Monday, 1 September 2014

algebra precalculus - Why does $2+2=5$ in this example?




I stumbled across the following computation proving $2+2=5$



calculation proving 2+2=5



Clearly it doesn't, but where is the mistake? I expect that it's a simple one, but I'm even simpler and don't really understand the application of the binomial form to this...


Answer



The error is in the step where the derivation goes from
$$\left(4-\frac{9}{2}\right)^2 = \left(5-\frac{9}{2}\right)^2$$
to

$$\left(4-\frac{9}{2}\right) = \left(5-\frac{9}{2}\right)$$



In general, if $a^2=b^2$ it is not necessarily true that $a=b$; all you can conclude is that either $a=b$ or $a=-b$. In this case, the latter is true, because $\left(4-\frac{9}{2}\right) = -\frac{1}{2}$ and $\left(5-\frac{9}{2}\right) = \frac{1}{2}$. Once you have written down the (false) equation $-\frac{1}{2} = \frac{1}{2}$ it is easy to derive any false conclusion you want.


No comments:

Post a Comment

real analysis - How to find $lim_{hrightarrow 0}frac{sin(ha)}{h}$

How to find $\lim_{h\rightarrow 0}\frac{\sin(ha)}{h}$ without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...