I really must know the answer to this important mathematical question

Discussion in 'The Spam Zone' started by What?, Aug 8, 2012.

  1. Amber PLUR

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    To explain my earlier answer of 10...

    Binary. The expanded form would be 00000001 + 00000001 = 00000010.
     
  2. Mixt The dude that does the thing

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    I did it first, under the same logic. :P /childish competing
     
  3. Makaze Some kind of mercenary

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    I thought so. I didn't want to look it up, though, so thanks for explaining.
     
  4. Judge Sunrose Destiny Islands Resident

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    Remember how people would only accept one of these?

    Well, I believe there will be a time in which mine will be one of the dominant arithmetic axioms, right next to Peano's.
     
  5. rikusorakairiown Contributor

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    0/0 does not equal 1.

    also,

    1+1 = 1.999...

    *the more you know.
     
  6. Te Deum Hollow Bastion Committee

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  7. Mixt The dude that does the thing

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    It could. The problem is that it could be anything else too. 0, 184, -16, 39 + 42i, anything. In this case it isn't hard to see it is actually 1/2. But it is best to avoid the indeterminate stuff in algebra, it is just bad.
     
  8. Nate_River Hollow Bastion Committee

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    [video=youtube;RMjcqT-_ZHI]http://www.youtube.com/watch?v=RMjcqT-_ZHI&feature=related[/video]
     
  9. Makaze Some kind of mercenary

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    I proved this incorrect. Infinite remainders are impossible and can only exist as a defiance of logic. They are unstable numbers because they continue on infinitely. Let's say that you have 1.9999999999999999999999. Well, if it is infinite, then it isn't actually 1.9999999999999999999999, it is 1.99999999999999999999999. But then it still isn't 1.99999999999999999999999. You have to add another. 1.999999999999999999999999. And another, 1.9999999999999999999999999. And on, and on forever. It is not an integer and as such can never be used in an equation. Similarly:

    1 / ∞ != 0

    It will grow very close to zero, but it will never reach zero, and 1.9999... will grow very close to two, but never reach two.

    Fletcher's paradox in math.
     
  10. Mixt The dude that does the thing

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    I can prove to you that 1.9999... is 2 if you will agree that the limit of (1/10)^x as x approaches infinity is 0. That is the one part of the proof that I could see you debating.
     
  11. Makaze Some kind of mercenary

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    I do not agree. It will never reach zero. It will just get as close to zero as it can, infinitely. The number keeps getting smaller and smaller, and there is no cap to how small it can go. The .111111... will repeat forever, adding one more 1 each time. It will never revert to zero.
     
  12. Mixt The dude that does the thing

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    It does not need to reach zero at any actual point for the limit to be 0, but other points must lead to that hypothetical "x = infinity" point being zero.
     
  13. Makaze Some kind of mercenary

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    However, that logic only works in conjecture and not in practice. If you actually tried to reach zero with any number of trailing 1's, you would go on forever until you died. Therefore it is not provable.
     
  14. Mixt The dude that does the thing

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    How about this...
    Lets define infinity as a number that is greater than any rational number you could imagine up.
    Let's then examine a given expression of 10^x
    For any rational number you give me I could make that expression greater by providing a larger and larger value of x
    If I were to set x as the defined infinity, 10^x would then exceed any rational number.
    Therefor 10^x would be infinity.
    Therefor as x approaches infinity 10^x also approaches infinity.
    Do you agree with that logic?
     
  15. Makaze Some kind of mercenary

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    No, because infinity is not a number itself. There is no point at which a number becomes too large to notate. Infinity simply is not a specific number. It is an ever increasing value. If it were to stop increasing, it would be finite.
     
  16. Mixt The dude that does the thing

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    You say that it isn't a number and then you proceed to treat it per my definition (an abstractly large number that can't be reached).

    You are correct that if 10^x settled it would be finite, but without combining that statement with proof that 10^x settles it does not prove 10^x to be a finite expression. In fact line 3 of my above proof is to show that 10^x does not ever settle
     
  17. Makaze Some kind of mercenary

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    You cannot use a number without a finite value in an equation. It doesn't work.

    The problem with your third sentence is that in order to use it in an equation, you would have to use each finite state as its own value. For instance, x(subtext a) = 0.11111 (6th state), and x(subtext b) = 0.111111 (7th state). You would have to use each of these as a separate variable.

    For the same reason you cannot do this:

    x + 1 = 5

    Where:

    x = 2, 3, or 4

    Instead, you can do:

    x + 1 <= 5

    But not with an equal sign.
     
  18. Roaringflames Merlin's Housekeeper

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    The answer is obviously 2

    Did you not learn preschool math?
     
  19. Makaze Some kind of mercenary

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    [​IMG]
     
  20. Roaringflames Merlin's Housekeeper

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