So, I got this number $1731.9146$ and I need to round it to $2$ decimal places. The answer should be $1731.91$, but I had always thought, for some reason, that the answer (if precise) should be $1731.92$.

I might have a misconception, but I always thought that a precise rounding had to consider all the decimals, and not only the next integer. Shouldn't I start rounding decimal by decimal until I reach to the desired place? For example.

  1. $1731.9146$ → Round the penultimate decimal
  2. $1731.915$ → Follow to the next decimal
  3. $1731.92$ → End, since we reached the desired decimal place

Am I might wrong in my believe? If I am wrong, any idea where I could of have gotten this misconception? I been looking for a while and it seems I am wrong, but I don't know why I can shake the feeling I got this from some chemistry class in my Uni.

  • 2
    $\begingroup$ You are wrong in your method. That is what you should not do. Instead round all at once: if you have $1731.915 \le x \lt 1731.925$ then round to $1731.92$ while if you have $1731.905 \le x \lt 1731.915$ then round to $1731.91$. Here $x=1731.9146 < 1731.915$ so round to $1731.91$ $\endgroup$
    – Henry
    Jan 27 at 23:42
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    $\begingroup$ In particular, to "consider all the decimals" like that incorrectly rounds up all numbers with fractional part between 4/9 and 1/2. It may also be worth thinking about "rounding" in this way but in binary instead of decimal: then, any fractional part not exactly equal to zero will be "rounded" to 1. $\endgroup$ Jan 28 at 7:16

1 Answer 1


Generally speaking, "rounding" a number means mapping that number to its nearest neighbor in a smaller set of numbers. The goal is the minimize the error between the original number and the approximation obtained by rounding.

For example, if I want to round $\pi$ to the nearest integer, I might note that $\pi$ is between $3$ and $4$, and then further note that $\pi$ is closer to $3$ than to $4$: roughly speaking, $$ |\pi - 3| \approx 0.14 \qquad\text{and}\qquad |\pi - 4| \approx 0.86, $$ where $|a-b|$ (the absolute value of the difference) gives the distance between two numbers. Since $0.14 < 0.86$, $\pi$ is rounded down to $3$.

Something similar occurs in the example given in the original question. The goal is to round a number to the nearest hundredth. We might immediately note that $$1731.91 < 1731.9146 < 1731.92, $$ so it will get rounded to one of those two numbers. Then observe that $$ |1731.9146 - 1731.91| = 0.0046 \qquad\text{and}\qquad |1731.9146 - 1731.92| = 0.0054. $$ Since $0.0046 < 0.0054$, the number is rounded to $1731.91$, as this minimizes the error between the original number and the approximation.

As to why it is sufficient to look only at the digit that is directly to the right of the desired level of precision, the following may be helpful:

To simplify the discussion, suppose that the goal is to round a number between $0$ and $1$ to either $0$ or $1$. The same idea can be extended to rounding numbers to the nearest integer, or to other decimal places. On the number line, these are the numbers

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Any number $N < 1/2$ rounds down to $0$, and any number $N > 1/2$ rounds up to $1$ (if $N$ happens to be exactly $1/2$, we have to make a decision about how to round it—the usual convention is to always round $0.5$ up to $1$, but there are other conventions, e.g. banker's rounding).

Notice that if $N < 1/2$, then $N$ can be written in decimal form as $$ 0.dxxx\dotso, $$ where $d$ (the digit in the tenths place) is less than $5$. The remaining digits farther to the right are irrelevant, and could be anything at all. Thus if the digit in the tenths place is less than $5$, round down.

On the other hand, if $N > 1/2$, then $N$ can be written in decimal form as $$ 0.dxxx\dotso, $$ where $d$ is at least $5$ (greater than or equal to $5$). Thus if the digit in the tenths place is $5$ or greater, round up.

  • $\begingroup$ That's a great answer. Just to complement it: as far as the procedure goes, you just see that the digit is $4$ and therefore we round down. There's nothing more to it. The fact that in $.914$ the third digit is $4$ means we're going to be closer to $.91$ than to $.92$. $\endgroup$
    – Snaw
    Jan 27 at 23:41
  • $\begingroup$ @Snaw I included some discussion. $\endgroup$
    – Xander Henderson
    Jan 28 at 3:53
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    $\begingroup$ Technically, you can't round a decimal just by looking at finitely many digits. For example, if for each $k∈ℕ^+$ the $k$-th digit is $9$ if $[(k-1)^2,k^2]$ includes a prime and is $4$ otherwise, then the digits are $0.4999...$ and you (probably) cannot provably determine whether rounding to integer (i.e. $0$ decimal places) gives $0$ or $1$. But practically, this is irrelevant to the asker's conceptual issue so it's just for your laughs. =P $\endgroup$
    – user21820
    Jan 28 at 8:15

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