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Calculus Problem
3

Calculus Problem

Calculus Problem

(OP)
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. ponder  Thanks in advance for all responses.

Mike McCann
McCann Engineering

RE: Calculus Problem

The one I vaguely remember contains division by zero. Don't remember the proof though.

RE: Calculus Problem

Just check whether the function is discontinuous in the applied range.

RE: Calculus Problem

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 remember the problem but calculus was far from my strong point...all the many times i retook it. thank god my pay is based on my performance and not on my college GPA (at least i did well in my civil class...most of them anyway).

RE: Calculus Problem

2
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.

RE: Calculus Problem

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.

RE: Calculus Problem

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.

RE: Calculus Problem

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

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