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I found this other question that deals with this somewhat, but I am still unclear as to why the rules for adding/subtracting and multiplying/dividing significant figures are the way they are.

In the linked question above, the response to why addition rule was true was:

If you are trying to add x and y but are in fact adding $x+δ$ and $y+ϵ$ then the absolute error will be $δ+ϵ$, which will be dominated by the larger of the absolute errors, most likely by the absolute error in the number with the fewer decimal places. Subtraction is similar.

However, here I do not understand why the number with the fewer decimal places contributes the largest error. Why can't it be the number with the fewest significant figures?

Similarly, in proving multiplication the answerer stated

If you are trying to multiply $x$ and $y$ but are in fact multiplying $x×(1+δ)$ and $y×(1+ϵ)$ then the relative error will be $(1+δ)(1+ϵ)−1=δ+ϵ+δϵ$, which will be dominated by the larger of the relative errors, most likely by the by the relative error in the number with the fewer significant figures. Division is similar.

However, here why is $x×(1+δ)$ used instead of using $x+δ$ which was used in the addition part? Further, in this case why is the largest relative error contributed by the number with the fewest significant figures, not fewest decimal places as in addition?

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For addition, imagine adding $1.3$ to $1.0002$ The $1.3$ has a potential error of $\pm 0.05$ (assuming rounding), while the $1.0002$ has a potential error of $\pm 0.00005$ So the sum is $(1.3 \pm 0.05) + (1.0002 \pm 0.00005)=2.3002 \pm 0.05005$ We then notice that the last $0.0002$ on the sum and the last $0.00005$ on the error are insignificant, so report $2.3 \pm 0.05$. The error is precisely the error on the least accurate term-the one with fewer decimal places. It can't always be the one with fewer significant figures-go through the same with $123456.3$ and $1.0002$ The former has more significant figures, but it also dominates the error.

For multiplication, using $x(1+ \delta)$ is quite different from $x \pm \delta$. The two $\delta$s are very different. Generally we expect (using fractional errors) that $\delta , \epsilon \ll 1$. For absolute errors, there is no reason to expect that $\delta, \epsilon \ll 1$, you just expect that $\delta \ll x, \epsilon \ll y$.

Then $x(1 \pm \delta)y(1 \pm \epsilon)=xy(1\pm \delta \pm \epsilon \pm \delta \epsilon)$. The last term is quadratic in small numbers and can be ignored. When you use absolute errors (the $x \pm \delta$).

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  • But why do you switch from absolute error to fractional error when doing multiplication? Why can't you use absolute error there also? – 1110101001 Aug 27 '14 at 04:36
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    Because relative error is more useful to get the final result. If the factors have very different magnitudes it becomes hard to compare $x \epsilon$ and $y \delta$ to $xy$ but you can just add the relative errors to get the error in the product. – Ross Millikan Aug 27 '14 at 04:39
  • Ok, understood that. Now, for the multiplication part why is $δ+ϵ+δϵ$, which will be dominated by the larger of the relative errors, determined by the relative error in the number with the fewer significant figures instead of the one with the fewest decimal places? – 1110101001 Aug 27 '14 at 04:53
  • Think about multiplying $1.3 \cdot 1.0002$ The relative error on $1.3$ is $\frac {0.05}{1.3} \approx 0.04$ The relative error on $1.0002$ is much smaller. It would still be true if we were multiplying $0.0000013$ by $10002$ The relative error on the first is much larger. – Ross Millikan Aug 27 '14 at 13:39