Yes, there is an implicitly "hidden" bit, so if your mantissa is 1100000000000000000000, then that corresponds to the base 2 number 1.11. Now, if you want to compute the value of .11 part, you just do the same thing as you would do to convert a normal binary number to decimal: multiply each bit by a power of 2. In binary that means the mantissa must be "1". Since the mantissa of a normalized binary floating point number is always 1, we don't need to store the 1. The first mantissa bit is hidden in the sense that it always exists, but we don't actually store the bit, because we know its value is 1. The DEC PDP-10 had the same mantissa with "hidden 1 bit", offset binary exponent to the left of the mantissa, but when the numbers were negative, the PDP-10 represented that value as the 2's complement of the positive value. So there was no "-0" representation (just one 0) and the integer compare operation worked in any case. Rest bits are mantissa and the actual floating point number is 1.mantissa x 2^exponent where 1.mantissa is in binary. So our numbers are 1.00101010100101110110001 x 2^-2 and 1.01000101100001011100001 x 2^6 I suggest you use Wikipedia it has a very big section on floating point representation. Much more then we can explain here to you. \$\endgroup\$ – Oldfart Apr 25 '18 at 9:50 \$\begingroup\$ Preface it by the "hidden bit" unless the exponent is 0. \$\endgroup\$ – Brian Drummond Apr 25 '18 at 9:53
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