# Problem in understanding the derivation of relationship between relative error and the number of significant numbers.

The following is excerpted from Numerical Analysis by K. Mukherjee where he discusses a theorem relating the relative error with number of significant figures:

Theorem: If the first significant figure of a number is $k$ and the number is correct to $n$ significant figures, than the relative error is less than $1/\left(k \times 10^{n-1}\right)\;.$

Proof: Let $N$ be the exact value of the number, $n$ be the number of significant figures, $m$ be the number of correct decimal places.

Three cases must be distinguished, namely $(a) ~m\lt n, ~(b)~m= n$ and $(c) ~m\gt n\;.$

## Case I:

Here the number of digits in the integral part of $N$ is $n-m\;.$ Denoting the first significant digit of $N$ by $k,$ we have \begin{align}E_\textrm{a} &\equiv \textrm{absolute-error} \leq \frac12 \cdot 10^{-m},\\ N & \geq \color{red}{k\cdot 10^{n-m-1}-\frac12 \cdot 10^{-m}} \;.\end{align}

Hence \begin{align}E_\textrm r &\equiv \textrm{relative-error}\leq \frac{\frac{1}{2}\cdot 10^{-m}}{\color{red}{k\cdot 10^{n-m-1}-\frac12 \cdot 10^{-m}}} = \ldots \end{align}

## Case II:

$N$ is a pure fraction and $k$ is the first digit after the decimal point.

Then \begin{align}E_\textrm r& \leq \frac{\frac{1}{2}\cdot 10^{-m}}{\color{red}{k\cdot 10^{-1}-\frac12 \cdot 10^{-m}}} = \ldots \end{align}

## Case III:

In this case, $k$ occupies the $(m-n+1)$th decimal place and therefore \begin{align}E_\textrm r& \leq \frac{\frac{1}{2}\cdot 10^{-m}}{\color{red}{k\cdot 10^{n-m-1}-\frac12 \cdot 10^{-m}}} = \ldots \end{align}

[...]

I'm having some problem in getting through the red marked terms above.

$\bullet$ How did the author deduce $$N \geq k\cdot 10^{n-m-1}-\frac12 \cdot 10^{-m}\;?$$

Also, he showed the application of the theorem by providing illustrations; one of them is as follows:

Example: Let the number $271.37$ be correct to five significant figures.

$$E_\textrm a \leq 0.01\times 0.5 = 0.005\;.$$

For relative error, we have $$E_\textrm r = \frac{E_\textrm a}{\textrm{True Value}}\leq \frac{0.005}{271.37 \color{red}- 0.005}\ldots$$

$\bullet$ I didn't get why there is $-$ sign in the denominator above; after all, $$\textrm{True Value} =\textrm{Corrected value} + E_\textrm a$$ which implies $$\textrm{True Value} \leq 271.37 + 0.005\;.$$ Isn't it so? Or, am I mistaking somewhere in my thinking?

• Has the OP now got answer to the question that he asked above? – user587389 Oct 14 '19 at 14:28

## 1 Answer

In case absolute error means $$E_\textrm a:=|\textrm{True Value} -\textrm{Corrected value}|$$ thus considering either rounding up or down, it must be $$\textrm{True Value} =\textrm{Corrected value} \pm E_\textrm a$$ so we have

$$E_\textrm r = \frac{E_\textrm a}{\textrm{True Value}}=\frac{E_\textrm a}{\textrm{Corrected value} \pm E_\textrm a}\leq\frac{E_\textrm a}{\textrm{Corrected value} - E_\textrm a}$$

• Ah! Thanks for the answer; I didn't know $E_\textrm a$ is defined so; my book only mentioned the $-$ one. I'm getting a bit of it then. +1. – user142971 Sep 8 '16 at 9:29