Do you round off insignificant digits in the middle of a calculation? I have a question... Do you round with significant digits during each subcalculation of a problem or only when the entire problem is complete?
Example:
multiply the following number:
$$1.8 \times 2.01 \times 1.542$$
saving rounding until the end:
$$(1.8 \times 2.10) \times (1.542) = (3.78)\times(1.542) = (5.82876) \to 5.8$$
rounding at each sub-calculation:
$$(1.8 \times 2.10) \times (1.542) = (3.8)\times(1.542) = (5.8596) \to 5.9$$
I also have the strong feeling that if you round at each sub-calculation then multiplication is no longer commutative (although after experiencing matrices that no longer seems to be too much of a problem)
 A: The purpose of rounding your sig figs is so that you don't miscommunicate to other people the precision of your result. Intermediate results aren't going to be communicated to anyone else, so that reason for rounding doesn't apply to them. You don't want to round too much at intermediate steps, because rounding errors can accumulate.
Sometimes people will say not to round at all at intermediate steps. This is wrong and in fact usually impossible. A calculator only has a finite number of digits of precision. If you calculate $\sqrt{17}$ at some intermediate step, you have to round it, because it can't be expressed as an exact decimal. It's also ridiculous to write down intermediate results on a piece of paper with 8 or 10 sig figs when you're doing a 2-sig-fig problem. It's a waste of time, because almost all of those digits are illusory precision.
The correct advice is not to round too much at intermediate steps. Usually it makes sense to maintain one or two extra sig figs during a calculation.
For example, suppose you're doing a calculation that involves 5 multiplications in a row. If you round off the result of each multiplication to one more sig fig than you intend to keep at the end, then each rounding error is about 1/10 of the size you expect to be significant at the end. Accumulating 5 of these rounding errors (probably some positive and some negative) is still a smaller error than what you expect to be significant at the end.
I suppose the nonsensical advice not to round at all comes from the assumption that the entire calculation will be done by punching buttons on a hand-held calculator, without ever writing anything down at all. Even in this style of calculation, you're still rounding -- you're just doing very little rounding. And this style of calculation is not typically very smart for a long, complex calculation, because there's no way to check for errors. It would be smarter to write down at least some intermediate results and check them. Check that they are of the right order of magnitude, have the right sign, have the right units, match up with reality, etc.
A: Significant digits is a convention that only affects how you write numbers, not what the numbers actually are. So you only round when you are asked to drop down to a given number of significant digits - that is, at the end.
Think of it like this: there's a difference between a number, which is an abstract idea, and a written representation of a number. Some numbers have exact written representations; all numbers have approximate written representations, which represent another, nearby number. For example, the notation $5.82876$ is an exact representation of a particular number, and $5.8$ is an approximate written representation, to two significant figures, of the same number. $5.8$ is also an approximate written representation (to two significant digits) of many other numbers, such as $5.810394$ and $5.79928129$. This is the idea behind uncertainty, and significant digits: if you are given the written representation $5.8$, you don't know which actual number it represents - it could be anything between $5.75$ and $5.85$. The only exception is if you are told that $5.8$ is an exact representation, which uniquely specifies which number you are supposed to take it to mean.
When you calculate the product $1.8\times 2.01\times 1.542$, you start with three written representations which you are supposed to assume are exact. Then you multiply the first two of them, and get a number which is exactly represented by the notation $3.78$. Now, it's true that $3.8$ is an approximate written representation of that number. But does that fact change what the number is? No. If you do the intermediate rounding, you're effectively deciding to replace one number, the one which is exactly represented by $3.78$, which another number, the one which is exactly represented by $3.8$. And the operation "replace one number with another number" is not part of the mathematical expression you are supposed to simplify. So don't do it.
A: In Numeric Methods, for algorithm development, at each step, the value is rounded to significant digits.
Round-off errors, experimental errors, and truncating errors are the reasons why final results of computations of unknown quantities are "approximations". But let's say, we take an example from Kreyszig's Advanced Engineering Mathematics, if you use 4S in the computation to find the roots of x^2 - 40x + 2 = 0, x2 = 1/2a * (-b - square root((b^2) - 4ac)) and x1 = 1/2a * (-b + square root((b^2) - 4ac)). x2 = 20 - 19.95 = 0.05. The rounding to 4S is implemented at each step. Otherwise the answer would have been 0.05006. The solution to this round off error is not to stop rounding at each step. The solution is to change the algorithm. For instance, once x2 is set equal to c/ax1, the result is 2.000/39.95 = 0.05006. 
A: When I see something as
$$1.8 \times 2.01 \times 1.542$$
in a physics problem, I don't assume that those three numbers are exact. At contrary I would apply the usual assumption that the last digits are uncertain. That is, that in reality the problem is
$$(1.8 \pm 0.1) \times (2.01 \pm 0.01) \times (1.542 \pm 0.001)$$
This generates an interval of possibles values, and both $5.8$ and $5.9$ fall within the interval, because the highest uncertainty is $\pm 0.1$.
