Algebraic demonstrations that 1=2
5 Comments Published August 24th, 2008 in Life, universe and everything Tags: .Let a=b, with a≠0
a=b a2=ab a2+a2=a2+ab 2a2-2ab=a2+ab-2ab 2a2-2ab=a2-ab 2(a2-ab)=1(a2-ab) 2=1
...or...
a=b ab=bb -ab=-bb a2-ab=a2-b2 a(a-b)=(a+b)(a-b) a=a+b a=a+a 1=2
Another nice one, with numers instead of letters
| 16=16 | Reflexive property of "=" |
| 4+12=16 | From definition of "+" |
| 4-12=16-24 | Subtract 24 from both sides |
| 4-12+9=16-24+9 | Add 9 to both sides, completing squares |
| (2-3)2=(4-3)2 | Factor, rewriting as squares |
| 2-3=4-3 | Sqrt of each side |
| 2=4 | Add 3 to each side |
| 1=2 | Divide by 2 |
This means basically that everything you know is wrong :)
5 Responses to “Algebraic demonstrations that 1=2”
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Nice try!
(Hint: division per 0 is not allowed)
i think u proved 0=0
Great work guys, you got it. But… Cuius rei demonstrationem mirabilem sane detexi hanc marginis exiguitas non caperet. :)
Refactored version: 2(0)=1(0) thus 2=1.
Two hints :)
1) division per 0 is not allowed (how is say later)
2) When you rewriting as squares you get a modulus values, ie
…
(2-3)^2=(4-3)^2 Factor, rewriting as squares
|2-3|=|4-3|
and only one way to remove modulus this:
1=1
:)