1
$\begingroup$

My book mentions:

The proper use of significant figures in addition and subtraction involves a comparison of only the absolute uncertainties of numbers. This means that only as many digits are retained to the right of the decimal point in the answer as the number with the fewest digits to right of the decimal

But I don't understand that why must the answer be rounded off after addition? Is the answer not accurate enough to be retained as it is ?

$\endgroup$
4
$\begingroup$

No. The answer isn't accurate enough. Say, for example I wish to add two numbers: $1.23$ and $2.367$.

I proceed like this, but I've no idea what the third digit after the decimal in $1.23$ is and so I represent it as $?$.

$$\begin{align*} 1 &.23? \\ + 2 &.367 \\ \hline 3 &.59? \\ \hline \end{align*}$$

This is why the end result is rounded off with the number of digits equal to the number having the fewest number of digits you initially set off with.

$\endgroup$
  • $\begingroup$ But in addition we add the extra digit to the answer like 1kg is added to 1kg and 2g we get 2kg and 1g $\endgroup$ – AksaK May 14 '17 at 8:19
  • 2
    $\begingroup$ @AksaK Ah, in that case it was because the weight "1kg" was accurately known. $\endgroup$ – Berry Holmes May 14 '17 at 8:20
  • $\begingroup$ Let one rod is measured 1.23m and another rod is measured 2.367m. We combine them into a single rod of the length which is combination of both i.e 3.597m so why do we need to round off this number which gives the accurate length? $\endgroup$ – AksaK May 14 '17 at 8:33
  • 3
    $\begingroup$ As @Barry Holmes has explained, in the 1.23 m rod we do not know if it is 1.231 m, 1.239 m or 1.235 m etc. but we do know the longer one is 2.367 m and so we have to do something about our ignorance of the true length of the shorter rod, and this is rounding. $\endgroup$ – porphyrin May 14 '17 at 9:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.