The precision of a calculator is largely influenced by the number of significant figures used for computation. It is normally greater than the number of digits displayed.
For example, if you compute $\pi$ and it displays $3.141592654$, the actual number stored in its RAM might be $3.14159265\color{blue}{359}$, where the digits in $\color{blue}{\text{blue}}$ denotes the extra figures which are not displayed.
That's the reason why you do not get $1$ from $1349303770/(2^{32} \pi)$: $1349303770$ was the displayed number which you read off from previous result, while the calculator used slightly more significant figures to compute $2^{32}\pi$.
There is an indirect way to count the total no. of significant figures used for computation:
- Compute $\sqrt{2}$
- Subtract $1$ from previous result (do not compute $\sqrt{2} - 1$ directly: you must make use of the previous result)
- Compute $1/\text{Previous result}$. This can be tricky because the method to do this depends on the type of calculator. Again, do not type the previous answer manually.
- Subtract $2$ from previous answer
- Repeat Steps 3-4, until you notice a discrepancy between the fractional part of the result from Step 3 and that from Step 1
- The approximate number of extra significant figures would be the total number of subtractions you have made before you notice a discrepancy
- The total number of significant figures is then
$$N = \text{no. of digits displayed} + \text{no. of extra significant figures}$$
Once this is known, you can sort of predict the result of computation by using a more precise calculator, and round off the result to $N$ figures. You might need an advanced calculator for this, such as PARI/GP.
Using my scientific calculator, these are the results:
- $1.414213562$ (Step 1)
- $0.414213562$ (Step 2)
- $2.414213562$ (Step 3)
- $0.414213562$ (Step 4)
- $2.414213562$ (Step 3)
- $0.414213562$ (Step 4)
- $2.41421356\color{red}{3}$ (Step 3)
- $0.41421356\color{red}{3}$ (Step 4)
- $2.4142135\color{red}{59}$ (Step 3)
Hence the approximate number of extra significant figures would be $3$ or $4$.
Why this method works: each subtraction causes catastrophic cancellation.