calculus problem

huangyhg

calculus problem
ok guys, my grandaughter is taking calculus this year, and i remember a problem from second quarter freshman calculus where there was a "proof" of 1 + 1 = 3 and we had to find what was wrong with it.  after 40+ years, i no longer have my notes, and was wondering if this jogs anyone's memory.  
does anyone remember the "proof" and what was wrong with it?  this is not a joke.   thanks in advance for all responses.
mike mccann
mccann engineering

the one i vaguely remember contains division by zero. don't re  
just check whether the function is discontinuous in the applied range.
you might find it in here:
it hasn't been nearly as long for me and i still don't recall the exact procedure. i believe quark is probably correct. i do re  
i believe the proof you are thinking of begins with a + b = 1.  later in the proof the variables a = b = 1 are established.  and the result, a + b = 1 then becomes the confusion.  in this proof, there is a step where they divide by 0, but you have to recognize it because it is expressed as a step where the formula says something like (a - b) x a = something else.  later they cancel the (a - b) term from both sides of the equals sign.  at this point they divided by (a - b) which is zero.
the proof i enjoy more ends up with "women = evil".  ever hear of that one?
to have a woman you have to expect to spend time and money ...
women = time x money
but we all know that time is money ...
time = money
which may be substituted into our original equation to obtain ...
women = money x (money)
which simplifies to ...
women = (money)^2
we also know that money is the root of all evil, which may be expressed mathmatically as ...
money = sqrt(evil)
and we then obtain by substitution ...
women = (sqrt(evil))^2
and we may simplify this to ...
women = evil
maybe your grandfather would enjoy that proof.
i may have this in one of my martin gardiner books- will check when i get home.  seems like they factor out a term which is actually zero.  of course, there's bound to be lots of ways to work the details.
proof that any number a is equal to a smaller number b:
let a = b + c
multiply both sides by (a-b) to get
a^2 - ab = ab + ac -b^2-bc
move ac to the left side:
a^2 - ab - ac = ab -b^2 - bc
factor:
a(a-b-c) = b(a-b-c)
divide each side by a-b-c to get:
a = b
note that from the definition, a-b-c = 0, so that last step divides each side by zero.
this is from p. 143 of "hexaflexagons and other mathematical diversions" by martin gardner.
a = b
a a = a b
a^2 = a b
a^2 - b^2 = a b - b^2
(a + b)(a - b) = b (a - b)
(a + b) = b
since a = b
(a + a) = a
2 a = a
2 = 1