Why do negative integers not equal their corresponding positive integers?

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

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

• If you owe the bank a hundred bucks, is that the same thing as having a hundred bucks in savings? – Deepak Sep 21 '16 at 10:46
• You should write this flow in the exact opposite direction to what you wrote (i.e., start with $x=x$, not with $x=-x$). – barak manos Sep 21 '16 at 10:47
• I see what you mean and I know this logic must be wrong but where? – Michael Lee Sep 21 '16 at 10:47
• Okay now I understand. – Michael Lee Sep 21 '16 at 10:49
• 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. – Michael Lee Sep 21 '16 at 10:59

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.
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.