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Prove that the crocodile is longer than it is wide.

Lemma 1. The crocodile is longer than it is green: Let's look at the crocodile. It is long on the top and on the bottom, but it is green only on the top. Therefore, the crocodile is longer than it is green.

Lemma 2. The crocodile is greener than it is wide: Let's look at the crocodile. It is green along its length and width, but it is wide only along its width. Therefore, the crocodile is greener than it is wide.

From Lemma 1 and Lemma 2 we conclude that the crocodile is longer than it is wide.

Theorem: 1 = 1/2

Proof:
We can re-write the infinite series 1/(1*3) + 1/(3*5) + 1/(5*7) + 1/(7*9)+... as 1/2((1/1 - 1/3) + (1/3 - 1/5) + (1/5 - 1/7) + (1/7 - 1/9) + ... ).

All terms after 1/1 cancel, so that the sum is 1/2.

We can also re-write the series as (1/1 - 2/3) + (2/3 - 3/5) + (3/5 - 4/7)+ (4/7 - 5/9) + ...

All terms after 1/1 cancel, so that the sum is 1.

Thus 1/2 = 1.

Theorem: e=1

Proof:
2*e = f
2^(2*pi*i)e^(2*pi*i) = f^(2*pi*i)
e^(2*pi*i) = 1
Therefore:
2^(2*pi*i) = f^(2*pi*i)
2=f
Thus:
e=1

Theorem: log(-1) = 0

Proof:
a. log[(-1)^2] = 2 * log(-1)
On the other hand:
b. log[(-1)^2] = log(1) = 0
Combining a) and b) gives:
2* log(-1) = 0
Divide both sides by 2:
log(-1) = 0

Theorem: 1 = -1

Proof:
1 = sqrt(1) = sqrt(-1 * -1) = sqrt(-1) * sqrt(-1) = 1^ = -1

Also one can disprove the axiom that things equal to the same thing are equal to each other.
1 = sqrt(1)
and -1 = sqrt(1)
Therefore 1 = -1

As an alternative method for solving:

Theorem: 1 = -1

Proof:
x=1
x^2=x
x^2-1=x-1
(x+1)(x-1)=(x-1)
(x+1)=(x-1)/(x-1)
x+1=1
x=0
0=1
=> 0/0=1/1=1

Theorem: All numbers are equal.

Proof: Choose arbitrary a and b, and let t = a + b. Then
a + b = t
(a + b)(a - b) = t(a - b)
a^2 - b^2 = ta - tb
a^2 - ta = b^2 - tb
a^2 - ta + (t^2)/4 = b^2 - tb + (t^2)/4
(a - t/2)^2 = (b - t/2)^2
a - t/2 = b - t/2
a = b
So all numbers are the same, and math is pointless.

Theorem: 4 = 5
Proof: -20 = -20
16 - 36 = 25 - 45
4^2 - 9*4 = 5^2 - 9*5
4^2 - 9*4 + 81/4 = 5^2 - 9*5 + 81/4
(4 - 9/2)^2 = (5 - 9/2)^2
4 - 9/2 = 5 - 9/2
4 = 5

Theorem: 1 + 1 = 2
Proof:
n(2n - 2) = n(2n - 2)
n(2n - 2) - n(2n - 2) = 0
(n - n)(2n - 2) = 0
2n(n - n) - 2(n - n) = 0
2n - 2 = 0
2n = 2
n + n = 2
or setting n = 1
1 + 1 = 2

Theorem: n=n+1
Proof:
(n+1)^2 = n^2 + 2*n + 1
Bring 2n+1 to the left:
(n+1)^2 - (2n+1) = n^2
Substract n(2n+1) from both sides and factoring, we have:
(n+1)^2 - (n+1)(2n+1) = n^2 - n(2n+1)
Adding 1/4(2n+1)^2 to both sides yields:
(n+1)^2 - (n+1)(2n+1) + 1/4(2n+1)^2 = n^2 - n(2n+1) + 1/4(2n+1)^2
This may be written:
[ (n+1) - 1/2(2n+1) ]^2 = [ n - 1/2(2n+1) ]^2
Taking the square roots of both sides:
(n+1) - 1/2(2n+1) = n - 1/2(2n+1)
Add 1/2(2n+1) to both sides:
n+1 = n

Theorem: 1$ = 10 cent
Proof:
We know that $1 = 100 cents
Divide both sides by 100
$ 1/100 = 100/100 cents
=> $ 1/100 = 1 cent
Take square root both side
=> squr($1/100) = squr (1 cent)
=> $ 1/10 = 1 cent
Multiply both side by 10
=> $1 = 10 cent