# Machine numbers in a 32-bits (binary digits) computer?

I'm currently studying Numerical Analysis with the book "Numerical Analysis: Mathematics of Scientific Computing" by Kincaid.

In this book, the authors have introduced a computer called "Marc-32" which is a 32-bits computer representing a nonzero real number with the form: x = ±q * 2^m

with the allocation:

• sign of the real number x: 1 bit
• biased exponent (integer e): 8 bits
• mantissa part (real number f): 23 bits

Now a problem in the book asks to consider different given machine numbers and whether they are contained in Marc-32. For example the number 2^-1 + 2^-26.

My problem is that I honestly do not know how to determine this. I've tried reading the section about "marc-32" multiple times, but the closest I've got was that in the book, the authors write that "...23 bits means that our machine numbers have a limited precision of rougly six decimal places".

Therefore I thought that 2^-1 + 2^-26 = 0.10000000000000000000000001 being a number with 26 decimal places would not be contained in marc-32, but I'm really not sure if this is correct at all, and if it is the method I'm supposed to be using.

Therefore I would like to reach out to the math community here for any pieces of advice on how to tackle this problem. It would be much appreciated.

• contained here means exactly representable. Why don't you understand 23 bits as exactly 23 binary places? (exclude first one implicitly defined) – user202729 Sep 7 '16 at 10:42

Be careful about the difference between "separation between the first and last nonzero digit" and "the position of the last nonzero digit". For example, although $2^{-1}+2^{-26}$ is not representable in your number system, $2^{-26}$ is (because $-26$ can be stored as a 9 bit signed integer).
• @NAstuden Exactly representing $2^{-1}+2^{-26}$ would require a 25 bit mantissa, with 24 zeros and a 1 (the leading $1$ is not in the mantissa). Since you only have a 23 bit mantissa, those last two digits of the 25 bit mantissa will get dropped, leaving 1.(23 zeros) with an exponent of $-1$. So that's $2^{-1}$. – Ian Sep 7 '16 at 11:04