Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Possible Duplicate:
Finding the error in a proof

  1. $a=b$
  2. $ab=a^2$
  3. $ab-b^2=a^2-b^2$
  4. $b(a-b)=(a+b)(a-b)$
  5. $b= a+b$

Reminder the first step where $b = 2b$ So,

$1=2$

In this case in my opinion the wrong step is that in the third, because you can´t subtract $b^2$ from both sides.

This question it was taken from a math exam.

share|improve this question

marked as duplicate by Douglas S. Stones, Martin Sleziak, Norbert, Belgi, Pedro Tamaroff Oct 21 '12 at 14:51

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

    
I wonder why the most common questions can't be researched on [a search engine] before being asked. –  Parth Kohli Oct 21 '12 at 14:24

3 Answers 3

$a=b \implies a-b=0$

from step 4 to 5 you are dividing through by $a-b=0$.

share|improve this answer
    
Where did you see question like this Alex? –  Vinicius L. Beserra Oct 21 '12 at 14:08
    
@ViniciusL.Beserra a friend has asked me this before. –  Alex Oct 21 '12 at 14:17

You divided by $a-b$ in step 5), which is zero by assumption 1).

share|improve this answer

The transition from step 4 to step 5 is wrong.$$\rm a = b\qquad\Rightarrow \qquad a - b = 0$$If $\rm \,a - b = 0\, $, then $\rm (a + b)(a - b) = b(a - b) $ can be written as $\rm 0(a + b) = 0b$

While what we do in the transition is, we divide both sides by $0$, hence breaking the so-called “fundamental rule”: never divide by zero!

share|improve this answer
    
Thanks for you commentary. –  Vinicius L. Beserra Oct 21 '12 at 19:01

Not the answer you're looking for? Browse other questions tagged or ask your own question.