error in Rounding and Significant Digits?

[bellefonte]

The question is written thusly:

Round 0.07284 to four, three, and two significant digits:
0.07284 (four significant digits)
0.0728 (three significant digits)
0.073 (two significant digits)

THe rules are:
1) All nonzero digits are significant.
2) All zeroes between significant digits are significant.
3) All zeroes which are both to the right of the decimal point and to the right of all non-zero significant digits are themselves significant.

Rule 3 states that the zero to the right of the decimal is Significant. Then would 4 significant digits be .0728?

[bellefonte]

and again, this same issue appears later in the lesson. It is written thusly:

Find the product of 0.00435 and 4.6 to the appropriate number of digits.
First I multiply:

0.00435 × 4.6 = 0.02001

Looking at the original numbers, I see that 4.6 has only two significant digits, so I will have to round 0.02001 to two significant digits. The 2 is the first significant digit, so the 0 following it will have to be the second significant digits. In other words, I must report the answer as being:

0.00435 × 4.6 = 0.020

The answer should not be 0.02, because 0.02 has only one significant digit; namely, the "2". The trailing zero in 0.020 indicates that "this is accurate to the thousandths place, or two significant digits", and is therefore a necessary part of the answer.

If all zeros to the right of the decimal are significant, why then do these examples state that it isnt so?

[stapel_eliz]

3) All zeroes which are both to the right of the decimal point and to the right of all non-zero significant digits are themselves significant.[/color]

Rule 3 states that the zero to the right of the decimal is Significant. Then would 4 significant digits be .0728?

Rule 3 states that a zero to the right of the decimal point and to the right of all non-zero significant digits is significant. Since the first zero after the decimal point is to the left of those digits, it is not significant.

[bellefonte]

ahh, ok. i thought the "and" was inclusive, not a requirement. Gotcha, thanks!