0
$\begingroup$

My textbook says

"Significant figures are the digits conveying meaningful information. It includes all certain and first uncertain digit. If you are given a no. like 11.2 (sf=3), then uncertainity of +_1 is understood in the last digit."

I understand that it means that first two digits '1','1' are certain and the last digit '2' is uncertain. i.e the number lies between 11.1 and 11.3.

Now, let me take some other eg. Let it be 12.0(sf=3). It means that first two digits are certain and the last one is uncertain. According to the rule, the no. should lie between 11.9 and 12.1. Also, according to the rule, first two digits are certain i.e they have fixed value(we are sure about them). They are '1'&'2' respectively, but here they have changed(12 was changed to 11). Then, there are two uncertain digits. Isn't it? It is not in accordance to my textbook rule that if there are 3 sf, then 2 digits are certain and the last sf is uncertain. How would you explain this? Is my textbook rule incomplete or wrong or am i taking it some other way?

Simple and basic answer is requested. No maths please!!

Improve this question
$\endgroup$
1

1 Answer 1

Reset to default
0
$\begingroup$

Your textbook is incorrect, both in its categorisation of certain and uncertain digits (as you have noted) and in the value it assigns. The rule is $\pm$ half if no further information is given. I.e. you treat every reported value as if it has been rounded to the precision expressed. E.g. 0.12 is $\pm 0.005$ and 0.120 is $\pm 0.0005$

Improve this answer
$\endgroup$
8
  • $\begingroup$ So, if i have 11.2(sf=3), then all the digits are certain? $\endgroup$
    – Jatin Gakhar
    Sep 22, 2019 at 5:52
  • $\begingroup$ So, all the sf are certain? $\endgroup$
    – Jatin Gakhar
    Sep 22, 2019 at 6:07
  • $\begingroup$ No. The term certain and uncertain figures is misleading and should not be used. I don't see how it adds any value to understanding what is going on. 11.2 could be between 11.15 and 11.25. In this case the 2 is not certain. $\endgroup$
    – Paul Childs
    Sep 22, 2019 at 6:20
  • $\begingroup$ So, if i have a no. 12.7, it should be 12.65,12.66,12.67,12.68,12.69,12.70,12.71,12.72,12.73,12.74 or 12.75.Right? Here the digits at ones and tens place were certain. Now, take 12.0. It means it could be 11.95, 11.96, 11.97, 11.98, 11.99, 12.00, 12.01, 12.02, 12.03, 12.04, 12.05. Here, only tens digit was certain and ones was uncertain including tenths. In all other cases, 12.1, 12.2, 12.3, etc. ones and tens, both are certain. But, in 12.0, ones place is uncertain, not very convincing! $\endgroup$
    – Jatin Gakhar
    Sep 22, 2019 at 13:43
  • $\begingroup$ Yes. It's not a very useful way of thinking of it as it is dependent on base boundaries. $\endgroup$
    – Paul Childs
    Sep 29, 2019 at 20:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.