- last post: 01.01.0001 12:00 AM PDT
i know this is offtopic, but the reason i post it here is because i doubt anyone looks in the flood... and i am positive the most mathematically minded people are in the pc forum, so dont flame me, i am posting this interesting tidbit among those who will appreciate it.
that said, i have proof that 2=1
check it out
given a=b, you can conclude that a^2=ab, correct? (a^2 is a squared, cause there is no way to write in superscript in this forum)
next you can conclude that a^2-b^2=ab-b^2.
from here you can use reverse distributed property of multiplication, and find that a^2-b^2=b(a-b).
then, you factor the left side of the equation, and get (a+b)(a-b)=b(a-b).
after this, you divide both sides by (a-b), and get (a+b)=b
because a=b, you can substitute b in for a, and you get b+b=b
you add the two b's together and get 2b=b
then you divide both sides by b, and get 2=1
if you write this as a proof it would look something like this:
a=b |given
a^2=ab |multiply both sides by a
a^2-b^2=ab-b^2 |subtract b^2 from both sides
a^2-b^2=b(a-b) |reverse distributive property
(a+b)(a-b)=b(a-b) |factor
a+b=b |cancellation
b+b=b |substitution
2b=b |addition
2=1 |cancellation
i assure you that i have used nothing but legitimate mathematical laws, so is this true? whoever can tell me what is wrong first gets....umm nothing...
the satisfaction of knowing that they won!
once again, i politely ask you not to flame me for posting this where i feel it is in appropriate context