Can you explain this Math.Log10 vs. BigInteger.Log10 behavior?

Question

Can somebody explain the following System.Numerics.BigInteger behavior?

Console.WriteLine(Math.Log10(100));       // prints 2
Console.WriteLine(Math.Log10(1000));      // prints 3 (as expected)

Console.WriteLine((int)Math.Log10(100));  // prints 2
Console.WriteLine((int)Math.Log10(1000)); // prints 3 (as axpected)

var bi100 = new BigInteger(100);
var bi1000 = new BigInteger(1000);

Console.WriteLine(BigInteger.Log10(bi100));       // prints 2
Console.WriteLine(BigInteger.Log10(bi1000));      // prints 3 (as axpected) 

Console.WriteLine((int)BigInteger.Log10(bi100));  // prints 2
Console.WriteLine((int)BigInteger.Log10(bi1000)); // prints 2 ???????

Console.WriteLine(Math.Floor(BigInteger.Log10(bi100)));   // prints 2
Console.WriteLine(Math.Floor(BigInteger.Log10(bi1000)));  // prints 2 ???????

Console.WriteLine(Math.Round(BigInteger.Log10(bi100)));  // prints 2
Console.WriteLine(Math.Round(BigInteger.Log10(bi1000))); // prints 3 (as expected)

EDIT: Please note that I know that it's a rouding problem. I want to know why the behavior of Math.Log10 and BigInteger.Log10 differs.

Answer 1

It is due to precision and rounding.

This line:

Console.WriteLine((int)BigInteger.Log10(bi1000)); 

is rounding down the value 2.9999999999999996 to 2, whereas Console.WriteLine is writing this out as 3

You can verify this using an intermediate double variable, and inspecting its value:

double x = BigInteger.Log10(bi1000);
Console.WriteLine((int)x);