# Determine the number of correct digits in the number $x$ given its relative error $E_r$

Determine the number of correct digits in the number $x$ given its relative error $E_r$

(a): $x=0.4785, E_r=0.2\times 10^{-2}$

(b) : $x=386.4, E_r=0.3$
(c): $x=86.34, E_r=0.3$

For the problem (b), we have
$E_r=0.3<0.5=\frac{1}{2}<\frac{1}{2} \frac{10}{4}=\frac{1}{2}\frac{1}{(3+1)\times 10^{0-1}}$. So, using the following result, we can say that in this case answer is 0. Is my process correct? Please tell me.
Result : If $E_r$ be the relative error in rounding off true value $V_T$ to Approximate value $V_A$ and $$E_r<\frac{1}{2(k+1)10^{n-1}}$$ where $k$ is the first significant digit in $V_A$, then $V_A$ is correct to $n$ significant figures($n$ is a positive integer).
Thanks

• Your process is not entirely clear. – Justpassingby Dec 14 '15 at 20:25

Part I : Abs Error $=E_r\times x = 0.002\times 0.4785 = 0.000957$, then $|{x-\hat x }|= 0.000957$ =>$\hat x = .000957+.4785 = 0.4794$. Thus the number of correct digits in the number x is 2. Similarly proceed with other examples. I still don't get your convoluted logic of finding the number of correct digits.