(no subject)

[plonq]

Assume         a = b
therefore      a - b = 0
additionally  2a - 2b = 0
therefore      a - b = 2a - 2b, or
             1(a - b) = 2(a - b)


Now if we factor out the (a - b) from this equation...


                 1          1
             1(a - b) = 2(a - b)
               -----------------
                   (a - b)
                      1

Ergo           1 = 2 

Most people spot the fallacy right away, but it's surprising how many people I've caught with this one.

Comments

[pierrekrahn.livejournal.com]

First off, a - b = 0, not 1.

And second of all, this equation is undefined as you cannot divide by zero

:P

[plonq.livejournal.com]

First off, a - b = 0, not 1.

Where did I claim that a - b = 1?

[pierrekrahn.livejournal.com]

the three areas that are struck out are (a - b) and have been replaced by a 1

[plonq.livejournal.com]

That's because that's the step where I'm factoring the expression out of the equation. It's not suggesting that a-b = 1, but that a-b/a-b = 1, which would normally work if a-b didn't equal 0.

[pierrekrahn.livejournal.com]

Ah ok, the way I read it, it seemed that you were claiming that a-b=1.

In that case, my first comment should read as follows:


First off, a - b = 0, not 1.

And second of all, this equation is undefined as you cannot divide by zero

:P

[plonq.livejournal.com]

his equation is undefined as you cannot divide by zero

Puh. Certainly you can divide by zero. A real-life example is when there are two slices of pizza left, and we're trying to decide which slice is yours... unless you go all spastic and lick your hands and paw the pieces before we can come to an equitable division I mean, then things get a bit undefined again.

[pierrekrahn.livejournal.com]

LOL

Your saliva tastes good.

[tzisorey.livejournal.com]

Are black-holes where God divided by zero?