Jump to content
Science Forums

Recommended Posts

Posted

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.

Posted
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....

Posted

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.

Posted
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.

Posted

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".

Posted
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.

Posted
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.

Posted
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?
Posted
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.

Posted
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?

Posted

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

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

Guest
Reply to this topic...

×   Pasted as rich text.   Paste as plain text instead

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

Loading...
×
×
  • Create New...