PROVE 1 + 1 = 3 AND IF YOU FIND IT DIFFICULT, SEE MY SOLUTION ATTACHED

$-2 \times -1$ = $2$

taking log to the base ten on both sides we get

$log -2 + log -1$ = $log 2$

$-log \frac{1}{2} + log -1$ = $log 2$ SINCE $log -a$ = $- log \frac{1}{a}$

$log -1$ = $log 2 + log \frac{1}{2}$

this can be written as

$log -1$ = $log \frac{2}{2}$

$log -1$ = $log 1$

this implies $log -1^{3}$ = $log -1^{2}$ since $-1^{3}$ = $-1$ AND $-1^{2}$ = $1$

$3$ $log -1$ = $2$ $log -1$

dividing by log -1 to the base ten on both sides,we get

$3$ = $2$

$3$ = $1$ + $1$

sorry this is a wrong proof

Easy Math Editor

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Comments

Note that the logarithm formula is $-\log \frac {1}{a} = \log a$ , not "SINCE $\log -a = - \log \frac {1}{a}$ " as you claimed.

Calvin Lin Staff - 8 years, 3 months ago

thanks for correcting my mistake

Revankumar Gnanavel - 8 years, 3 months ago

i appreicate ur attempt for doing this,i request u think more like this....... (eppadi da room pottu yosipeengaloooo...)-just for fun

Pradeep Ravichandran - 8 years, 3 months ago

but you're using a number system where 1 plus 1 DOES equal 2 to prove something that contradicts this system... so no.

Alan Liang - 8 years, 3 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$