# 0=-1 (Calculus Puzzle)

Here’s a neat puzzle I’ve come up with/discovered… See if you can find the error/oversight.

∫tan(x) dx = ∫sin(x)/cos(x) dx = ∫sin(x)sec(x) dx

integration by parts:
u = sec(x)
du = sec(x)tan(x) dx

dv = sin(x) dx
v = -cos(x) dx

∫tan(x) dx = -cos(x)sec(x) - ∫-cos(x)sec(x)tan(x)dx

∫tan(x) dx = -1 + ∫tan(x)dx

Now subtract ∫tan(x) dx from both sides and…

0 = -1

∫tan(x)dx - ∫tan(x)dx is not equal to 0.

It’s an arbitrary constant.

Yep… that’s basically it…

I had a feeling you’d be the first one to respond to this thread, and you’d have the answer… //Andrew

Well, and you left off the “+ C” that you always need…
(though that may very well be what Ether meant)

now make 1=2 0 = -1

2 = 1

Let a and b be equal non-zero quantities

``````a = b
``````
1. Multiply through by a

a^2 = ab

2. Subtract b^2

a^2 - b^2 = ab - b^2

3. Factor both sides

(a - b)(a + b) = b(a - b)

4. Divide out (a - b)

a + b = b

5. Observing that a = b

b + b = b

6. Combine like terms on the left

2b = b

7. Divide by the non-zero b

2 = 1

1/3=.33333…
2/3=.66666…
1/3+2/3=.99999…
1=.99999…

Divide by zero.

The ellipsis means infinitely repeating decimal so

.99999… means limit(sum(9/10^i,i,1,n),n,inf) = 1

Has this turned into a thread of trying to find a math trick that Ether doesn’t know how to work it? If so, i want a chance to disprove it before he dose :).

Peck, sorry to tell you but i doubt ull win haha

Although ether, can you prove or disprove Riemann Sums? Now that would be cool.

167 = -53

No illegal operations were performed, [strike]slightly[/strike] non-standard

13 * 7 = 28

Here is the explanation (BTW, the guy in the glasses is Ether as a young man )

fas est dividere cuius rei demonstrationem mirabilem sane detexi. Hanc marginis exiguitas non caperet

ingles por favor sinor

1. *Inglés or favor señor.

2. according to google translate: It is a wonderful demonstration of what share of corruption. Shortness of this margin does not take

but that seems like it was lost a bit in the translation:p

It is actually taken from Fermat who wrote about a theorem (a^x + b^x = c^x has no integral solution set a,b,c for any integer value for x greater than 2) in the margins of a book in 1637. It is actually the description of the proof for this theorem which he came up with and it translates more accurately to

I have discovered a truly marvelous proof of this, which this margin is too narrow to contain.

unfortunately his proof seems to have disappeared off the face of the earth, assuming he even wrote it down.
Over the years people proved that it held true for specific exponents. It wasn’t until 1995 that it all came together in an extremely complex proof that was probably not what Fermat had in mind, but a general proof of the theorem non the less.

Sounds like the classic “I have discovered a truly marvelous proof of this, which this margin is too narrow to contain” from Fermat. Which, on investigating the Wikipedia entry on Fermat’s last theorem, is exactly what it is.

Well played, Ether.

Thank you my computer is annoying whenever i try to insert alt symbols. also, i was using spanish, not latin and por was correct.