# Proving 2=1

1. Dec 14, 2004

### T@P

Hey can everyone please post any/all ways they know of "proving" 2 =1? obviously they are all wrong, but some are more "convinving" than others.

For example, heres an old one to start:

e^(2*pi *i) = 1, so ln (e^(2*pi *i)) = ln (1) or 2 * pi * i = 0

clearly pi != 0 and i != 0 so 2 = 0. Although this dosent *prove* 2 = 1, from 2= 0 you can really show that almost anything is true.

any other ideas?

2. Dec 14, 2004

### Nylex

I haven't seen that version. The only "proof" I've seen is this one.

If you can't spot what's wrong with it, it's the dividing by a^2 - ab as you're dividing by a(a - b), which is 0 as a = b.

Last edited: Dec 14, 2004
3. Dec 14, 2004

### Zurtex

But of course in actuallity:

e^(2*pi*k*i) = 1

And when you take the log of both sides it's the k that is equal to 0.

It's no diffrent from saying sin(2*pi) = 0, so 2=0

4. Dec 14, 2004

### jcsd

ln z is a multivalued function in C

ln z = |ln z| + iarg(z)

therefore ln e2&pi;i = 2&pi;i + n2&pi;i for all n in Z.

5. Dec 16, 2004

### T@P

anyother ways? there are more such as:

1/(-1) = (-1) /1
taking the square root of both sides yeilds:
1/i = i/1 or -1 = 1, 2 = 0

6. Dec 16, 2004

### theCandyman

You seem to enjoy these kind of "proofs" T@P. I think most of the ones here were already posted here , with a bit of changing around. I love the last one in that thread, it actually stumped my Calculus I teacher, but not my TA though, not even for a second (he's working on his doctorate, my instructor has been teaching for about 25 years). These problems are fun for me now, I finally understand enough math to get what they are saying! :rofl:

7. Dec 17, 2004

### suresh_jeans

Question: not a Answer....consider the honey bee cell, regular hexagon, with side length is 'R'. we can find the distance between the adjacent cells, regular hexagons, centre's are sqroot3*R. how we can find the distance between the centre's of a cell's which are not adjacent

8. Dec 17, 2004

### suresh_jeans

Give any group of order infinite, every element of that group is of finite order?

9. Dec 17, 2004

### NateTG

For any particular infinite cardinality $C$, take the subgroup of $S_C$ which only contains permutations that permute a finite number of elements.

10. Dec 21, 2004

### nicot

(1) X = Y Given
(2) X^2 = XY Multiply both sides by X
(3) X^2 - Y^2 = XY - Y^2 Subtract Y^2 from both sides
(4) (X+Y)(X-Y) = Y(X-Y) Factor both sides
(5) (X+Y) = Y Cancel out common factors
(6) Y+Y = Y Substitute in from line (1)
(7) 2Y = Y Collect the Y's
(8) 2 = 1 Divide both sides by Y

11. Dec 23, 2004

### T@P

thats fancily dividing by 0, since x = y you cannot divide by (x-y)

i was looking more for proofs that rely on little known rules, but thanks for your input anyway :)

12. Dec 23, 2004

### daster

Here's one:

(-x)^2 = x^2
log[(-x)^2] = log(x^2)
2 log(-x) = 2 log(x)
log (-x) = log (x)
-x = x
-1 = 1