What is Relative Error of equation $2.3(4.18-3.24)$ I've trying to solve it and here's my procedure:
$e1\:=\:\frac{1}{2}\cdot 10^{-1}\:=\:5\cdot 10^{-2}$
$e2\:=\:\frac{1}{2}\cdot 10^{-2}\:=\:5\cdot 10^{-3}$
$e3\:=\:\frac{1}{2}\cdot 10^{-2}\:=\:5\cdot 10^{-3}$
$Relative\:Error\:=\:\frac{e1}{2.3}\:+\:\frac{e2}{4.18}\:+\:\frac{e3}{3.24} = 0.024478512$
But the solution on my book is: 
$Relative Error: 0.03238$
Where am I missing?
 A: Hint
The exercise is done in order you combine absolute errors and relative errors.
You are considering $$x=a\times(b-c)=a\times d$$
$$d=b-c=(4.18\pm 5\times 10^{-3})-(3.24\pm 5\times 10^{-3})=0.94\pm 2\times 5\times 10^{-3}=0.94\pm 10^{-2}$$ So, now $$\frac{\Delta x}x=\frac{5\times 10^{-2}}{2.3}+\frac{ 10^{-2}}{0.94}=0.03238$$ 
Edit
With regard to the problem you wrote in comments, now, considering $$y=\frac{b-c}a=\frac da$$ $$d=b-c=(1.3384\pm 5\times 10^{-5})-(2.038\pm 5\times 10^{-4})=-0.6996\pm 5.5\times 10^{-4}$$ Now, the same as before $$\frac{\Delta y}y=\frac{5\times 10^{-4}}{4.577}+\frac{5.5\times 10^{-4}}{0.6996}=0.0008954$$ (this is exactly your result); this makes which makes $$\Delta y=0.00013686$$ which is exactly your result.
Either the book answer is wrong or we are both wrong (and I do prefer the last hypothesis !). More than likely, there is a typo in the book answer (it should be $0.13\times 10^{-3}$ and not $0.23\times 10^{-3}$).
Good to see that you made it correctly !
