0
$\begingroup$

I just saw this equation and I can't find out where's the problem:

$$25-45 = 16-36$$

$$25- 2 \cdot 5 \cdot \frac{9}{2} = 16- 2\cdot4\cdot\frac{9}{2}$$

$$25 - 2\cdot 5\cdot \frac{9}{2} + \frac{81}{4} = 16 - 2\cdot 4 \cdot \frac{9}{2} + \frac{81}{4}$$

$$\left( 5-\frac{9}{2} \right) ^2 = \left (4-\frac{9}{2} \right) ^2$$

$$5-\frac{9}{2} = 4 - \frac{9}{2}$$ $$5=4$$

$\endgroup$
  • 4
    $\begingroup$ you could try to solve the left and right side on each line and see where it starts to differ $\endgroup$ – ratchet freak Jul 14 '12 at 15:00
  • $\begingroup$ Besides what has been said in the answers, there's actually another mistake: your brackets in the third line are not right. $\endgroup$ – Raskolnikov Jul 14 '12 at 15:17
  • $\begingroup$ The brackets were made wrong by an edit from Joe L. I have submitted a correction. $\endgroup$ – user12861 Jul 14 '12 at 15:46
  • $\begingroup$ Related, same fallacy. $\endgroup$ – Daniel Fischer Nov 19 '16 at 13:03
  • $\begingroup$ Not the same, but rather similar approach: $2+2 = 5$? error in proof $\endgroup$ – Martin Sleziak Nov 19 '16 at 13:20
11
$\begingroup$

$a^2 = b^2$ does not imply that $a = b$.

$\endgroup$
  • 3
    $\begingroup$ It only implies $|a|=|b|$, i.e. $a=\pm b$. $\endgroup$ – Martin Sleziak Jul 14 '12 at 10:48
8
$\begingroup$

$5-9/2$ is positive, $4-9/2$ is negative. They are not equal, although their squares are equal.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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