1 = 2 ...?

[Jay-qu]

x = 2

x(x-1) = 2(x-1)

x2-x = 2x-2

x2-2x = x-2

x(x-2) = x-2

x = 1

 

it looks right but there is a mistake in there -> can you see it?

[Tormod]

Well, this is apparently not a progression but just a lot of equations. So there is nothing wrong. The last line is just as isolated as the rest.

 

Edit: I misread the equations, sorry.

 

Still, the final line does not make sense at all. I don't see how you derive x=1 from this.

[Tormod]

  Jay-qu said:

x2-x = 2x-2

x2-2x = x-2

 

This should be

x2-x = 2x-2

x2 = 3x-2

x=(3x-2)/2

x=(6-2)/2

x=2

 

And the last line is simply wrong...nowhere does x equate 1.

[Jay-qu]

i see where your coming from but if you sub in x = 1 then the equations also work except the one you pointed out...

[Tormod]

Maybe I'm not getting the point. I thought you somehow resolved x=2 to become x=1... ;)

 

I am NOT a math wiz...

[C1ay]

  Jay-qu said:

x = 2

x(x-1) = 2(x-1)

x2-x = 2x-2

x2-2x = x-2

x(x-2) = x-2

x = 1

 

it looks right but there is a mistake in there -> can you see it?

 

x(x-1) does not equal 2(x-1) or x2-x

x(x-1) = x^2-x

 

x2-2x does not equal x-2 either

 

The whole thing is riddled with errors....

[Qfwfq]

The real point is that:

 

x = 2

 

and

 

x(x-1) = 2(x-1)

 

are only equivalent if x is not 1. This resolves the paradox.

 

There's nothing wrong with the rest of the algebra.

[C1ay]

  Qfwfq said:

The real point is that:

 

x = 2

 

and

 

x(x-1) = 2(x-1)

 

are only equivalent if x is not 1. This resolves the paradox.

 

There's nothing wrong with the rest of the algebra.

 

Huh? Doesn't x(x-1) = 2(x-1) resolve to x^2 - x = 2x - 2? I don't see how these can be equivalent except for x = 1 and x = 2.

[zadojla]

x = 2

x(x-1) = 2(x-1)

x2-x = 2x-2

x2-2x = x-2

x(x-2) = x-2

x = (x-2)/(x-2) <===== implied step

x = 1

 

The implied step is the problem. Because x = 2, this is equivalent to dividing by 0. This is a common trick in this kind of algebraic "paradox".

[Qfwfq]

  C1ay said:

Doesn't x(x-1) = 2(x-1) resolve to x^2 - x = 2x - 2?

Yes.

 

The trouble is not in that step but in the previous one. Or the one Zadojla points out. It's a matter of which way you go, or IOW the equivalence fails because, if x = 1 then (x - 1) is 0, if x = 2 then (x - 2) is 0. Multiplying both sides by zero doesn't give an equivalent equation, dividing both sides by zero doesn't make sense.

[C1ay]

  Qfwfq said:

Yes.

 

The trouble is not in that step but in the previous one. Or the one Zadojla points out. It's a matter of which way you go, or IOW the equivalence fails because, if x = 1 then (x - 1) is 0, if x = 2 then (x - 2) is 0. Multiplying both sides by zero doesn't give an equivalent equation, dividing both sides by zero doesn't make sense.

 

I realize why the fallacy in the math. I was questioning your statement, "...are only equivalent if x is not 1."

 

I've seen similar 1=2 problems. They most always rely on division by 0 somewhere in the stack.

[Qfwfq]

  C1ay said:

I was questioning your statement, "...are only equivalent if x is not 1."

I don't get your point, all the equations obtained after multiplication by (x - 1) are 0 = 0 in the x = 1 case. Hence they aren't fully equivalent to the equation x = 2. What, then, was your reason for questioning the statement?

[C1ay]

  Qfwfq said:

I don't get your point...

 

Nothing, it was just a nitpick on wording. IMO the statements x=2 and x(x-1) = 2(x-1) are only equivalent for x = 2. It didn't make since to me to say, "they are equivalent only if x does not equal 1" because they are not equivalent for other values of x either.

[freQ]

x-"my butt" = ("my butt"/ pi) x

 

;)

 

>_>

 

o_Õ

 

LMAO

[C1ay]

  freQ said:

x-"my butt" = ("my butt"/ pi) x

 

;)

 

>_>

 

o_Õ

 

LMAO

 

While all of that makes for an amusing equation I fail to see how it is relevant to the discussion. BTW, Welcome to the forum. Have you read the FAQ yet?

[Aki]

This is very interesting. At first when I read this, I was like ??????, but now it makes sense

[Bo]

interesting this all; especially the amount of discussion it stirs up ;)

 

It is indeed impid that x = (x-2)(x-2).

This equation means in simple terms: x=1, unless it is 2. since we required x to be 2, this equation does not imply that x is 1.

 

Bo