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let S=1+2+4+8+16+32.....-----(1)
i.e. S-1=2+4+8+16+.....
i.e. S-1=2*(1+2+4+8+16+....)
from (1)
S-1=2S
therefore....
S=-1 ??????????????????????????????????????????

2006-08-11 19:07:37 · 13 answers · asked by navin 1 in Science & Mathematics Mathematics

13 answers

The flaw is that the original series diverges.
You cannot perform the subtraction as noted.
Consider the even more improbable:
X= 1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + ...
You can show X equals anything, lets say N>2.
First, add up the odd terms until the sum is just greater than N.
Then subtract 1/2.
Then add enough of the rest of the odd terms until the sum is greater than N again,
then subtract 1/4
Continuing in this fashion, X should be equal to N in the end!
(p.s., I'm really tired of people who say that 2 times infinity is infinity. INFINITY IS NOT A NUMBER FOLKS! You dont multiply, add, subtract, or divide with it!)

2006-08-11 19:34:51 · answer #1 · answered by Scott R 6 · 0 3

There are two ways to think about this problem.
It depends on weather S is limited or not. For example, if S=1+2+4+8+16+32+64, then S-1=2+4+8+16+32+64. It means that S=127 and S-1=126. Therefore, S-1=2S is wrong. On the other hand, if S is unlimited, S=1+2+4+8+16+32+......., it means that S is infinite and you know that infinite is not a number. It is just something which is more than every other number. It is just one imagination or an idea which helps to solve some problems in mathematics. Thus, you can't get result that "S-1=2S". You can just get this result that S is infinite.

2006-08-12 02:49:36 · answer #2 · answered by Arash 3 · 0 2

S=1+2+4+8+...
S-1=2+4+8+... or S-1=2(1+2+4...)
so S-1=2S
Subtract S from the both sides and you are left with
-1=S
There is no flaw.

2006-08-12 07:09:16 · answer #3 · answered by disgruntledpostal 3 · 0 1

you can only use this "trick" if the original series converges.

for the finite sum:

S = ∑ 2^n from n=0 to n=a

S = 2^(a+1) -1

note: finite sums ALWAYS converge (i.e. you stop counting)

to answer your question you need to write:

S = lim a→∞∑ 2^n from n=0 to n=a

S = lim a→∞ {2^(a+1) -1}

S = 2lim a→∞ {2^a} - 1

so we see that lim a→∞ {2^a} approaches ∞ so the infinite series diverges and you cannot do that trick.

2006-08-12 19:00:35 · answer #4 · answered by cp_exit_105 4 · 0 0

The first thing to notice is that S = infinity.

I must point out that it makes perfect sense to say things like:
*** infinity + 1 = infinity
*** 2 * infinity = infinity
Since these have a single solution on the extended real number line. This can easily be proved using real analysis.

It is common for people to be afraid of infinity and say things like "infinity is a concept not a number."

It does not make sense is to say the following:
*** infinity - infinity
*** infinity / infinity
Since these have multiple solutions. Notice that in your "proof" you had the line S-1=2S, and subtracted S = infinity from both sides to give you your solution. This is the 'infinity - infinity' case - this is the flaw.

2006-08-12 07:03:39 · answer #5 · answered by Anonymous · 0 1

S is a divergent infinite series, so S = 2S = 127 S = ...

In other words, infinity times 2 is still infinity.

2006-08-12 02:27:38 · answer #6 · answered by bpiguy 7 · 0 1

If S-1=2S,
S DOES NOT NEED TO BE -1
It can be infinity, as:
infinity-1=infinity,
2*infinity=infinity
infinity=infinity!!!

2006-08-12 02:22:23 · answer #7 · answered by Anonymous · 1 1

You are not allowed to rearrange the additions in the infinite sum S, ( step 2 )
Because the sum doesnot converge absolute.

2006-08-12 02:11:55 · answer #8 · answered by gjmb1960 7 · 0 2

The flaw is I don't know how to do this math. But at once point in time I did. I think.

2006-08-12 02:09:29 · answer #9 · answered by Scarlet 3 · 0 2

The flaw is attempting to use 'S' to represent both a series and a number.

Nothing is better than freedom.
Prison is better than nothing.
Therefore, prison is better than freedom.

2006-08-12 02:16:45 · answer #10 · answered by Frank N 7 · 2 2

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