I have:

$$\frac{(10.3) + (0.01345) }{ (10.3) \cdot (0.01345)}$$

The answer need to be given in scientific notation. I know the rules for addition/subtraction and multiplication/division, however, something is messing me up. Do I round before putting it in scientific notation or after?

For example, $10.3 + 0.01345$ is $10.3145$. Rounding it before would yield $10.3$ thus $1.03 \cdot 10^1$. Rounding it after would yield $1.0 \cdot 10^1$. Depending on what numbers I get here will affect my answer, and so far I've gotten $3$ different answers.

  • $\begingroup$ Depends on what your teacher has told you. "Significant figure" rules aren't really used in real science. If this were a "real" question then you would be trying to fit the following data into the following formula: $$ f(x,y) = \frac{x + y}{xy} $$ You would then figure out significant digits from the formula: $$ \Delta f \approx \left|\frac{\partial f}{\partial x}\right|\Delta x + \left|\frac{\partial f}{\partial y}\right|\Delta y $$ $\endgroup$ – Jared Oct 5 '16 at 2:10

You should round according to the rules of significant figures in the order of operations after performing each operation. For the numerator, you should add them and then round to the appropriate number of significant figures. I can't speak for actual scientists, but that's how they want you to do it in school most of the time.

  • $\begingroup$ My issue with that (and you are probably "right", btw) is that this then means you would be doing $\frac{1}{0.01345}$ and then taking three significant figures because adding $10.3$ to $0.01345$ means that the entire $0.01345$ is dominated by by $10.3$ (it's more significant). $\endgroup$ – Jared Oct 5 '16 at 2:14
  • $\begingroup$ In reality, you would probably want at least some error bounds. In school, however, you only need the rules. $\endgroup$ – AlgorithmsX Oct 5 '16 at 2:23

If you use what I said...which is almost certainly not what you "should" do...then we would have:

$$ f(x) = \frac{x + y}{xy} = \frac{1}{x} + \frac{1}{y}\\ \frac{\partial f}{\partial x} = -\frac{1}{x^2} \\ \frac{\partial f}{\partial y} = -\frac{1}{y^2} $$

This would give (assuming that each last significant digit is correct to $\pm 0.5$ of the next):

$$ \Delta f \approx \frac{1}{10.3^2}(0.05) + \frac{1}{0.01345^2}(0.000\ 005) \approx 0.028\ 110\ 495\ 86 $$

The actual result would then be:

$$ f(x) \approx \frac{10.3 + 0.01345}{10.3*0.01345} \approx 74.446\ 529\ 757\ 8 $$

This would suggest that the result is accurate to the "second" decimal:

$$ f(x) \approx 74.4(4) $$

This notation means that the results is approximately $74.44$ but that the final digit is uncertain. If, instead, you use the "rules", you will find that:

$$ 10.3 + 0.01345 \approx 10.3 $$

Then that $10.3 * 0.1345 \approx 0.139$ and finally that:

$$ \frac{10.3}{0.139} \approx 74.1 $$

If, instead, we were to use: $\frac{1}{x} + \frac{1}{y}$, we would find that:

$$ \frac{1}{10.3} \approx 0.0971\\ \frac{1}{0.01345} \approx 74.35 $$

We would then go ahead and add them, to get:

$$ 74.35 + 0.0971 \approx 74.45 $$


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