numerics - Finding differences between Pi with varying number of decimals

I have the following code

In[32]:= N[Pi, 2]

Out[32]= 3.1


In[33]:= N[Pi, 1]

Out[33]= 3.

In[34]:= N[Pi, 2] - N[Pi, 1]

Out[34]= 0.*10^-1

Why can't Mathematica find the difference between the two numbers?

Answer

N does not round numbers or truncate them.

The internal forms of

N[Pi, 1] // FullForm
N[Pi, 2] // FullForm

are respectively

(*
  3.1415926535897932384626433832795028842`1.
  3.1415926535897932384626433832795028842`2.
*)

Note that the numerical values are identical although the precisions differ. Their difference is zero and all significant digits are lost. The Front End typesets numerical results in StandardForm by showing only the significant digits determined by their precision.

This may be closer to what you're after:

Table[
 N[Floor[Pi, 10^n] - Floor[Pi, 10^(n + 1)]],
 {n, 0, -5, -1}]
(*
  {3.,
   0.1,
   0.04,

   0.001,
   0.0005,
   0.00009}
*)

---

Update to incorporate some of the related points made in the comments:

According to the tutorial NumericalPrecision, a nonzero approximate number x with precision p is defined to have uncertainty Abs[x] 10^-p. Likewise, the precision of a nonzero approximate number x with uncertainty d is given by -Log10[d/Abs[x]].

If approximate numbers are added or subtracted, the uncertainties are added. There are similar such rules for other arithmetic operations and the evaluation of functions. Consequently given two numbers $x_1,x_2$ with uncertainties $d_1,d_2$, the precision of $x_1\pm x_2$ is given by

$$p = -\log_{10}\left({d_1+d_2 \over |\,x_1\pm x_2|} \right)\,.$$

In the OP's case the precision would be negative (negative infinity at that). In cases where the precision is negative, Mathematica returns zero with corresponding computed accuracy.

Here is an illustration from the old comments. The function precDiff[{x1, p1}, {x2, p2}] computes the precision of the difference of numbers x1, x2 with precisions p1, p2.

precDiff[{x1_, p1_}, {x2_, p2_}] :=
   -Log10[(10^-p1 Abs[x1] + 10^-p2 Abs[x2])/Abs[x1 - x2]]

This shows that the computed precision of N[34/10, 2] - N[Pi, 1] is negative.

N @ precDiff[{34/10, 2}, {Pi, 1}]
N[34/10, 2] - N[Pi, 1] // FullForm
(*
  -0.129473
   0``0.4582220427094728

*)

This shows that the computed precision of N[35/10, 2] - N[Pi, 1] agrees with Mathematica's .

N @ precDiff[{35/10, 2}, {Pi, 1}]
N[35/10, 2] - N[Pi, 1] // FullForm
(*
  0.0113533
  0.35840734641020676153735661672049580005`0.011353331924907081
*)