Finding the error in a proof that 2 = 1

[jrdunaway]

Homework Statement

2. What is wrong with the following "proof"? Let x = y. Then
x² =xy,
x² - y² =xy - y ,
(x + y)(x -y) = y(x - y),
x + y = y,
2y = y,
2= 1.

Homework Equations

It's obvious that x + y =/= y, but I do not know how to "prove" this, i.e. which proof from algebra is applicable here. My hunch is that it is the Distributive Property?

The Attempt at a Solution

I honestly am not sure where to start with this. All I can see is that dividing (x-y) out from step 3 to step 4 is what causes the inequality.

 

[tiny-tim]

welcome to pf!

hi jrdunaway welcome to pf!

(x + y)(x -y) = y(x - y),
x + y = y,
…
All I can see is that dividing (x-y) out from step 3 to step 4 is what causes the inequality.

that's right!

if x = y, that means that you divided by 0, which isn't allowed …

so (if x = y) that step is not valid​

 

[HallsofIvy]

Homework Statement

2. What is wrong with the following "proof"? Let x = y. Then
x² =xy,
x² - y² =xy - y ,

You mean xy- y² on the right.

But you still can't divide by 0.

(x + y)(x -y) = y(x - y),
x + y = y,
2y = y,
2= 1.

Homework Equations

It's obvious that x + y =/= y, but I do not know how to "prove" this, i.e. which proof from algebra is applicable here. My hunch is that it is the Distributive Property?

The Attempt at a Solution

I honestly am not sure where to start with this. All I can see is that dividing (x-y) out from step 3 to step 4 is what causes the inequality.

5*0= 0 and 3*0= 0 so 5*0= 3*0. It does not follow that 5= 3!

 

[Mentallic]

What is wrong with the following "proof"? Let x = y. Then
x² =xy,
x² - y² =xy - y ,
(x + y)(x -y) = y(x - y),
x + y = y,

What instead should have been done was

$$(x+y)(x-y)=y(x-y)$$

$$(x+y)(x-y)-y(x-y)=0$$

$$(x-y)(x+y-y)=0$$
(factorized out (x-y) from both factors)

$$(x-y)x=0$$

So either x-y=0, thus x=y, or x=0

This is common practice when solving quadratics and such. If you end up with $$x^2+x=0$$ for example, you don't divide through by x to obtain $x+1=0$ because then you lose the solution of x=0. What you instead do is factorize into the form $x(x+1)=0$ which allows you to find all the solutions.

 

[jrdunaway]

PhysicsForums,

you guys have exceeded my expectations! Thank you! I'll be back soon I'm sure :)