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2013-08-30T13:06:02.233
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2013-08-30T14:32:15.540
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2013-08-30T14:32:15.540
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Assuming you are using the IEEE standard, the formula for the representation of numbers is: <pre><code>number = sign*(1+2^(-m)*significand)*2^(exponent-bias) </code></pre> where <code>m</code> is the number of bits used to store the (integer) significand (or mantissa), and <code>bias</code> is equal to <code>2^(e-1) - 1</code> where <code>e</code> is the number of bits used to store the exponent. Let's see what we can derive from that. Note that <ul> <li>The value of <code>significand</code> ranges between <code>0</code> and <code>2^m - 1</code> (in your case: between 0 and 1048575).</li> <li>The value of <code>exponent</code> ranges between <code>0</code> and <code>2^e - 1</code>. However, both extremal values are reserved for exceptions (subnormal numbers, infinites and NANs), called unnormalized numbers.</li> </ul> Consequently, <ul> <li>The smallest value for the fractional part <code>(1+2^(-m)*significand)</code> is 1, the biggest value is <code>2-2^(-m)</code> (in your case 2-2^(-20), approximately 1,999999046).</li> <li>The smallest non-exceptional value for the total exponent <code>exponent-bias</code> is <code>-2^(e-1)+2</code> (in your case -14), the biggest is <code>2^(e-1)-1</code> (in your case: 15).</li> </ul> So it turns out that: <ul> <li>The smallest (positive) normalized number that can be represented is <code>2^(-2^(e-1)+2)</code> (in your case 2^(-14), approximately 0,000061035)</li> <li>The biggest is <code>(2-2^(-m))*(2^(2^(e-1)-1))</code> (in your case (2-2^(-20))*(2^15), approximately 65535,96875).</li> </ul> As for "machine precision", I'm not sure what you mean, but one calls <code>m+1</code> (21 here) the binary precision, and the precision in terms of decimal digits is <code>log10(2^(m+1))</code>, for you this is approximately 6.3. I hope I didn't get anything wrong, I'm no expert about this.
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