
Please read the following:
Let x = 2 + 4 + 8 + 16 + 32 + ...
So
x = 2 + 4 + 8 + 16 + 32 + ...
x = 2 ( 1 + 2 + 4 + 8 + 16 + ... )
x = 2 ( 1 + x )
x = 2 + 2x
x  2x = 2
x = 2
x = 2
Therefore
2 + 4 + 8 + 16 + 32 + ... = 2
Yay. I'm bored. Sometimes this gets into the weirdest discussions. Probably even abstract as well.
10 Frih$ for good reason as to what I did wrong. Sometimes this is quick, that'll make me sad :[
WOWWWWWWWWW Really interesting problem. Let me consult a Phd
(Working hard to resolve this prolem)
Well i have found the actual problem (within 1 minut) that the number assigned to X an infinite value.
So, if you will try to find a solution of a problem that contains 1/0 or other infinite values, the answer will always be un predictable.
Ask more questions in this regard or tell me if you are sure that i am wrong.
How did you get from here:
Quote:  x = 2 ( 1 + 2 + 4 + 8 + 16 + ... ) 
to here:
I believe by using the ellipse you mean that the sequence continues. If however we stop it at 32 the total is 62, therefore x = 62. Thus 2(1 + x) = 126 and not 62. This proves that this stage is incorrect for any number of numbers added on in this sequence. Am I right?
I think that's all wrong with this.
ninjakannon wrote:  How did you get from here:
Quote:  x = 2 ( 1 + 2 + 4 + 8 + 16 + ... ) 
to here:
I believe by using the ellipse you mean that the sequence continues. If however we stop it at 32 the total is 62, therefore x = 62. Thus 2(1 + x) = 126 and not 62. This proves that this stage is incorrect for any number of numbers added on in this sequence. Am I right?
I think that's all wrong with this. 
no you are absolutely wrong, the question is true only for a continuing expression like in the question. If you try to stop it at any stage, the next equalities will not hold true.
so the only way is let the value of X continue and add infinite times.
imagefree wrote:  no you are absolutely wrong, the question is true only for a continuing expression like in the question. If you try to stop it at any stage, the next equalities will not hold true. 
Well then the only problem with this is that x = infinity and infinity is not a number, or at least not one that follows the same rules in this situation.
imagefree wrote:  Well i have found the actual problem (within 1 minut) that the number assigned to X an infinite value.
So, if you will try to find a solution of a problem that contains 1/0 or other infinite values, the answer will always be un predictable.
Ask more questions in this regard or tell me if you are sure that i am wrong. 
lol yeah I would think you would be the one to know from other posts xP
but there's one specific reason as to why the infinite values do not work in this case.
Basically: Why does working with infinity not work in this case?
But since you got the correct reason I'll give you the 10 frih$
I love this one because it has one flaw that many people don't understand: infinity. I actually stumped my calculus teacher with this. My physics teacher got it quickly since, he's a physics teacher so he understands infinity better than a precalc teacher. If I ever decide to teach as a career I'll probably teach Algebra 2 and show this to the students because it's a great thing to start understanding when basic laws of math start to break down. There are always exceptions.
ninjakannon wrote:  imagefree wrote:  no you are absolutely wrong, the question is true only for a continuing expression like in the question. If you try to stop it at any stage, the next equalities will not hold true. 
Well then the only problem with this is that x = infinity and infinity is not a number, or at least not one that follows the same rules in this situation. 
infinity is a funny thing T_T imagefree had this thought first though.
DjinniFire wrote:  lol yeah I would think you would be the one to know from other posts xP 
Other Posts???
imagefree wrote:  DjinniFire wrote:  lol yeah I would think you would be the one to know from other posts xP 
Other Posts??? 
sorry i meant the other mathy contests
Infinity as a concept defines a "number" of a magnitude that we cannot comprehend.
So, in this case, when you take
x = 2 ( 1 + x )
On the RHS, you have 2 (1 + infinity)
Let's multiply them:
= 2 + 2(infinity)
But any number multiplied by infinity itself is infinity
Therefore, = 2 + infinity.
Addition or subtraction of any finite value from infinity once again gives infinity
therefore, = infinity
THis implies x = infinity, which is true.
tidruG wrote:  Infinity as a concept defines a "number" of a magnitude that we cannot comprehend.
So, in this case, when you take
x = 2 ( 1 + x )
On the RHS, you have 2 (1 + infinity)
Let's multiply them:
= 2 + 2(infinity)
But any number multiplied by infinity itself is infinity
Therefore, = 2 + infinity.
Addition or subtraction of any finite value from infinity once again gives infinity
therefore, = infinity
THis implies x = infinity, which is true. 
the main problem with infinity and wrong use of 0 is that you can take your solution to any direction.
You can derive more than 1 possible solutions.
tidruG wrote:  Infinity as a concept defines a "number" of a magnitude that we cannot comprehend.
So, in this case, when you take
x = 2 ( 1 + x )
On the RHS, you have 2 (1 + infinity)
Let's multiply them:
= 2 + 2(infinity)
But any number multiplied by infinity itself is infinity
Therefore, = 2 + infinity.
Addition or subtraction of any finite value from infinity once again gives infinity
therefore, = infinity
THis implies x = infinity, which is true. 
Ahh but all your infinities are of different magnitudes.
2+2(infinity) is technically going to be bigger than the original infinity I set x to be. I know that sounds strange but in the world of infinity, one infinity may be greater than another, that is why technically infinity  infinity is not always 0.
DjinniFire wrote:  Ahh but all your infinities are of different magnitudes.
2+2(infinity) is technically going to be bigger than the original infinity I set x to be. I know that sounds strange but in the world of infinity, one infinity may be greater than another, that is why technically infinity  infinity is not always 0. 
An infinite number is one which is not finite, a finite number has a limited number of digits, therefore an infinite number has no end to its digits and could never be written down. Thus multiplying infinity by 2 would change nothing because the number would still go on forever. While it may seem that the result of this equation should be larger than it was before it was multiplied it is in fact true that the number cannot get any larger.
That's what I would have thought, at any rate. However, I'm no mathematician (or not yet, at any rate ).
Quote:  Ahh but all your infinities are of different magnitudes. 
So you're basically assigning a value to infinity...
Please tell me what that value is :)
... see, whether you add, subtract or multiply any finite, positive number to/from infinity, you still have infinity. It has no magnitude. It's just a concept.
x = 2 + 4 + 8 + 16 + 32 + ... = infinity
x = 2 ( 1 + 2 + 4 + 8 + 16 + ... )
x = 2( 1 + [(x/2)1])
x = 2 + 2(x/2) 2
x = 2 + x  2
x = x
Given a finite problem, it is much easier to see how the above is driven.
x = 2 + 4 + 8 + 16 + 32 = 62
x = 2 ( 1 + 2 + 4 + 8 + 16)
x = 2 ( 1 + 2 + 4 + 8 + 16 +32 32)
x = 2 ( 1 + [(2 + 4 + 8 + 16 +32) 32])
x = 2 ( 1 + [(x/2)1] )
x = 2 ( 1 + [(62/2)1] )
x = 2 ( 1 + [311] )
x = 2 ( 31 )
x = 62
Since you used x instead of [(x/2)1]
x=(x/2)1
2x=x2
x=2
Which explains the x=2 you are getting.
I have Asperger's.... I can kind of "see" what you are talking about in my head, but I cannot do this kind of math... I LOVE math though...it is kind of like life isn't it? Things look as if they should always have a definite answer, but there are always "exceptions"...as was stated in this thread of discussion.
I am working on an equation, but it is an equation on boundaries. So far I have how religion, government, family, culture, and society affect us in the way that we give ourselves boundaries. The thinking people do "inside the box"...and how to get "outside the box"... I don't even know if this will work...I have been working on it for months... LOL It is something fun to do...and challenging, for me anyway, due to AS.
If you like this kinds of problems you should checkout taylor expansions. Once you learn taylor expansions, you'll look at calculus and alot of math in a whole new way.
It's easy. But to bored to resolve it.
DizzyDesign wrote:  It's easy. But to bored to resolve it. 
Thanks so much for digging this up, for THAT...
