What's the Flaw in this Algebraic Proof?

[dianzz]

Homework Statement

guys ,it says that what's wrong with following proof ? let's see if x=y ,then

x2=xy
x2-y2=xy-y2
(x+y)(x-y)=y(x-y)
x+y=y
2y=y
2=1

Homework Equations

The Attempt at a Solution

we know that if ax=ay x=y only and if only a is not zero ,but how to say this in "proof" language ? sory if look stupid ..im just kid in math ..hehe ,thanx

 

[Cyosis]

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

This is where it goes wrong. Tell me how do you go from the first to the second step, then tell me what x-y is.

 

[Mentallic]

know that if ax=ay x=y only and if only a is not zero ,but how to say this in "proof" language ?

This is satisfactory:

$$ax=ay$$

$$x=y , a\neq 0$$

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

This step is where your problem occurred. Since you restricted the algebra to y=x, you cannot do this step because it is dividing by 0.

 

[Pengwuino]

Well there's a problem with the proof. The third line (x-y), we know that is 0 so you have 0 = 0 immediately.

 

[dianzz]

thanx man ..sorry if I am not reply immediatly ..but for me the conclusion there is a restriction in algebra,right ?? ..how can the manipulation allow that ..

 

[Mark44]

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

To go from the first equation to the second, you divide both sides of the first equation by x - y. This is not allowed, because x - y = 0, by assumption ("Let x = y"). You cannot divide by zero. Period.

 

[dianzz]

its very clear right now ..biG thanks all of u

 

[Дьявол]

You should tried like:
(x+y)(x-y)=y(x-y)

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

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

(x-y)x=0

So x=0 or x-y=0

 

[Mark44]

This is one of a number of fallacious "proofs" that end in some obviously false conclusion. All of them have one or more steps that are not allowed (like division by zero) that make the erroneous conclusion seem valid. Other examples I've seen use integration or differentiation to arrive at a similar conclusion.