$$\\\mathbf{i}=\sqrt{-1}\\
\sqrt{ab}=\sqrt{a}\times \sqrt{b}
\\\textup{So:}\\
1+1=1+\sqrt{1}\\
=1+\sqrt{(-1)\times(-1)}\\
=1+\sqrt{-1}\times \sqrt{-1}\\
=1+\mathbf{i}\times \mathbf{i}\\
=1\times \mathbf{i}^2\\
=1+(-1)=1-1=0$$
Is there a miscalculation ? Or is 1+1 really equal to 0 ?
$$\\\:I'm\:possitive\:it\:equals\:Zero\\\You\:made\:LaTeX\:look\:like\:a\:walk\:in\:the\:park\\This\:my\:first\:time\:using\:it$$
.Well......I think that you've made a mistake....because
$${\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}} = {\mathtt{2}}$$
If you think that all that calculation are wrong, then find a miscalculation.
MathsGod1, you did a good job with that LaTex,
Here is another way you could have done it
$$\\$I'm positive it equals zero (it doesn't by the way)$\\
$made Latex look like a walk in the park$\\
$this is my first time using it$\\\\
OR\\\\
\\\text{I'm positive it equals zero (it doesn't by the way)}\\
\text{made Latex look like a walk in the park}\\
\text{ this is my first time using it}\\\\
OR\\\\
\\\mbox{I'm positive it equals zero (it doesn't by the way)}\\
\mbox{made Latex look like a walk in the park}\\
\mbox{ this is my first time using it}\\\\$$
HERE IS THE CODE :)
\\$I'm positive it equals zero (it doesn't by the way)$\\
$made Latex look like a walk in the park$\\
$this is my first time using it$\\\\
OR\\\\
\text{I'm positive it equals zero (it doesn't by the way)}\\
\text{made Latex look like a walk in the park}\\
\text{ this is my first time using it}\\\\
OR\\\\
\mbox{I'm positive it equals zero (it doesn't by the way)}\\
\mbox{made Latex look like a walk in the park}\\
\mbox{ this is my first time using it}\\\\
(you will have to take the extra line breaks out, I put them in so it is easier to read)
There are a lot of different ways of doing the same things in LaTex, I really do not know why. :/
mbox stands for message box.
$$\\\mathbf{i}=\sqrt{-1}\\
\sqrt{ab}=\sqrt{a}\times \sqrt{b}\\
\textup{So:}\\
1+1=1+\sqrt{1}\\
=1+\sqrt{(-1)\times(-1)}\qquad or \qquad 1+\sqrt{1}\times \sqrt{1}\\
=1+\sqrt{-1}\times \sqrt{-1}\qquad or \qquad 1+\sqrt{1}\times \sqrt{1} \\
=1+\mathbf{i}\times \mathbf{i}\qquad or \qquad 1+1\\
=1\times \mathbf{i}^2\qquad or \qquad 1+ 1\\
=1+(-1) \;\;or\;\; 2\\
=1-1 \;\;or\;\; 2\\
=0 \;\;or\;\; 2$$
0 can be discounted as an incorrect answer :)