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=16Reflexive property of "="
4+12=16From definition of "+"
4-12=16-24Subtract 24 from both sides
4-12+9=16-24+9 Add 9 to both sides, completing squares
(2-3)2=(4-3)2Factor, rewriting as squares
2-3=4-3Sqrt of each side
2=4Add 3 to each side
1=2Divide by 2

This means basically that everything you know is wrong :)


5 Responses to “Algebraic demonstrations that 1=2”  

  1. 1 Filippo

    Nice try!
    (Hint: division per 0 is not allowed)

  2. 2 tim

    i think u proved 0=0

  3. 3 Luigi

    Great work guys, you got it. But… Cuius rei demonstrationem mirabilem sane detexi hanc marginis exiguitas non caperet. :)

  4. 4 arley

    Refactored version: 2(0)=1(0) thus 2=1.

  5. 5 Ilzar

    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
    :)

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