Puzzler

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 ?

Davorak

[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?

The Bob

[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 ©)

K.J.Healey

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.

gerben

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