2 Rounding

In general, a computer is unable to store every decimal place of a real number. Real numbers are rounded . To round a number to $n$ significant figures we look at the ${\left(n+1\right)}^{\text{th}}$ digit in the decimal expansion of the number.

• If the ${\left(n+1\right)}^{\text{th}}$ digit is 0, 1, 2, 3 or 4 then we round down : that is, we simply chop to $n$ places. (In other words we neglect the ${\left(n+1\right)}^{\text{th}}$ digit and any digits to its right.)
• If the ${\left(n+1\right)}^{\text{th}}$ digit is 5, 6, 7, 8 or 9 then we round up : we add $1$ to the $n^{\text{th}}$ decimal place and then chop to $n$ places.

For example

$\begin{array}{ccccc}\hfill \frac{1}{3}& \hfill =\hfill & 0.3333\hfill & \hfill & \text{rounded to 4 significant figures,}\hfill \\ \hfill & \hfill \hfill & \hfill & \hfill & \hfill \\ \hfill \frac{8}{3}& \hfill =\hfill & 2.66667\hfill & \hfill & \text{rounded to 6 significant figures,}\hfill \\ \hfill & \hfill \hfill & \hfill & \hfill & \hfill \\ \hfill \pi & \hfill =\hfill & 3.142\hfill & \hfill & \text{rounded to 4 significant figures.}\hfill \end{array}$

An alternative way of stating the above is as follows

$\begin{array}{ccccc}\hfill \frac{1}{3}& \hfill =\hfill & 0.3333\hfill & \hfill & \text{rounded to 4 decimal places,}\hfill \\ \hfill & \hfill \hfill & \hfill & \hfill & \hfill \\ \hfill \frac{8}{3}& \hfill =\hfill & 2.66667\hfill & \hfill & \text{rounded to 5 decimal places,}\hfill \\ \hfill & \hfill \hfill & \hfill & \hfill & \hfill \\ \hfill \pi & \hfill =\hfill & 3.142\hfill & \hfill & \text{rounded to 3 decimal places.}\hfill \end{array}$

Sometimes the phrases “significant figures" and “decimal places" are abbreviated as “s.f." or

“sig. fig." and “d.p." respectively.

Example 1

Write down each of these numbers rounding them to 4 decimal places:

$0.12345$ , $-0.44444$ , $0.5555555$ , $0.000127351$ , $0.000005$

Solution

$0.1235$ , $-0.4444$ , $0.5556$ , $0.0001$ , $0.0000$

Example 2

Write down each of these numbers, rounding them to 4 significant figures:

$0.12345$ , $-0.44444$ , $0.5555555$ , $0.000127351$ , $25679$

Solution

$0.1235$ , $-0.4444$ , $0.5556$ , $0.0001274$ , $25680$

Task!

Write down each of these numbers, rounding them to 3 decimal places:

$0.87264$ , $0.1543$ , $0.889412$ , $-0.5555$

Answer

2.1 Rounding error

Clearly, rounding a number introduces an error. Suppose we know that some quantity $x$ is such that

$x=0.762143\text{6 d.p.}$

Based on what we know about the rounding process we can deduce that

$x=0.762143\pm 0.5\times 10^{-6}.$

This is typical of what can occur when dealing with numerical methods. We do not know what value $x$ takes, but we have an error bound describing the furthest $x$ can be from the stated value $0.762143$ . Error bounds are necessarily pessimistic. It is very likely that $x$ is closer to $0.762143$ than $0.5\times 10^{-6}$ , but we cannot assume this, we have to assume the worst case if we are to be certain that the error bound is safe.

Note that successive rounding can increase the associated rounding error, for example

2.2 Accumulated rounding error

Rounding error can sometimes grow as calculations progress. Consider these examples.

Example 3

Let $x=\frac{22}{7}$ and $y=\pi$ . It follows that, to 9 decimal places

1. Round $x$ and $y$ to 7 significant figures. Find $x+y$ and $x-y$ .
2. Round $x$ and $y$ to 3 significant figures. Find $x+y$ and $x-y$ .

Solution

1. To 7 significant figures $x=3.142857$ and $y=3.141593$ and it follows that, with this rounding of the numbers $$\begin{array}{rcll}x+y& =& 6.284450& \\ x-y& =& 0.001264.& \end{array}$$

   The outputs ( $x+y$ and $x-y$ ) are as accurate to as many decimal places as the inputs ( $x$ and $y$ ). Notice however that the difference $x-y$ is now only accurate to 4 significant figures.

2. To 3 significant figures $x=3.14$ and $y=3.14$ and it follows that, with this rounding of the numbers $$\begin{array}{rcll}x+y& =& 6.28& \\ x-y& =& 0.& \end{array}$$

   This time we have no significant figures accurate in $x-y$ .

In Example 3 there was loss of accuracy in calculating $x-y$ . This shows how rounding error can grow with even simple arithmetic operations. We need to be careful when developing numerical methods that rounding error does not grow. What follows is another case when there can be a loss of accurate significant figures.

Task!

This Task involves solving the quadratic equation

$x^2+30x+1=0$

1. Use the quadratic formula to show that the two solutions of $x^2+30x+1=0$ are $x=-15\pm \sqrt{224}$ .
2. Write down the two solutions to as many decimal places as your calculator will allow.
3. Now round $\sqrt{224}$ to 4 significant figures and recalculate the two solutions.
4. How many accurate significant figures are there in the solutions you obtained with the rounded approximation to $\sqrt{224}$ ?

Answer

Task!

In the previous Task it was found that rounding to 4 sig. fig. led to a result with a large error for the smaller root of the quadratic equation. Use the fact that for the general quadratic

$a{x}^2+bx+c=0$

the product of the two roots is $\frac{c}{a}$ to determine the smaller root with improved accuracy.

Answer