1=2
Moderators: jestingrabbit, Moderators General, Prelates
10 posts
• Page 1 of 1
by Heirtopendragon » Tue Oct 31, 2006 5:49 pm UTC
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
- Posts: 25
- Joined: Mon Sep 25, 2006 8:07 am UTC
Une See Fights - crayon super-ish hero webcomic!
by phlip » Wed Nov 01, 2006 7:37 am UTC
Heh, there's some funny stuff there... though I feel the need to point out the flaw in the square-root proof...
√x √y = ±√(xy)
√x / √y = ±√(x/y)
That is, √x √y equals either √(xy) or −√(xy), (but not both if x,y≠0).
The ± 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".
√x √y = ±√(xy)
√x / √y = ±√(x/y)
That is, √x √y equals either √(xy) or −√(xy), (but not both if x,y≠0).
The ± 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.
but how about watch phone?
but how about watch phone?
-
phlip - Restorer of Worlds
- Posts: 6740
- Joined: Sat Sep 23, 2006 3:56 am UTC
- Location: Australia
by EvanED » Sat Nov 04, 2006 1:39 am UTC
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?
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
- Location: Madison, WI
by phlip » Sat Nov 04, 2006 2:21 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"?
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.
but how about watch phone?
but how about watch phone?
-
phlip - Restorer of Worlds
- Posts: 6740
- Joined: Sat Sep 23, 2006 3:56 am UTC
- Location: Australia
by archgoon » Mon Nov 06, 2006 10:53 am UTC
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.
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 - Posts: 62
- Joined: Wed Jul 26, 2006 6:08 am UTC
- Location: Large (But Finite) Dimensional Hilbert Space
10 posts
• Page 1 of 1