$$\\\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 ?

EinsteinJr
May 2, 2015

#1**0 **

$$\\\:I'm\:possitive\:it\:equals\:Zero\\\You\:made\:LaTeX\:look\:like\:a\:walk\:in\:the\:park\\This\:my\:first\:time\:using\:it$$

MathsGod1
May 2, 2015

#2**+5 **

**Well......I think that you've made a mistake....because**

$${\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}} = {\mathtt{2}}$$

Hawkeye
May 2, 2015

#3**0 **

If you think that all that calculation are wrong, then find a miscalculation.

EinsteinJr
May 2, 2015

#4**+5 **

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.

Melody
May 3, 2015

#5**+5 **

$$\\\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 :)

Melody
May 3, 2015