# How to proof ?

1. Jun 20, 2009

### dianzz

1. The problem statement, all variables and given/known data
guys ,it says that whats wrong with following proof ? lets 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
2. Relevant equations

3. 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

2. Jun 20, 2009

### Cyosis

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.

3. Jun 20, 2009

### Mentallic

This is satisfactory:

$$ax=ay$$

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

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.

4. Jun 20, 2009

### 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.

5. Jun 29, 2009

### dianzz

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

6. Jun 29, 2009

### Staff: Mentor

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.

7. Jun 29, 2009

### dianzz

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

8. Jun 30, 2009

### 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

9. Jun 30, 2009

### Staff: Mentor

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.