Question

here's an interesting thing
a disprove question
find the error in the following proof

if we know that
2+2=4 then
$2+2=4- \frac{9}{2} + \frac{9}{2}$
$2+2= \frac{9}{2} + 4- \frac{9}{2}$
$2+2= \frac{9}{2} + \sqrt{(4- \frac{9}{2})^2}$
$2+2= \frac{9}{2} + \sqrt{16-2*4* \frac{9}{2}+( \frac{9}{2})^2}$
$2+2= \frac{9}{2} +\sqrt{16-36+( \frac{9}{2})^2}$
$2+2= \frac{9}{2} + \sqrt{-20+( \frac{9}{2})^2}$
$2+2=\frac{9}{2} + \sqrt{25-45+( \frac{9}{2})^2}$
$2+2= \frac{9}{2} + \sqrt{5^2-2*5*\frac{9}{2}+( \frac{9}{2})^2}$
$2+2= \frac{9}{2} + \sqrt{(5- \frac{9}{2})^2}$
$2+2= \frac{9}{2} +5- \frac{9}{2}$
$2+2= 5+\frac{9}{2}- \frac{9}{2}$
$2+2= 5$

Answer 1

In step 3, when you combine the two terms, square them, and square root them, the squaring and square rooting messes everything up.  Lets see what the right side is equal to throughout.
$4- \frac{9}{2} + \frac{9}{2} = 4$
$\frac{9}{2}+ 4- \frac{9}{2}=4$
Now, since the $\frac{9}{2}$ isn't going to change value throughout the whole problem, lets simply look at the other two numbers on that side.
$4- \frac{9}{2}=-.5$
$\sqrt{(4- \frac{9}{2})^{2} } = \sqrt{ (-0.5)^{2} } = \sqrt{0.25 }=0.5 \ or -0.5$
In this step, the process of squaring and square rooting takes the number from -0.5 and turns it into 0.5.  Since square roots are either positive or negative, this proof squares and square roots the initial number, allowing the value to be changed from the value we started with.  Thus, instead of picking the -0.5, which makes the statement true, the proof uses the number 1 larger, 0.5.  This is a very sneakily hidden way to increase the value of the number by one.

This is actually one of the smartest and the second hardest false proofs I have ever attempted to disprove.  Only one has been harder to solve, and I am yet to solve it.  I enjoyed the challenge!

Hope this was the answer you were looking for!

Answer 2

I hope this helps you