1 + 1 = 0: The Proof!

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Uploaded by on Jul 21, 2007

I've seen a few so-called "Maths tricks" on YouTube, most involving division by zero or basic algebra, so here's my contribution. You do need to know a bit about square roots and the complex number i, though. Of course, there is a flaw in the proof - the point is for you to spot it. The music is Random by Gary Numan (1979). I didn't know how to get the "√" symbol when I did this so "sqrt" means "√" throughout.

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  • 1 + 1 = 0 if you use condom right :D

  • sqrt(1)=sqrt[(-1)(-1)] is true

    sqrt[(-1)(-1)]=sqrt(-1) x sqrt(-1) is false

    The last rule only applies to non-negative numbers, where -1 is clearly negative.

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All Comments (8,587)

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  • Jesus christ I come here and I see big equations trying to prove this guy wrong. This is how I do it. 1 x One wholeNumber times One more Whole Number Equals two Whole Numbers.

    1 = One Whole Number.

    So, Add another Whole One to it. Viola you have two Whole Numbers.

    Math 101

  • @PykohYT I have seen the same thing, 1+i^2= 2

    i^2=1

    i= sqrt(-1).

    Please, delete your video LOL.

  • I am baffled.(period)

  • 1:54

    1+i=1i+i^2

    If i has a value other than 1, this is wrong.

  • sqrt[a] * sqrt[b] = sqrt[ab] is always true if and only if a and b are both positive. That's the problem with this equation.

  • @TheHarboe Actually, it applies for any positive number a such that a = b*c where b,c are both positive AND for any negative number a such that a = b*c where b,c are both negative. :)

  • WELL, WHEN I TRY TO SUM UP 1+1, I GET FREAKIN'

    404 Forbidden

  • I have one apple. I add another apple. I have two apples, not zero.

    I win.

  • @xouris123 you are absolutly right, but the definition still says that you can exchange the -1 in the sqrt with i^2, not that you can define i=sqrt(-1)

    Of course it started in another way and practically, when calculating with it, it doesn`t matter, but it is still not defined as sqrt(-1)^^

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