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Professor Egghead entered his classroom one morning when one of his undergraduate students boasted that he could prove that 2 is equal to 1. The student then showed the professor the following proof:

Given:

**x = 1** and

**y = 1** therefor:

**x = y**

1. Multiply each side by

**x**:

**x² = xy**

2. Subtract

**y²** from each side:

**x²-y² = xy-y²**

3. Factor each side:

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

4. Divide by the common term

**(x-y)**:

**x+y = y**

5. Put the initial values back in the equation:

**1+1 = 1**

or

**2 = 1**

Professor Egghead saw the problem right away, can you?

## 11 comments:

Something wrong at step 3 but I can't explain it

I think...

It has to do with the square of 1 still being equal to one...multiplying by 1 does not increase the value of the numbers.

The problem is you cannot divide by zero and x-y is 0!

you can't divide by zero

the problem is between step 3 and 4

The suy has taken s-y in both sides of eqn and we know that x=y so x-y is 0.there fore we cant cancel out the term x-y becoz 0 cant be cancelled in two side...

the reason for that is say:

0 x 3=0 x 5

it doesn not mean 5 = 3..

so i found out the mistake

i should be rewarded...

that is completly wrong!

x^2-y^2 is not equal to (x+y)(x-y) but to x^2+b^2-2xy!

dork

that last 'anonymous'....W*H*A*T*????

the problem is step 3: the right side of the equation is incorrect. xy-y2 does not equal y(x-y). So (x-y) is not the common denominator.

X=1 therefore, Y must equal something else. The point of using different letters for variables is that they represent different numbers. If that weren't true, then x=y and you stop there.

If we ignore the fact that x and y technically should represent different variables and carry on...

Step one is moot. If x = 1 then multiplying both x and y by x at step one still results in 1 on both sides of the equation.

please excuse my dear aunt sally, see is very old and did not mean to confuse everyone.

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