2
$\begingroup$

According to my logic, $1=(-1)$, $2=(-2)$, $3=(-3)$, etc. This can't be right but please tell me where I went wrong.

$$\begin{array}{l} {\rm{Suppose}}\\ x = \left( { - x} \right)\\ {x^2} = {\left( { - x} \right)^2}\\ {x^2} = {x^2}\\ \sqrt {{x^2}} = \sqrt {{x^2}} \\ x = x \end{array} % MathType!MTEF!2!1!+- % faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbqedmvETj % 2BSbqefm0B1jxALjharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0x % bbL8FesqqrFfpeea0xe9Lq-Jc9vqaqpepm0xbba9pwe9Q8fs0-yqaq % pepae9pg0FirpepeKkFr0xfr-xfr-xb9Gqpi0dc9adbaqaaeGaciGa % aiaabeqaamaabaabaaGceaqabeaacaqGtbGaaeyDaiaabchacaqGWb % Gaae4BaiaabohacaqGLbaabaGaamiEaiabg2da9maabmaabaGaeyOe % I0IaamiEaaGaayjkaiaawMcaaaqaaiaabofacaqGXbGaaeyDaiaabg % gacaqGYbGaaeyzaiaabccacaqGIbGaae4BaiaabshacaqGObGaaeii % aiaabohacaqGPbGaaeizaiaabwgacaqGZbaabaGaamiEamaaCaaale % qabaGaaGOmaaaakiabg2da9maabmaabaGaeyOeI0IaamiEaaGaayjk % aiaawMcaamaaCaaaleqabaGaaGOmaaaaaOqaaiaadIhadaahaaWcbe % qaaiaaikdaaaGccqGH9aqpcaWG4bWaaWbaaSqabeaacaaIYaaaaaGc % baWaaOaaaeaacaWG4bWaaWbaaSqabeaacaaIYaaaaaqabaGccqGH9a % qpdaGcaaqaaiaadIhadaahaaWcbeqaaiaaikdaaaaabeaaaOqaaiaa % dIhacqGH9aqpcaWG4baaaaa!5D90! $$

$\endgroup$
8
  • 1
    $\begingroup$ If you owe the bank a hundred bucks, is that the same thing as having a hundred bucks in savings? $\endgroup$ – Deepak Sep 21 '16 at 10:46
  • 2
    $\begingroup$ You should write this flow in the exact opposite direction to what you wrote (i.e., start with $x=x$, not with $x=-x$). $\endgroup$ – barak manos Sep 21 '16 at 10:47
  • $\begingroup$ I see what you mean and I know this logic must be wrong but where? $\endgroup$ – Michael Lee Sep 21 '16 at 10:47
  • $\begingroup$ Okay now I understand. $\endgroup$ – Michael Lee Sep 21 '16 at 10:49
  • $\begingroup$ I find this question comical and thus worthy of being posted; essentially I started from an equation that isn't true to begin with and thus anything I derive with it is not true as well. $\endgroup$ – Michael Lee Sep 21 '16 at 10:59
4
$\begingroup$

You start with assuming that $ x = (-x)$. That possible, but if we solve this we get $$x = -x$$ $$x + x = 0$$ $$2x = 0 $$ $$ x = 0 $$ So $x = 0$. Now you can see why the rest holds.

$\endgroup$
0
$\begingroup$

If you delete your third equation and adjust the following equations accordingly, you go from $x^2 = (-x)^2$ to $\sqrt{x^2} = \sqrt{(-x)^2}$. This makes it more obvious that you have been led astray by a false equivalence. Kind of like what is going on with the Trump and Clinton Foundations.

$\endgroup$

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.