Math problem

[kengr]

http://www.fugly.com/pictures/11232/math-problem.html

Ignore the ignorant comment about it being wrong from the start. How many of you can find the *actual* error in the problem?

Comments

[janetmiles.livejournal.com]

#DIV/0!

[griffen.livejournal.com]

I know it's wrong, but I can't spot the error.

Oh, wait.

They can't substitute in "a" for "b" in the third line from the bottom. You're solving for A. You can't do that if you monkey around with the variables. This is how it's supposed to fall out:

Given that A = B,

A² = AB

A² - B² = AB - B²

(A + B)(A - B) = B(A-B)

A + B = B

A = 0

Then solve for B:

A = B
A = 0

Therefore, B = 0.

This is why you don't monkey around with the variables.

[kengr]

You aren't solving for a particular variable.

The problem is

(A + B)(A - B) = B(A-B)

To reduce this, you divide by (A-B). But look at the given. A=B. Therefore, if you divide by (A-B) what are you dividing by? <eg>

[aurictech.livejournal.com]

[without looking at other comments]

Well, if a = b, then (a - b) = 0.

Dividing by zero (the part left out between steps #4 and 5) is a Mathematics Naughty^TM.

For Possible Bonus Points

[aurictech.livejournal.com]

Next question: Who can spot the pun in my response above?

[kengr]

I once came up with a much shorter proof that A=B for any value of A & B.

The trick in mine was that I divided 0 by 0. :-)

It just wasn't obvious that that was what I was doing. :-)