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Why is ∞+1=∞? 
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#1
Nov1812, 02:20 PM

P: 20

The proof that 1/2+1/4+1/8+...=1 goes like this:
X=1/2+1/4+1/8+... 2X=2/2+2/4+1/8+... 2X=1+1/2+1/4+... 2XX=1+(1/21/2)+(1/41/4)+... X=1 The assumption that goes with this is that we can pair up the first term of X with the second term of 2X and so on without having the smallest term of X leftover. This is because there are infinitely many terms and ∞+1=∞. The result seems intuitively satisfying, the term that is "ignored" is after all infinitely small. But lets look at another example where we "ignore" the largest term: X=1+2+4+... 2X=2+4+8+... 2XX=1+(22)+(44)+... X=1 So here we see that we get a negative number from adding up only positive numbers. Also we get a finite number from something that goes towards infinity. This is intuitively not very satisfying in my opinion and ought to be a proof that ∞+1=∞ doesn't work. After all X is also an infinity ∞, so: ∞+1=∞ 1+1=1 0=1 Also since: ∞=∞+1 We can add as many as we like ∞=∞+1+1+1+... ∞=∞+∞ Again we can keep adding ∞=∞+∞+∞+∞+.... ∞=∞*∞ ∞/∞=∞ This is not only not intuitive, it is completely useless, since anything "tainted" by this type of infinite, will be impossible to calculate anything with. Also lets try the pairing up method for 1...000 and 0 1...000=...000=0 Every zero in the large number can be paired up with every zero in 000... So we have both ∞=0 and ∞=1 again 0=1 If on the other hand we say: ∞≠∞+1 We get an infinite amount of infinities that are not equal. Further if we write the infinite in this way: ∞=X+Y+Z+... ∞/∞=1 because there will always be an equal amount of terms (infinite or not). That also means that X=2∞ X/∞=2 So one infinite divided by another can be any finite. Also ∞*∞=X X/∞=∞ That means that dividing infinities can also get you an infinite. Lets try to assume that an infinite divided by a finite X could yield another finite Y: ∞/X=Y ∞=YX Since the product of two finite numbers can never be infinite, we know this is incorrect. Also we can also say something about the infinities divisibility, like 2+2+2+... is divisible by 2. Anyway now we get different results for a lot of calculations: 1/2+1/4+1/8+...=1+1/∞ (where ∞ is the number of terms) X=1+2+4+... = 2^(∞1)1 1=0.999...+0.1^∞ (∞ being the number of nines) If you have an infinite hotel fully booked, you cannot fit any more people in. This makes much more sense to me. What do you think? 


#2
Nov1812, 03:39 PM

Sci Advisor
P: 6,056

The method for the first case works because the series converges. All your other examples violate this, so the manipulations are invalid.



#3
Nov1812, 03:45 PM

P: 894

I think the mistake you are making is to treat infinity as a number. I believe that infinity is not a number and thus can not be added or divided by itself or by another number. Also there is no such thing as a fully book hotel with an infinity of rooms.



#4
Nov1812, 04:37 PM

P: 20

Why is ∞+1=∞?



#5
Nov1812, 04:43 PM

P: 894




#6
Nov1912, 07:27 AM

P: 150




#7
Nov2012, 06:32 AM

P: 20

X=0.999... 10X=9.999... 10XX=9 X=1 The trick is here 10X should have the least significant digit 10 times as large as X. The reasoning goes that you can pair these up all the way to infinity and there is therefore no least significant digit. This is the same logic I used in my example, but because I disregard the most significant digit by doing this, I get a very different number, where as when we do it for the smallest significant number, we get an infinitely small difference. But I would rather argue that I just think this other way of looking at infinity gives us more options to calculate with it and seems to eliminate a lot of "paradoxes". I do not think my way of calculating with infinities is inherently more correct than the standard way, for the same reason that I do not think base 10 is more correct than base 3. I just think it is more useful. I know that Feynman had to use some non standard infinity calculation methods to derive his equations of quantum electrodynamics, I would love to know what it was. 


#8
Nov2012, 06:53 AM

P: 150




#9
Nov2012, 07:01 AM

P: 3,014

For a strictly valid infinite pairing method, use the partial sum:
[tex] S_n = \frac{1}{2} + \frac{1}{2^2} + \ldots + \frac{1}{2^n} [/tex] Now, if you multiply by 2, and subtract, you will have: [tex] 2 S_n  S_n = \left( 1 + \frac{1}{2} + \ldots + \frac{1}{2^{n  1}}\right)  \left( \frac{1}{2} + \ldots + \frac{1}{2^n} \right) [/tex] [tex] S_n = 1  \frac{1}{2^n} [/tex] There are 2 terms that do not get canceled in this case. But, if we take the limit [itex]n \rightarrow \infty[/itex], you prove that the partial sums have a limit (the infinite series is convergent), and that this limit is 1, because [itex]\lim_{n \rightarrow \infty}{1/2^n} = 0[/itex]. 


#10
Nov2012, 09:09 AM

P: 928

We call it a sum and for most purposes it behaves like a sum but technically speaking it is the limit of a sequence of partial sums. Ordinary addition is associative and commutative. And the distributive law applies as well. Using the principle of mathematical induction you can prove that for any _finite_ list of numbers, associativity, commutativity and the distributive law apply to the entire list. Proof by induction does not work when the number of terms is infinite  or when the thing you're looking at is not technically even a sum. It turns out that when a series is "absolutely convergent" that these familiar properties do turn out to apply. You can work through the epsilons and deltas and prove it. But when a series is not "absolutely convergent" those properties do not apply. For instance, rearranging the terms in the series 1/2  1/3 + 1/4  1/5 + 1/6  ... can produce an "infinite sum" (limit of the sequence of partial sums) that comes to any figure that you like. Briefly: An infinite sum is fixed under a finite rearrangement of terms. An infinite sum is fixed under any rearrangement if the underlying series is absolutely convergent. But an infinite sum is not neccessarily fixed under an infinite rearrangement when the series is not absolutely convergent. 


#11
Nov2012, 09:20 AM

Sci Advisor
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P: 12,016

Pay particular attention to what Dickfore tells you, JohnLuck.
Your "counterexample" with the sum of the powers of 2 evaporates when doing the perfectly legitimate operation, gaining: S_n=2^(n+1)1 Do you see that you cannot ignore the first term here? 


#12
Nov2012, 10:34 AM

Engineering
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P: 7,101

The arguments in your OP don't "prove" anything about the series. At most, they show that if the limit of the sum exists, then it must have a certain value.
Buit you also have to show that the limit does exist. In the case of 1+2+4+8... it does not. Proving that "If elephants laid eggs, they would be green with pink spots" doesn't prove that "Elephant eggs are green with pink spots", because elephants don't lay eggs. 


#13
Nov2012, 11:04 AM

Mentor
P: 18,209

I'm going to lock this thread. The answers given are all excellent and to the point. But I'm afraid JohnLuck doesn't have the knowledge to really understand them. I understand that things indeed seem fishy and illdefined, but they are really not. I would suggest you make an effort to learn calculus and specifically sequences and series.
We can't explain sequences and series to you as it would require too much information. But I'm sure that any good calculus book will be understandable enough for you. If you really want to understand what is going on here, then studying calculus is indispensable. Once you studied series a bit, we can continue the discussion. 


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