A calculator is out of order. For all number, before and after any arithmetic operation, the calculator will round up the numerical value to one decimal place if the value at the second decimal digit is 4 and above, or else it rounds down the value to one decimal place.
If the calculator displays 2.5. Find the maximum relative error for the true value of this number.
can anyone help me solve this ? im really stuck, i dont know how to do it.
The calculator would round up when the "true" value is $2.44..., 2.45..., \cdots, 2.49...$. If it were right on $2.50$, there'd be no relative error. The calculator would round down for true values $2.51..., 2.52,...2.53.$
Now, $2.44$, which would be rounded up to $2.5$ is the "furthest away" from $2.50$: at a distance of $|2.50-2.44| = 0.06\,$units. (Note that the only other candidate, $2.53$ gives us a distance of only $|2.50 - 2.53| = 0.03$ from $2.50$
Then the maximum relative error is equal to $$\left| \frac{ \text{calculator's return value} - \text{true value}}{\text{true value}}\right| = \dfrac{0.06}{2.44} \approx 0.0246 = 2.46\% \;\text{error}.$$
Hint $$ \text{Relative Error} = \left| \frac{ \text{Your Value} - \text{True Value} } {\text{True Value}} \right| $$
What is the largest difference of 2.5 away from what the true value should have been? Could the true value be 2.53? What about 2.55?
R.E = |0.005/2.5|
I have found something like this in my note so i tried to follow, is it related to this question or im totally off topic @gt6989b
– matin Feb 21 '14 at 16:42