Don’t be ridiculous. Everyone knows 1.9… is totally different to 2.
Part of the difficulty here lies in the fact that the example given is 100.0 yards. This has four significant figures, right? But we all know that the actual measurement could have been, say 99.984352 which they rounded off to 100.0. The underlying question is WHY did they round it off at the tenths instead of rounding it off at the hundredths? Presumably, they had their reasons, based on the accuracy of the measuring device. But we don’t know what their reasons were. All we know is that they chose to round off at the tenths, which implies that the true number is probably somewhere between 99.95 and 100.05. But look what happens if they decided to round off to hundredths instead. Take 99.84352 and round it off to 99.84. Now you’re implying that the actual number is between 99.835 and 99.845.
Here’s the crazy part:
100.0 has four significant figures.
99.84 also has four significant figures.
But the second is ten times more precise than the first.
Significant Figures are just a rough-and-ready way to tackle the problem of precision when you don’t actually know what the precision was. In the example given from that textbook, they want you to realize the following. [a] The measurement 100.0 has four significant figures. ** We will multiply it by 3 to get feet, but this is a precise number. [c] We will divide that answer by 5280, which is also a precise number. [d] We should round off our answer to 4 significant figures. But this is just a rough-and-ready answer. It’s not the best answer, really. It’s just a method which gives pretty good results most of the time.
So, 100 x 3 / 5280 = .0568181818181818181818181818181818181818181…
and we round it off to .05682 (the leading zero doesn’t count).
Unfortunately, .05682 implies that the actual number is somewhere between .056815 and .056825, which is an error of +/- 8.04672 mm (exactly). But our original number 100.0 yards implies an error of +/- 45.72 mm (exactly). So we are implying stronger precision than what is actually there. The person who measured the field was confident that their answer was less than 45.72 mm away from the actual number, but now we are saying our conversion is less than 8.04672 mm away from the actual number.
And both of those are correct roundings. It depends exactly on how accurate you believe your measurement to be.
But the alternative, using significant figures, is an implied error of +/-80 mm, and you’ve lost some of the precision you originally had. I’d rather keep “extra” precision than throw some away. If my real measurement is 99.84352 +/-0.034, give me 99.84, not 99.8.
You’re right that it’s just a rough and ready approach, but don’t damage the data.
Yes but the real alternative is to actually specify what your error is instead of just implying it by using significant figures. BTW, I actually said 99.984352; you dropped one of my 9s.
If the actual measurement was 99.984352 +/- .034 then we should divide both of those numbers by 1760 (number of yards in a mile) and get .056809291 +/- .000019 as the answer in miles.
Sure, just using significant figure is the best approach, but you still should keep at least enough digits to preserve your accuracy.
I was using the numbers from the end of the first paragraph in your earlier post:
is not the best approach…
Oops, looks like I used different numbers at the beginning the paragraph than I did at the end. It should have been:
Take 99.984352 and round it off to 99.98. Now you’re implying that the actual number is between 99.975 and 99.985.
Here’s the crazy part:
100.0 has four significant figures.
99.98 also has four significant figures.
But the second is ten times more precise than the first.
If the uncertainty is being conveyed only through significant figures, I would rather a degraded precision. Otherwise, I might be relying on precision that isn’t really there.
Sorry for the slow reply. (Life gets in the way of a follow-up post.) The answer given was 0.05682. I have come to understand the answer has 4 sig figs as the zero to the right of the decimal is not significant (and the zero to the left is merely a place holder). I’ll have to read through all the posts a little later to gain a better understanding.
I agree. That would make the point more clearly.
Hence the rule I quoted that says to assume four significant digits if it starts with a 1, and three significant digits if it starts with other than 1. That doesn’t make everything equally precise, but it avoids the extreme 10:1 situation.
You can’t avoid the 10:1 situation. Your rule still has problems with 199.98 vs. 200.0
The problem is - in real lab work, when you multiply or divide, you add the relative errors. the conversion values, being constants, have no relative error. the relative error in the measurement is =/- half the measuring scale, unless stated otherwise. Thus,
error = +/-0.05yd relative error, 0.05/100.0
converting, error +/- 2.84090909…E-07 miles.
But, errors contain only one significant digit - so error is +/-3*10-7
(If I can still do math)
BTW, if you assume results can randomly end in any digit, then you are rounding 0-4 down, 5-9 up. Of course, you may measure 4.49 as 4.5 and round the wrong way; but it’s so close then that it’s a tossup which way the rounding goes.
If the rounding makes a huge difference to the final result, then your answer is not very precise. Two digits is 'one in a hundred", most people if gifted a hundred dollars probably would not quibble if given $99 - especially if the odds are just as good that next time it’s $101.