Are Significant Figures Really That Important?

[nobahar]

Hello!

I've used significant figures without really thinking about them too much; and I have never really understood there use.
I found this thread: https://www.physicsforums.com/showthread.php?t=477786&highlight=significant+figures

Where it has been argued that they should be 'taken lightly', so to speak.

Here's where I have always been confused, and I hope someone can help!:

If I were to perform the multiplication, 2.09*3.52, I have three significant figures in both numbers, and so the answer would be 7.36 (from 7.3568).
I think this example shows one issue, and that is that the 6 is not necessarily 'known'.
Indeed, depending on the numbers in the multiplication, digits in higher powers of ten columns can be affected by 'unknown' numbers in smaller powers of ten columns. So when a number is given to x significant figures, these numbers may not be the actual numbers!, since they can be influenced by unknown numbers of smaller magnitudes of ten e.g. 1/10th column, 1/100th column, etc can affect the value in the 10⁰ column or the 10¹ column, etc.
So when I see some calculation using significant figures, I should not necessarily believe any of the numbers are absolutely certain?

I hope that makes sense!
Thanks in advance.

 

[NascentOxygen]

if we say x=1.97, then it is taken as being between 1.96 and 1.98

consider the equation, y=1/(2-x)

If x=1.96, then y=25
if x=1.98, then y=50

so, we have found, y=37 +/- 13

Summarizing, although x was known to 3 sig figs, y is known to not much better than 1 sig fig.

Does this illustrate what you were talking about?

 

[olivermsun]

if we say x=1.97, then it is taken as being between 1.96 and 1.98

consider the equation, y=1/(2-x)
...
Summarizing, although x was known to 3 sig figs, y is known to not much better than 1 sig fig.

Yes, but clearly the problem step was (2-x), which left you with 0.03, known to one sig fig. The OP's example had only multiplication, which tends to be less tricky.