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Feb10-05, 01:25 PM   #1
 

Puzzler


This a puzzle for me.

x^3 -x -x^3=0 Obviously x=0
but, divide both sides by x

x^2 -1 -x^2=0 except for when x=0 'cause undefined
but that means
-1=0

is this apparent conflict not in fact one because you can't rule out x=0 ?
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Feb10-05, 03:05 PM   #2
 
[QUOTE=regor60]This a puzzle for me.
x^2 -1 -x^2=0 except for when x=0 'cause undefined
but that means
-1=0
QUOTE]

Hmm looks to me like you would have
-1 = 0/x
But x = 0 so you have
-1 = 0/0
0/0 is undefined.

-1 = undefined is not well formed equation.
Any know better mathematical terminology for 0/0?
Feb10-05, 04:27 PM   #3
 
[tex]x^3 - x - x^3 = 0[/tex] then divide by x:

[tex]\frac{x^3 - x - x^3}{x} = \frac{0}{x}[/tex]

[tex]x^2 - x^2 = 1[/tex]

Even saying that [tex]\frac{0}{x}[/tex] is 0 you should see that it is not possible to get two different numbers to make 1 when the two x values will give the same to take away.

The Bob (2004 ©)
Feb10-05, 05:50 PM   #4
 

Puzzler


I thought that -1=0 is not a statement of equality, but more of a qualifier for the original equation. We want to know when say, x^2 - x^2 == 1, and that will only happen when 0 = 1, which is never. So that has no solution. I thought this meaning was always implied in mathematics. I mean I can say
1 + 1 =5, but how can that be?! I mean I just wrote it, but its not supposed to be possible! Just becasue yo ucan write down an equation doesnt mean its true.
Feb11-05, 11:36 AM   #5
 
Quote by regor60
This a puzzle for me.

x^3 -x -x^3=0 Obviously x=0
but, divide both sides by x

x^2 -1 -x^2=0 except for when x=0 'cause undefined
but that means
-1=0

is this apparent conflict not in fact one because you can't rule out x=0 ?
You are simply breaking the rule of "not dividing by zero".
You could state your "problem" simpler:
x = 0
divide both sides by x
x/x = 0/x <-> 1 = 0
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