Random joke from this category
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
=> $1 = 10 cent
Theorem: 1$ = 1c.
Proof:
And another that gives you a sense of money disappearing.
1$ = 100c
= (10c)^2
= (0.1$)^2
= 0.01$
= 1c
Here $ means dollars and c means cents. This one is scary in that I have seen PhD's in math who were unable to see what was wrong with this one. Actually I am crossposting this to sci.physics because I think that the latter makes a very nice introduction to the importance of keeping track of your dimensions.
