Round the following number to $4$ significant figures: $794834.$

Question

Round the following number to $4$ significant figures:
$X=794834.$

For the number $X=794834$, I simply observed that all the digits in here are significant. The $4^{th}$ significant digit in this case, is $8$ and the $5^{th}$ digit is $3$. Now, as $3 \lt 5$ so, we can represent the given number as $X \approx 7948 \times 10^2.$ This is the scientific representation and $7948 \times 10^2$ has $4$ significant digit.

My question, is, if we were given a number like $Y=794894,$ then should I round it off as $Y \approx 7949 \times 10^2.$ In this scientific representation $7949 \times 10^2$ has $4$ significant digits. I also changed the $4^{th}$ significant digit in $7949 \times 10^2$ to $9$ as, the $5^{th}$ digit in $794894$ was $9$ and $9 \gt 5.$

I am much new in working with rounding off numbers upto some number of significant digits and hence, I am not sure whether I have understood the concept or not.

Till now, I am only used to rounding off numbers like $Z=45.67$ upto the $3^{rd}$ significant figures. So, this is my first time for rounding off a whole number.

Answer 1

Easiest way to think about this is to write $X=794834$ in the form $X=7.94834\times10^5$ & then make it $4$ significant figures to get $X \approx 7.948\times10^5$ , using your "known" way to handle fractional numbers.

With $Y=794894$ , we get $Y=794894=7.94894\times10^5$ , hence when we make it $4$ significant figures , we will get $Y \approx 7.949\times10^5$ , using your "known" way to handle fractional numbers.

Essentially , we will get Equivalent values with either way.
It is a "standard" format to write using the numbers between $1$ & $10$ , where the Power of $10$ (Either Positive or Negative) will fix the Decimal Point.