Math Contest #19 Results and Solution

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Solution

The problem of this contest was to solve an infinite sum:
Untitled.png
There are in general 3 different intuitive ways to find a solution to this:

  1. 1-1+1-1+1-1+… = (1-1)+(1-1)+(1-1)+… = 0+0+0+… = 0
  2. 1-1+1-1+1-1+… = 1 + (-1+1)+(-1+1)+(-1+1)+… = 1+0+0+0+… = 1
  3. 1-1+1-1+1-1+… = x
    1-(1-1+1-1+1-1+…) = 1-x
    1-1+1-1+1-1+… = 1-x
    x = 1-x
    2x = 1x = ½

But there are also infinitely many more ways to find a solution that you can explain. you could even reorder the sum(by taking some 1's from infinity and placing them between them at some point sooner in the sequence. This is possible, because before and after that operation the total number of 1's and -1's is still the same as before(∞)) to:
1+1-1+1+1-1+1+1-1+… = (1+1-1)+(1+1-1)+… = 1+1+1+… = ∞

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List of participants with their entries:

Namesolutions foundcomment
@sparkesy430 or 1
@crokkon0 or ½ or 1
@golddeck0 or 1
@rxhector-∞That's not really intuitive, but still you can get there as shown above for +∞.
@fullcoverbetting0
@tonimontana½But how can the sum of integers be no integer :P

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Winner draw:

Screenshot from 2019-10-24 19-40-18.png

Congratulations @rxhector you won 1 SBI!

Of course everyone also got 10 STEM for participating(check your steem engine)!
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The next contest starts in 2 days. Don't miss it!



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6 comments
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Hi @quantumdeveloper, a small upvote and a tip from the ADDAX trading game! Round 2 has closed - please sell your tokens asap if you still hold some.
$trendotoken

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(Edited)

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That's what one would think would happen, but the sum here is infinite. So you can repeat the pattern that of (1+1-1) for ever without having a zone of (1-1-1).
It may fell like some -1's are lost during this process, but if you would(and could) count them you would still get the same number of 1's and -1's: infinitely many!

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