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If I have some values to use in a calculation, which all have 3 significant digits, then I know that the result will also have no more than 3 significant digits.

Am I allowed to round up/down to 3 digits during intermediate steps in the calculation? If not (which it seems is the case), then why not? Isn't any more digits just "random" inaccurate digits anyways since we started off with only 3-digit values? I do not see how these extra digits will make a difference.


If the question is not entirely clear, here is the reason I ask.

I am looking at an easy vector problem from a university physics book. My result is exactly the answer given, BUT is off by $1$ at the last (third) digit.

The question:

Follow a path $180\,\mathrm{m}$ straight west ($\vec A$), then $210\,\mathrm{m}$ in a direction $45 ^\circ$ east of south ($\vec B$), and then $280\,\mathrm{m}$ at $30^\circ$ east of north ($\vec C$). A fourth displacement $\vec D$ brings you back to start. Determine the magnitude of $\vec D$.

My answer for the magnitude is $D=144\,\mathrm{m}$. The answer in the book says $D=143\,\mathrm{m}$.

My procedure is to set up the vectors' coordinates:

$$\vec A=(-180,0)\,\mathrm{m}\\ \vec B=(B\sin 45^\circ,B\cos 45^\circ)\,\mathrm{m}=(148,-148)\,\mathrm{m}\\ \vec C=(C\sin 30^\circ,B\cos 30^\circ)\,\mathrm{m}=(140,242)\,\mathrm{m}$$

then to add them:

$$\vec D=-(\vec A+\vec B+\vec C)=-((-180,0)+(148,-148)+(140,242))\,\mathrm{m}\\ =\underline{-(108,94.0)\,\mathrm{m}}$$

and then to find the magnitude:

$$D=\sqrt{(-180\,\mathrm{m})^2+(94.0\,\mathrm{m})^2}=\underline{\underline{144\,\mathrm{m}}}$$

The issue is that I have messed with an intermediate step: I have rounded the coordinates for $\vec B$ and $\vec C$ up/down to 3 digits. If I don't and instead keep all digits in my calculator all the way and don't remove it until the end, it is correct.

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  • $\begingroup$ why do you mind the 1 last digit? IMO it doesn't even matter unless you are dead serious about significant digits. $\endgroup$ – RE60K Apr 3 '15 at 16:26
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    $\begingroup$ @ADG Sure, it doesn't matter much. But do you know that feeling of frustation that you should know such a simple rule but apparently don't? If my kidbrother ever asks, I will not know what to say... That scares me. $\endgroup$ – Steeven Apr 3 '15 at 16:34
  • $\begingroup$ the young fredman one? $\endgroup$ – RE60K Apr 3 '15 at 16:35
  • $\begingroup$ @ADG Yes, that is the physics book. $\endgroup$ – Steeven Apr 3 '15 at 16:38
  • $\begingroup$ Generally I keep all intermediate calculations one digit more accurate than my required accuracy for the final answer (in this case 4 significant figures during intermediate steps). It's really hard to introduce an error this way, so it's worked for all my classes. Essentially, the idea is that rounding the value during intermediate steps will cause an error that can compound as you do things to it. For example, $10^{440}$ is very different from $10^{400}$, even though you're only rounding down a digit. $\endgroup$ – onetoinfinity Apr 3 '15 at 16:38
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So, you would not be able to round it. It is a good rule of thumb to always let the original number limit your certainty.

Example: Add $3.1416+(2/7)$:

$3/7 = 0.4285714$

The correct way to solve would be:

$3.141(6) + .428571(4)$, making sure all other numbers are certain past the original numbers certainty (by at least 1 digit, but the more you keep the better).

Then add to give you: $3.570171$

Now if you did not do that, you would get the following: $3.1416+(2/7)$

= $3.1416+(.4286)$

= $3.5702$

which are not equivalent to each other. Generally, significant figures are prevalent in science than math . But again, just keep the rule of thumb in the beginning in mind when you do problems like that.

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