Round off the following numbers to four places of decimal.

Question

7.141552 → 7.1416
6.814259 → 6.8142
7.864651 → 7.8646
11.464752 → 11.4648

I have followed, "If this discarded digit is exactly half a unit in the n th place, increase the n th digit by unity if it is odd, otherwise leave it unchanged.".

Is this correct? I have exposed this question because of my obscurity in the round-off theories especially when a discarded digit is five.

Answer 1

I mostly summarize what's there: https://www.ibm.com/support/pages/how-does-rounding-work-calculation

In your examples the discarded $5$ is followed by other digits, thus the rule you cited doesn't apply :

• $0.000051,\ 0.000059,\ 0.000051,\ 0.000052>0.00005$

And you need always to increase the $n^{th}$ digit for these figures.

The reason for the odd/even rule is to get $50\%$ of digits rounded above and $50\%$ rounded below when the last discarded digit is $5$, else there would be a bias always in the same direction ($n^{th}$ digit either always increasing, either always unchanged) and that would accumulate errors, while we hope that with the alternating method it will statistically compensate.

Answer 2

The "discarded digit is exactly half a unit in the n th place, increase the n th digit by unity if it is odd" exists for the rare cases where then number being rounded is (so far as we can tell) exactly half way between the result of rounding up and the result of rounding down. In base $10$ this will only occur if the digit in the position you are rounding is $5$ and if there are any digits following it they are all $0$.

None of your cases does this occur.

In all other cases you simply round up or down according to which ever is closer.

Example 1: $7.141\color{blue}5\color{red}{52}$. In this case, the $\color{red}{52}$ must either be rounded up to $\color{red}{100}$ or down to $\color{red}{000}$ but as $52 > 50=\frac 12\times 100$ we round up to $7.141\color{orange}6$

And so on. $6.814\color{blue}2\color{red}{59} \to 6.814\color{orange}3$ because $\color{red}{59}$ is closer to $\color{red}{100}$ than to $\color{red}{000}$.

and so on.

(Hmm, in none of the cases cited to you round down. I guess that is because they you deliberately trying to trip you up with the "five" rule. So all of these cases then next digit is $5$. But as the all have non-zero digits later the are all more then $50$ so they all round up.)

The only times this won't work is if you have something like $9.183\color{blue}6\color{red}{50}$. In this case you have $\color{red}{50}$ is EXACTLY half way between $\color{red}{000}$ and $\color{red}{100}$ and you have utterly no way of knowing which direction to go.

So what do you d? It doesn't matter as neither is better or worse. So you flip a coin.

..... or some guy in an office in the 1940s arbitrarily makes up a rule that if the number in being looked at is odd you round up, and if it is even you round down? Why? No reason at all, it's this way you round up half the time and down half the time. That's all.

So $9.183\color{blue}6\color{red}{50}\to 9.183\color{orange}6$. And $6.927\color{blue}3\color{red}{50}\to 6.927\color{orange}4$.