Ok all You Math Wannabes..prove that 2 equals 1

[TillEulenspiegel]

No Message ...go on, savage logic !

[Pragmatist]

Originally posted by TillEulenspiegel
No Message ...go on, savage logic !

2 * 0 = 0
1 * 0 = 0

Therefore 2 * 0 = 1 * 0

Cancel the zeros.

2=1

:D

I know, I know, but you did ask! :D

[Pantastic]

Posted it before but....

a = b + c

multiply both sides by (a-b)

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

(a^2) - ab = ab + ac - (b^2) - bc

take ac from both sides

(a^2) - ab - ac = ab - (b^2) - bc

factor out

a(a - b - c) = b(a - b - c)

cancel out

a = b

works better with a=3 b=2 c=1 but can have a=2 b=1 c=1
Dodgy working is easy to spot, though.

[TillEulenspiegel]

Funny chit but let's get creative!

[rppa]

-1/1 = 1/(-1)

Take the square root of both sides:
sqrt(-1)/sqrt(1) = sqrt(1)/sqrt(-1)

i/1 = 1/i

Divide by 2: i/2 = 1/(2i)
Add 3/(2i): i/2 + 3/(2i) = 1/(2i)+ 3/(2i)
Multiply by i:
i^2/2 + 3/2 = 1/2 + 3/2
-1/2 + 3/2 = 1/2 + 3/2
1 = 2

I can't claim credit for making this up. There's another one I've seen that uses complex numbers and square root, but I don't recall it.

[Cabbage]

Here's one I like that's a little different from the ones you commonly see:

Let "S" denote the integral symbol.

Evaluate the indefinite integral S (1/x)dx by parts, setting u=1/x, dv = dx. You'll get:

S (1/x)dx = 1 + S (1/x)dx

Subtract S (1/x)dx from both sides, and we have 0=1.

[T'ai Chi]

Originally posted by TillEulenspiegel
No Message ...go on, savage logic !

d/dx (x+x+x+x+...) = 2x
(that summation happens x times)

However, d/dx (x+x+x+x+...) =
d/dx x + d/dx x + d/dx x + d/dx x +... =
1+1+1+1+... = x.
(these summations happen x times too)

So 2x = x, or 2 = 1.

;)

[TeaBag420]

What was it that Jeff Spicoli said about his surfing opponents? "Those guys are fags!!!!"

2 is defined as equal to 1.

This was hardly worth using both hands for.

[phildonnia]

The series
1/1 - 1/2 + 1/3 - 1/4 + 1/5 + ...

= sum [k=1 to infinity] (-1)^k / k = ln 2.

Rearranging the terms,

(1/1 - 1/2 - 1/4) + (1/3 - 1/6 - 1/8) + (1/5 - 1/10 - 1/12) + ...

= sum [k=1 to infinity] (1/(2k-1)-1/(4k-2)-1/(4k))
= sum [k=1 to infinity] (1/(4k(2k - 1))
= (ln 2) /2

(ln 2) /1 = (ln 2) /2

1 = 2

[Soapy Sam]

I kinda like Teabag's version. What it lacks in elegance it makes up in confidence.

[Bruce]

Let's use some circular reasoning:

p = (pi)

A = pr^2 [area of a circle]

Define area of the circle as 1

1=pr^2

Take square root of both sides.

1 = sqrt(p) * r

Define radius of the circle as 2/sqrt(p)

1 = sqrt(p) * 2/sqrt(p)

1 = 2

[Bruce]

Staw Man:
TeaBag says 1=2, and I disagree with that.

Slippery Slope:
If Pragmatist thinks he can kill two birds with one stone, then he obviously thinks 1=2.

Non-Sequitur:
If A then B. Not A, therefore not B.
If 2=2 and 1=1, then 0=0
If 2/2=2/2 and 1/1 = 1/1, then 0/0 = 0/0
You cannot divide by zero, therefore
2/2 does not equal 2/2, and 1/1 does not equal 1/1
What we are saying is 1 does not equal one,
therefore
1=2

[Igopogo]

2-sided piece of paper = 1-sided mobius strip

or

apply pressure to 1 toothpick until it magically transforms into 2 toothpicks before your very eyes.

[T'ai Chi]

I can prove that 1.999... = 2. ;)

[phildonnia]

1^1 = 1
1^2 = 1
1^1 = 1^2
ln (1^1) = ln (1^2)
1 * ln (1) = 2 * ln (1)
1 = 2.

[Brown]

From George Carlin:

If you break a crumb in half, you don't get two half-crumbs; you get TWO CRUMBS!

Ergo, 1 = 2 (at least when the units are crumbs).

[uruk]

If two items are grouped as one set. The set is a single entity.

If the symbol we use to signify the number "2" is, by some astronomical coincidence the same the alpha centarian symbol for the number "one".

How about if we are in bizzaro world? No?

[TillEulenspiegel]

Some good'ns there .
So we did algebra, logic and calculas....I wonder can someone come up with either a statistical or quantum mechanical approach? :)

[Dr Adequate]

A probabalistic approach. Let's caculate the probability that a natural number is "square-divisible" (divisible by an integer square greater than 1).

(0) Partition the natural numbers > 1 according to the number of prime factors they have, eg 2*2*2*2 goes in the same set as 3*5*7*11. If we look at any of these sets, each number in them is a choice of a finite number of primes from the infinite set of primes, and so the chances that any two of these primes are identical is 0. Hence, the probability that a number is square-divisible is 0.

(1) Partition the natural numbers > 1 according to the number of distinct prime factors they have, eg 2^1*3^1 goes in the same set as 2^100*3^256. If we look at any of these sets, each number is of the form p1 ^ n1 * p2 ^ n2 * ... * pk ^nk, where n1 ... nk are taken from the infinite set of natural numbers, and so the chances that all of these natural numbers are equal to one is 0. Hence, the probability that a number is square-divisible is 1.

We conclude that 0 = 1.

[Renfield]

Originally posted by Pantastic
Posted it before but....

a = b + c

multiply both sides by (a-b)

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

(a^2) - ab = ab + ac - (b^2) - bc

take ac from both sides

(a^2) - ab - ac = ab - (b^2) - bc

factor out

a(a - b - c) = b(a - b - c)

cancel out

a = b

works better with a=3 b=2 c=1 but can have a=2 b=1 c=1
Dodgy working is easy to spot, though.

dividing by (a - b - c ) is a no no right? since a = b + c, you actually dividing by 0.

[Renfield]

Originally posted by Brown
From George Carlin:

If you break a crumb in half, you don't get two half-crumbs; you get TWO CRUMBS!

Ergo, 1 = 2 (at least when the units are crumbs).

Reminds me of the movie Rain Man, where Raymond is complaining that he doesn't have enough fish sticks, so Tom Cruise cuts each of them in half.

Great flick.

[Crossbow]

Originally posted by TillEulenspiegel
No Message ...go on, savage logic !

Uhem!

The word ONE has three letters,
The word TWO has three letters,
Therefore,

ONE = TWO

:p

[Elind]

Originally posted by TillEulenspiegel
No Message ...go on, savage logic !

2^0=1^0

2=1

:c2: