# Infinite(ly) fun

Discussion in 'Math' started by Mark44, Apr 1, 2008.

1. ### Mark44 Thread Starter Well-Known Member

Nov 26, 2007
626
1
Let S = 1 + 2 + 4 + 8 + ...
Then 2S = 2 + 4 + 8 + 16 + ...
And 2S + 1 = 1 + 2 + 4 + 8 + ... = S
Now, since 2S + 1 = S, it follows that S = -1
In other words, 1 + 2 + 4 + 8 + ... = -1

2. ### Dave Retired Moderator

Nov 17, 2003
6,960
145
The weird world of infinities.

It is the same as:

0.99999.... = 1

Consider the case to be true:

0.99999.... = 1 (Eqtn 1)

Multiply by 10:

9.9999... = 10 (Eqtn 2)

(Eqtn 2) - (Eqtn 1)

9 = 9

Yes they are equal!

Alternatively:

1/3 = 0.33333...

2/3 = 0.66666...

So:

0.33333... + 0.66666... = 0.99999...

And:

1/3 + 2.3 = 3/3 = 1

Yup, 0.99999... = 1

Dave

3. ### thingmaker3 Retired Moderator

May 16, 2005
5,072
6
Except 2S - 1 also = S. Fun, isn't it?

How about taking a number increasing without bound and raising it to the power of itself?

4. ### Mark44 Thread Starter Well-Known Member

Nov 26, 2007
626
1
Of course, that result isn't quite as astonishing as 1 + 2 + 4 + ... = -1

5. ### cumesoftware Senior Member

Apr 27, 2007
1,330
11
Basically, you cannot assert that 2S + 1 = S, since S is a different sum than 2S + 1. For example, if we admit that S is a finite sum of 4 members, 2S + 1 would equal to a sum of 5 members and not equal to S.

S(n) = 1 + 2 + 4 + 8 + ... + 2^n

Therefore:
2S(n) + 1 = 2 x (1 + 2 + 4 + 8 + ... + 2^n) + 1 = 1 + 2 + 4 + 8 + 16 + ... + 2 x 2^n

Thus:
1 + 2 + 4 + 8 + ... + 2^n is not equal to 1 + 2 + 4 + 8 + 16 + ... + 2 x 2^n, for every value of n.

But it is a fun math problem, though.

6. ### Mark44 Thread Starter Well-Known Member

Nov 26, 2007
626
1
In fact, it is entirely reasonable and legitimate to assert that 2S + 1 and S are equal. They have exactly the same cardinality.

I don't admit this at all. S is not a finite sum, and this is alluded to by the ellipsis (...), meaning "continuing in the same pattern."

I'll admit that S(n) = 1 + 2 + 4 + 8 + ... + 2^n + ...
I.e., that this is an infinite sum. Here the ellipsis indicates that the next term in the sum would be 2^(n+1), followed by 2^(n+2), and so on.
You're omitting lots and lots of terms here.
Mark

7. ### Dave Retired Moderator

Nov 17, 2003
6,960
145
Mark, are you not astounded that two completely different numbers are in fact the same?!

Dave

8. ### Mark44 Thread Starter Well-Known Member

Nov 26, 2007
626
1
Always!

But if you're referring to 1 and .9999999..., then I'm not astounded at all, since they aren't different.

Granted, 1 and .9 are different, as are 1 and .99. And 1 and .999...9 are also different, as long as there is that final 9 digit.

1 and .9999999..., or more formally $\sum$9/(10^k), with k running from 1 to $\infty$, are the same.

Mark

9. ### thingmaker3 Retired Moderator

May 16, 2005
5,072
6
But 2S, 2S - 1, and 2S + 1 are also not different. Nor are 2S * S.

So I ask again, what about S^S? Or, more properly phrased, (x$\rightarrow\infty$)^(x$\rightarrow\infty$)

10. ### Mark44 Thread Starter Well-Known Member

Nov 26, 2007
626
1
That's a whole 'nother story--much larger.

To get back to the original conundrum, when you're dealing with infinity, the ordinary rules of arithmetic don't all apply. Where the calculations at the beginning of this thread break down is right after the equation
2S + 1 = S​
in subtracting S from each side.
Here, both 2S + 1 and S are infinitely large, so subtracting an infinitely large quantity from another infinitely large quantity yields an indeterminate result. The quantity 2S is not twice as large as S, so 2S - S can't be calculated.

Somewhat related to this is the theoretic concept of sets with infinitely many members, such as {1, 2, 3, 4, ...}, the set of positive integers. One might intuitively think that this set, {101, 102, 103, 104, ...} is a smaller set, given that it is missing the numbers 1 through 100, but in fact both sets have exactly the same number of members (and therefore have the same cardinality). While you can't count the members of either set directly, you can do the next best thing, which is to establish a one-to-one pairing between the two sets. The most obvious pairing is 1 <--> 101, 2 <--> 102, and so on. No matter which number you choose in one set, it's very easy to find the corresponding number in the other set.

Clearly this is some esoteric stuff, and very much in the realm of pure mathematics. I've always found it very fascinating and thought-provoking. I hope you do, too.

More later...
Mark

11. ### flubbo Member

Apr 21, 2008
25
0
I saw a similar math exercise that "proved" 1 + 1 = 1, showing the problem with division by zero

let A = B = 1

then

A^2 = AB (Multiply both sides by A)

A^2 - B^2 = AB - B^2 (Subtract B^2 from both sides)

(A + B)(A - B) = B(A - B) (Factor both sides)

Divide out the common term (A - B), and you're left with

(A + B) = B

-or-
(1 + 1) = 1

Basically, Mark44 is saying that "Infinity" equals "Infinity", which makes perfect sense to me, because the limit of any of the equations mentioned is infinity.

12. ### kinkyknob New Member

Apr 25, 2008
5
0
omg ROFLMAO 1+1=2

13. ### Mark44 Thread Starter Well-Known Member

Nov 26, 2007
626
1
Everyone knows that 1 + 1 = 10

14. ### flubbo Member

Apr 21, 2008
25
0
And 10 + 10 = 100

Of course, don't forget that 6 x 9 = 42

15. ### Caveman Active Member

Apr 15, 2008
471
0
I think your calculator's circuitry has been contaminated. Perhaps by hairdressers...

16. ### Dave Retired Moderator

Nov 17, 2003
6,960
145
Is this the same as "There are only 10 types of people in the world: Those who understand binary, and those who don't"

Dave

17. ### flubbo Member

Apr 21, 2008
25
0
@Caveman: Nope...the calculator works fine as long as you don't mind converting from base 13 all the time.