# Are zeros to the right really significant? [duplicate]

I was completely shocked. Both scientific notation numbers when converted to integer would represent same quantity of $$1200 \text{kg}$$ or $$1.2 \text {metric tonnes}$$.

Question is, How can same number be represented with different amount of significant figures?

And second question is How much really are significant figures in these examples?

I would opt for 2 in both cases, because for me it doesn't matter if zeros are leading or trailing. In both cases they can be reduced choosing appropriate scale.

• The two extra zeros in 1.200 are telling you about the precision of the measurement. Both 1.201 and 1.200 will be rounded off to 1.2 (if one wants to round off to one decimal place), but they are distinct numbers and hence the difference. Commented May 18 at 16:34
• Do not change the conditions, it's just 1.200 without 1 in the end. Your case would be different sure. Commented May 18 at 16:37
• The problem, as I see it, is that for values with no decimal part in a given unit, the "trailing* zeroes (to the right) have ambiguous significance. For me, the difference between 1.2 kg and 1.200 kg is crystal clear. But if you just tell me 1200 g, I don't know if the two zeroes on the right are significant or just an artefact of how we write numbers in the "place system" of decimal numbers. Commented May 18 at 16:53
• You might be misunderstanding the entire point of significant digits. Measurements and measurement precision are central ideas in empirical science. If you had a readout which displayed 5 digits that read $1.2XYZ$ where $XYZ$ were constantly fluctuating, that would mean something very different than if you had a readout that read $1.200X$ where $X$ was constantly fluctuating. Commented May 18 at 17:01
• Well, the reverse. But actually quantifying that uncertainty in a precise way requires more information. My point is that while 1.2 and 1.200 are of course mathematically equal, reporting a value to be 1.200 implies (and requires) a higher degree of precision than reporting a value to be 1.2, and it is this difference which is the focus when counting significant digits. Commented May 18 at 20:01

Zeroes to the right are significant in error analysis. This is because when you make a measurement, you make an error (usually because your measurement device is not precise enough). The number of significant digits tells you how many digits correspond to the real weight of the object. The real weight might not be the one you measure, as real measurement devices are not perfect ;)

In the first case, the measured mass is $$1.200\times 10^{3}\,\text{kg}$$. This means that the incertitude in the measurement is of order $$0.001\times 10^{3}\,\text{kg}$$.

Instead, if you measure $$1.2\times 10^{3}\,\text{kg}$$, then the error is of order $$0.1\times 10^{3}\,\text{kg}$$. This means that in this second case the uncertainty is a hundred times bigger!

An extreme case:

Let's say that you use a house scale to measure how much a hair weights. It will tell you that it weights $$0.0\,\text{g}$$. Of course, this doesn't mean that the hair doesn't have a mass! It just means that your scale is not precise enough. But one thing is sure, the mass of the hair will be bounded by $$0 \leq m_{hair} \leq 0.1 \,\text{g},$$ where the $$0$$ is the real zero.

• Would you consider substituting uncertainty (or incertitude!) for error in the last paragraph of 'your case'? Commented May 18 at 16:39
• Sure, I meant "error in the measurement". But it might be that that's not the word in english :( Commented May 18 at 16:42
• So, if I will measure 1200 kg and I will write it in the paper as $1.2 \times 10^3 kg$, then it would mean that I have not really measured 1200 kg, even if $1.2\cdot 10^3 kg$ is same 1200 kg? Have you spot a nonsense? Commented May 18 at 16:42
• No. In both cases you would have measured 1200kg. The difference is by how much your measurement is wrong. In the $1,2$ case, the actual weight of the thing is bounded in between $1100 kg < \text{real weight}<1300 kg$, while in the $1,200$ case, the actual weight of the thing is bounded in between $1199 kg < \text{real weight}<1201 kg$. Commented May 18 at 16:45
• Yes. But the order of magnitude of the error (which is basically what we care abount) is encoded in the number of significant digits. Commented May 18 at 16:55