# How many insignificant digits need to be used to calculate a number accurate to the same number of significant digits?

Not sure if this question is right for this site, but I hope it will be okay. As an example, say I calculate a value of $\sim1.32755552$ that has three significant digits (say I divided two three-significant digit numbers together). How many insignificant digits (after the $1.32$) would I need to keep from this number if I wanted to multiply it with another number (say, with infinite significant digits) and get the same answer when rounded to the right number of significant digits as if I had a perfect measurement of a value?

Just to clarify, this comes up if you were to multiply $1.7$ (1 sig dig) by $3$ (infinite sig digs). If you didn't keep any insignificant digits, you would get $2\times3=6,$ but if you kept the $.7,$ you would get $1.7\times3=5.4,$ or $5$ when rounded one significant digit. Finally, the question is: How can one determine how many insignificant digits are needed to have an answer that is the same as a perfect measurement of the value, comparing up to the number of significant digits?

Given a decimal approximation, like $$3.14,\qquad 0.000873,\quad 123\,315.3,\quad12\,000,$$ of a mathematical or physical constant, the number $n$ of "significant digits" is not a clear cut mathematical entity. It is just an indication that the relative error of the given decimal has order of magnitude $10^{-n}$.
If you perform operations with decimals having $n$ "significant digits" the number of correct digits in the result may be drastically lower. This happens in particular if you take the difference of two almost equal such decimals, or if you subtract $0.000873$ from $1.000217$.
The number of significant digits is the max number used in the equation. The best you could do is multiply a number by $1.33$ and get a number with 3 significant digits. The reason for this is the number of significant digits is a measure of accuracy of our estimated $1.33$ value. We don't really know if its closer to $1.327$ or $1.328$ in our calculation because things were only estimated "measured" so exactly (accurately) so we have to round.