Difference between revisions of "User:Tonypr"
| Line 4: | Line 4: | ||
Proof <math>2=1</math>: | Proof <math>2=1</math>: | ||
| − | |||
| − | + | Suppose that <math>a=b</math> | |
| − | + | <math> | |
| − | + | \begin{eqnarray*} | |
| − | + | a&=&b\\ | |
| − | + | a^2&=&ab\\ | |
| − | + | a^2-b^2&=&ab-b^2\\ | |
| − | + | (a+b)(a-b)&=&b(a-b)\\ | |
| − | + | a+b&=&b\\ | |
| − | + | b+b&=&b\\ | |
| − | + | 2b&=&b\\ | |
| − | + | 2&=&1 | |
| − | + | \end{eqnarray*} | |
| + | </math> | ||
Revision as of 22:39, 23 May 2009
Awesome math solver from Puerto Rico, soon to be International Math Olympiad.
Proof 
:
Suppose that 
$\begin{eqnarray*} a&=&b\\ a^2&=&ab\\ a^2-b^2&=&ab-b^2\\ (a+b)(a-b)&=&b(a-b)\\ a+b&=&b\\ b+b&=&b\\ 2b&=&b\\ 2&=&1 \end{eqnarray*}$ (Error compiling LaTeX. Unknown error_msg)
