## 1=2

A forum for good logic/math puzzles.

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### 1=2

Clay

Posts: 9
Joined: Tue Sep 05, 2006 3:40 pm UTC

He divides by zero, therefore he loses at elementary math.

Gelsamel
Lame and emo

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Location: Melbourne, Victoria, Australia

specifically when he divides by x - y
Linoc

Posts: 10
Joined: Wed Oct 25, 2006 2:14 am UTC

ab = ac → b = c or a = 0

People always forget the "or a = 0" bit.
While no one overhear you quickly tell me not cow cow.

phlip
Restorer of Worlds

Posts: 6740
Joined: Sat Sep 23, 2006 3:56 am UTC
Location: Australia

Gelsamel wrote:He divides by zero, therefore he loses at elementary math.

Yeah if x-y = 0 then it's not 1/1 it's actually 0/0 which, well, I don't need to explain the problem there.
Heirtopendragon

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Joined: Mon Sep 25, 2006 8:07 am UTC

Une See Fights - crayon super-ish hero webcomic!
Spoiler:
Nullcline wrote:What a colossal waste of stupidity.
fjafjan wrote:I got quite a lot of "batter" left
natraj wrote:skydiving is p fun (in this respect it is almost exactly unlike centipedes)

LE4dGOLEM
is unique......wait, no!!!!

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Location: :uoıʇɐɔol

Heh, there's some funny stuff there... though I feel the need to point out the flaw in the square-root proof...

√x √y = Â±√(xy)
√x / √y = Â±√(x/y)

That is, √x √y equals either √(xy) or −√(xy), (but not both if x,y≠0).

The Â± isn't necessary in the real numbers, since the square root is defined to be the positive root of the number... but such a definition can't be made with the complex numbers, as there is no such thing as "positive".
While no one overhear you quickly tell me not cow cow.

phlip
Restorer of Worlds

Posts: 6740
Joined: Sat Sep 23, 2006 3:56 am UTC
Location: Australia

I've got a "higher-math" version of this:

x^2 = x*x = x + x + x + ... + x [x terms]

d(x^2)/dx = 2x
d(x^2)/dx = d(x + x + ... + x)/dx = 1 + 1 + 1 + ... + 1 = x
2x = x

So what this says is that either 2 = 1 (if x != 0) or that one of the above equalities only holds for x = 0. But all of the above equalities (except the last) hold for any x.

Where's the problem?
EvanED

Posts: 3767
Joined: Mon Aug 07, 2006 6:28 am UTC

The problem is that the expansion:
x*x = x + x + x + ... + x [x terms]
Is only valid when x is an integer (which is implicit in the expansion itself)

Derivatives require a function that's defined over all the real numbers.

In particular, the derivative is just shorthand for:
lim_dx->0 {((x + dx) + (x + dx) + ... + (x + dx) [(x + dx) terms]) − (x + x + ... + x [x terms])}/dx
And how would you define "x + dx terms"?
While no one overhear you quickly tell me not cow cow.

phlip
Restorer of Worlds

Posts: 6740
Joined: Sat Sep 23, 2006 3:56 am UTC
Location: Australia

Replace

x*x = x+x+x... x [x terms]
with

Sum[x,{n,1,Floor[x]}}+x*(x-Floor[x])

This is now defined for real x (and is equal to x^2 for x>0). When x is not an integer, the sum doens't change its bound and its derivative is Floor[x]. The second term gives us 2x - Floor[x], the Floor[x] cancel out.

archgoon

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Joined: Wed Jul 26, 2006 6:08 am UTC
Location: Large (But Finite) Dimensional Hilbert Space