Should I keep significant figures constant when multiplying a number with infinite significant figures by a number with finite significant figures? I want to multiply $\frac{1}{2}$ with $1.88 \times 10^{-2}$. The $\frac{1}{2}$ is exact and the $1.88 \times 10^{-2}$ is not.
Should I write the answer as $9.4 \times 10^{-3}$ or $9.40 \times 10^{-3}$?
I know that when multiplying by $\frac{1}{2}$, the uncertainty should also halve, but my gut instinct tells me that $9.40 \times 10^{-3}$ is not right. What is confusing me further is that when I find $6 \times 1.88 \times 10^{-2}$, where 6 is known exactly, then the answer $1.128 \times 10^{-1}$ seems like its not right to me (again, gut instinct), and that $1.13 \times 10^{-1}$ is correct.
So, is $\frac{1}{2} \times 1.88 \times 10^{-2} = 9.4 \times 10^{-3}$
or $9.40 \times 10^{-3}$?
And
is $6 \times 1.88 \times 10^{-2} = 1.128 \times 10^{-1}$ or $1.13 \times 10^{-1}$?
And why the two correct answers (correct as per the rules of significant figures) are correct and why the other two are wrong? Clarification would be very helpful. Thanks!
 A: It depends on how rigorous you want to be. Significant figures are only ever a rule of thumb.
Suppose we just use the rules of significant figures, and apply them to the first example. We could ask: What happens if we replace $\frac{1}{2}$ with $0.500$? Or $0.5000$? Or $0.50000000000000000$? Or $0.5000000000000000000000000000000000000$? The answer is always the same: A product has the same number of significant figures as the factor with the least number of significant figures, which, in this case, is $1.88 \times 10^{-2}$. In other words, the product is $9.40 \times 10^{-3}$, with three significant figures.
When we make the factor $\frac{1}{2}$ exact (infinitely many significant figures), this reasoning still applies, with the same result.
With the second example, the same reasoning also applies, giving an answer of $1.13 \times 10^{-1}$.
But if we want to be more rigorous, we can apply the reasoning in Toffomat’s comment (spelling and values corrected):

You probably want to express that your value is larger than $1.87 \times 10^{-2}$ and smaller than $1.89 \times 10^{-2}$, so half of it is presumably larger than $0.935 \times 10^{-2}$ and smaller than $0.945 \times 10^{-2}$.

This gives an answer of $9.4 \times 10^{-3}$, with two significant figures.
Why the difference? To understand this, we must first understand how significant figures work as a rule of thumb.
When a value has three significant figures, it basically means that the third digit has an error of about $1$, ie the relative error is about $10^{-2}$. (For $1.00 \pm 0.01$, the relative error is exactly $10^{-2}$.) Multiplying by an exact value does not change the relative error, so the product’s relative error should also be about $10^{-2}$, and the product should also have three significant figures.
In this case, the mantissa we were given ($1.88$) is close to $1$, so this rule of thumb works fairly well here. But the mantissa of the product ($9.40$) is much bigger than $1$; in fact, we could say it is a bit less than $10$. So applying the same reasoning as above, writing it with three significant figures would be like saying the relative error is about $10^{-3}$. Since the relative error is actually about $10$ times bigger than this, the more rigorous analysis concludes that there should be one less significant figure.
For the second example, both mantissas are close to $1$, so we expect the more rigorous analysis to give the same answer as our significant figures rules of thumb. Let us check:
\begin{align}
6 \times (1.88 \pm 0.01) \times 10^{-2} &= (1.128 \pm 0.006) \times 10^{-1} \\
&\approx (1.13 \pm 0.01) \times 10^{-1} \\
&= 1.13 \times 10^{-1}
\end{align}
