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copied!<p>If group operation is multiplication then if n is the highest bit, then r1=1/power(2,n-1) is the least decimal that you can operate and the set [r1,2 * r1,4 * r1,8 * r1...1] union [-r1, -2 * r1, -4 * r1,....-1] union [0] will be the group that you are expecting. For integer [1,0,-1] is the group. </p> <p>if Group operation can be any thing else, then to form n set of valid groups,</p> <p>A(r)=cos(2*Pi*r/n) from r=0 to n-1</p> <p>and group operation is </p> <p>COS(COSINV(A1)+COSINV(A2))</p> <p>I don't know whether you want this.....</p> <p>or if you want INFINITY set as a valid group then</p> <p>simple answer : GROUP OPERATION = AVG(A1,A2) = (A1+A2)/2 </p> <p>or some functions exists F which has FINV as it's inverse and then FINV(F(A1)+F(A2)/2) Example of F is Log, inverse, square etc .. </p> <pre><code> double a = drand(); while ( forever ) { double b = drand(); a = (a+b)/2 //invariant - a is a valid floating point number } </code></pre> <p>or if you want INFINITY set of DIGITAL format as a valid group then Let L be the lowest float number and H be highest float number</p> <p>then GROUP OPERATION = AVG(A1,A2, L, H) = (A1+A2+L+H)/4</p> <p>this operation will always be within L & H for all Positive numbers.</p> <p>You can take L as four times the lowest decimal number and H as the (highest decimal number /4) for practical purpose.</p> <pre><code>double l = (0.0000000000000000000000000//1) * 4 double h = (0xFFFFFFFFFFFFFFFFFFFFFF///F) / 4 double a = abs(drand()) / 4; while ( forever ) { double b = abs(drand()) / 4; a = (a+b+l+h)/4 //invariant - a is a valid floating point number } </code></pre> <p>this has a subset of all possitive float number / 4.</p>

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