Infinite(ly) fun

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

  1. Mark44

    Thread Starter Well-Known Member

    Nov 26, 2007
    626
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    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 :confused:
     
  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
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    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
    10
    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.

    Lets admit that:
    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
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    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 \sum9/(10^k), with k running from 1 to \infty, are the same.

    Mark
     
  9. thingmaker3

    Retired Moderator

    May 16, 2005
    5,072
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    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" :D

    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. :)
     
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