certainly can not be true.


This riddle shows why it's important to write proofs the proper way, doing all the necessary checks, annoying as it may seem. And it might also explain why math teachers stress that you should always use 'OR' and write concluding sentences, and freak out if you just leave x=1, x=2, x=7, etc. scattered all over the place :D
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#17 Smithsps

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Posted 09 December 2007 - 05:58 AM

Its not wrong but 1 and 2 work for the last formula.

The problem is that the problem was not worked out until
the problem had the variable by itself.
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#18 a person

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Posted 13 January 2008 - 07:11 PM

It seems to me that there are two possible problems, depending on how you look at the question. The first line of the proof states that x = 2. Not x = 1, not x = 2 or 1, only x = 2. In the next line of the proof, we multiply both sides of the equation by (x - 1). Well, this is where we go wrong. We've added a solution. That isn't legitimate. When you solve equations, you can do any operation you want, as long as you do the same thing to both sides of the equation. We've done that by multiplying both sides by (x - 1). But you must also go back to the original equation and make sure that all of the solutions you find satisfy the initial conditions. Well, our initial conditions say that x = 2. So when we get that x can be 1 or 2, we go back: do both of these solutions fit the original equation, x = 2? Well, of course x = 1 doesn't work.

If this, to some, is evidence that 1 = 2, because x = 1 and x = 2 so of COURSE 1 and 2 must be equal, look at it this way: After we added the factor (x - 1), we have a quadratic. If we write this quadratic in factored form, as (x - a)(x - b) = 0, we get the solution x = a OR x = b. The given proof simply confuses a conjunction and a disjunction, and states that x = a AND x = b rather than or.
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#19 yomero7

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Posted 14 May 2008 - 06:15 PM

Here is one possible "quick" proof that does not involve the "Sigma" notation (i.e., the notation Sum [n=1 to infinity]).

Everyone agrees that

1/3 = 0.333333333.......

Multiply both sides by 3 to obtain:

1 = 3 * ( 1/3 ) = 3 * ( 0.33333333........) = 0.99999999...................

Thus 1= 0.999999........


And by the way: 0.999...... + 0.9999..... = 1.99999999..... = 2

The reason being the same
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#20 sparx

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Posted 15 May 2008 - 03:26 PM

the very original problem in this post is actually perfectly right once you realize that x= is just asking for the zeros of the quadratic equation y=x^2-3x+2

the second one is more intersting but here's my take on it: we assume correctly that .999999999...is a nonterminating decimal with infinite 9s. the problem comes from multiplying that decimal by 10. thinking it through, .9 x10 = 9.0 .99x10 =9.90 etc... when u multiply a decimal by 10 think of it as moving the first zero to the left of the decimal to behind the last non-zero numeral to the right of the decimal. in other words 10x.9999... will actually = 9.9999.....99990. this zero is disputable but if u think of it as being there, then 10x-x will actually equal 8.99999....99999 because 10 x has a zero in the very last place where x has a 9 resulting in infinite carrying over of 10s (reference to how your teacher taught u to subtract 7 from 16 in the first grade). at this point x doesn’t equal 1 because if u divide this number by 9 you simply get .9999999999 and thus this paradox is also solved.

As for the next idea that 10/3= 3.33333…. and multiplying that by 3= 9.99999 by arithmetic and 10 by algebra is actually just an exploitation of the fact that math is usually done in base 10. basically the less numerals in a system the less numbers are expressable through decimal. Think about the same situation in binary code. 1010/11 the closest u can get to this is 11.111111… which when converted back to base 10 = 3+ the summation of 1/(2^n) when n from 1 to infinity which is less accurate than 3.33333 (it will have a mixture of numbers not just 3. but in base 12 10 would be a 1 digit number which when divided by 3 would simply yield 3.4 (because instead of .1 being 1/10 it will be 1/12 and 3.3333 would still be a non-terminating decimal but written differently.

but this is just my take on it.
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