Decimal floating point to binary floating point

[Dan]

(12-26-2018 11:14 AM)Thomas Klemm Wrote:  What exactly are you trying to gain by using \(1.25^N\) instead of \(10^N\)?
The mantissa in binary is the same, but now you have to deal additionally with powers of 8.

It's one way that came to mind to change 10^N to 2^n, because IEEE 754 binary32 format is (p,e,m) where M*10^N = (-1)^p * (1+m) * 2^e.

I thought to store numbers in IEEE 754 binary64 format, i.e. double precision and write algorithms for +.-,*,/ using the single precision hardware instructions, but I may drop that idea.