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Sig figs

So in my chemistry class, and I've been wondering,"aren't sig figs a little inaccurate"

\(\frac{2 mi}{3hr}=0.\overline{6} mph\)

Which by sig figs

\(\frac{2 mi}{3hr}=0.7 mph=\frac{7}{10} mph\)

Or even more absurdly

\(35J\cdot 1 s=40J\cdot s\)

\(\therefore 35=40\)

I do see the reason why we change numbers like \(\dfrac{1.3739}{1729.36}\) into decimal form.

But why can't we leave the pretty things in fractional form like \(\frac{2}{3}\)

Note by Trevor Arashiro
2 years, 11 months ago

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Yes, I am also posting this in protest to my last chem test where. All 4 points I got off were because of sig figs. 😕

Trevor Arashiro - 2 years, 11 months ago

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Here's an example to illustrate why we use sig figs:

Suppose you were to find the speed of your average walking in miles per hour, and you got that in \(2\) miles, you walked \(3\) hours.

Thus, you have to calculate \(2\text{ mi}/3\text{ hr}\)

This is where you are wondering why not just leave it in a pretty fraction form. This is also the part where science diverges from theoretical mathematics.

In theory, the \(2\) miles and \(3\) hours are exact. Thus, it is actually \(2.000\ldots\) miles and \(3.000\ldots\) hours.

However, in the problem, we are given only one sig fig: \(2\) miles and \(3\) hours.

Therefore the actual exact value could range from anything like \(1.55\text{ mi}/3.4\text{ hr}=0.46\text{ mph}\) to \(2.45\text{ mi}/2.5\text{hr}=0.98\text{ mph}\)

Thus, saying that the miles per hour is \(\dfrac{2}{3}=0.666\ldots\) is incorrect because the value is too precise, and does not take into regard the actual possible values.

This is why the answer is \(2\text{ mi}/3\text{ hr}=0.7\text{ mph}\) with only one sig fig.

An important thing to note here is that \(0.7\ne \dfrac{7}{10}\). Usually, when we use a fraction not as a division but as a numeral, the value is exact [citation needed], which is not the case here.

Daniel Liu - 2 years, 11 months ago

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Well, for 2/3, 0.7 is still closer to the true value than 0.6. Remember, sig figs are used for approximation, and they help preserve accuracy when conducting measurements that could otherwise become incredibly ambiguous given the different degrees of accuracy in measuring values.

Sharon Lin - 2 years, 11 months ago

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