-0.000 000 000 000 000 000 108 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 000 000 108(10) to 64 bit double precision IEEE 754 binary floating point representation standard (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

What are the steps to convert decimal number
-0.000 000 000 000 000 000 108(10) to 64 bit double precision IEEE 754 binary floating point representation (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. Start with the positive version of the number:

|-0.000 000 000 000 000 000 108| = 0.000 000 000 000 000 000 108


2. First, convert to binary (in base 2) the integer part: 0.
Divide the number repeatedly by 2.

Keep track of each remainder.

We stop when we get a quotient that is equal to zero.


  • division = quotient + remainder;
  • 0 ÷ 2 = 0 + 0;

3. Construct the base 2 representation of the integer part of the number.

Take all the remainders starting from the bottom of the list constructed above.

0(10) =

0(2)


4. Convert to binary (base 2) the fractional part: 0.000 000 000 000 000 000 108.

Multiply it repeatedly by 2.

Keep track of each integer part of the results.

Stop when we get a fractional part that is equal to zero.


  • #) multiplying = integer + fractional part;
  • 1) 0.000 000 000 000 000 000 108 × 2 = 0 + 0.000 000 000 000 000 000 216;
  • 2) 0.000 000 000 000 000 000 216 × 2 = 0 + 0.000 000 000 000 000 000 432;
  • 3) 0.000 000 000 000 000 000 432 × 2 = 0 + 0.000 000 000 000 000 000 864;
  • 4) 0.000 000 000 000 000 000 864 × 2 = 0 + 0.000 000 000 000 000 001 728;
  • 5) 0.000 000 000 000 000 001 728 × 2 = 0 + 0.000 000 000 000 000 003 456;
  • 6) 0.000 000 000 000 000 003 456 × 2 = 0 + 0.000 000 000 000 000 006 912;
  • 7) 0.000 000 000 000 000 006 912 × 2 = 0 + 0.000 000 000 000 000 013 824;
  • 8) 0.000 000 000 000 000 013 824 × 2 = 0 + 0.000 000 000 000 000 027 648;
  • 9) 0.000 000 000 000 000 027 648 × 2 = 0 + 0.000 000 000 000 000 055 296;
  • 10) 0.000 000 000 000 000 055 296 × 2 = 0 + 0.000 000 000 000 000 110 592;
  • 11) 0.000 000 000 000 000 110 592 × 2 = 0 + 0.000 000 000 000 000 221 184;
  • 12) 0.000 000 000 000 000 221 184 × 2 = 0 + 0.000 000 000 000 000 442 368;
  • 13) 0.000 000 000 000 000 442 368 × 2 = 0 + 0.000 000 000 000 000 884 736;
  • 14) 0.000 000 000 000 000 884 736 × 2 = 0 + 0.000 000 000 000 001 769 472;
  • 15) 0.000 000 000 000 001 769 472 × 2 = 0 + 0.000 000 000 000 003 538 944;
  • 16) 0.000 000 000 000 003 538 944 × 2 = 0 + 0.000 000 000 000 007 077 888;
  • 17) 0.000 000 000 000 007 077 888 × 2 = 0 + 0.000 000 000 000 014 155 776;
  • 18) 0.000 000 000 000 014 155 776 × 2 = 0 + 0.000 000 000 000 028 311 552;
  • 19) 0.000 000 000 000 028 311 552 × 2 = 0 + 0.000 000 000 000 056 623 104;
  • 20) 0.000 000 000 000 056 623 104 × 2 = 0 + 0.000 000 000 000 113 246 208;
  • 21) 0.000 000 000 000 113 246 208 × 2 = 0 + 0.000 000 000 000 226 492 416;
  • 22) 0.000 000 000 000 226 492 416 × 2 = 0 + 0.000 000 000 000 452 984 832;
  • 23) 0.000 000 000 000 452 984 832 × 2 = 0 + 0.000 000 000 000 905 969 664;
  • 24) 0.000 000 000 000 905 969 664 × 2 = 0 + 0.000 000 000 001 811 939 328;
  • 25) 0.000 000 000 001 811 939 328 × 2 = 0 + 0.000 000 000 003 623 878 656;
  • 26) 0.000 000 000 003 623 878 656 × 2 = 0 + 0.000 000 000 007 247 757 312;
  • 27) 0.000 000 000 007 247 757 312 × 2 = 0 + 0.000 000 000 014 495 514 624;
  • 28) 0.000 000 000 014 495 514 624 × 2 = 0 + 0.000 000 000 028 991 029 248;
  • 29) 0.000 000 000 028 991 029 248 × 2 = 0 + 0.000 000 000 057 982 058 496;
  • 30) 0.000 000 000 057 982 058 496 × 2 = 0 + 0.000 000 000 115 964 116 992;
  • 31) 0.000 000 000 115 964 116 992 × 2 = 0 + 0.000 000 000 231 928 233 984;
  • 32) 0.000 000 000 231 928 233 984 × 2 = 0 + 0.000 000 000 463 856 467 968;
  • 33) 0.000 000 000 463 856 467 968 × 2 = 0 + 0.000 000 000 927 712 935 936;
  • 34) 0.000 000 000 927 712 935 936 × 2 = 0 + 0.000 000 001 855 425 871 872;
  • 35) 0.000 000 001 855 425 871 872 × 2 = 0 + 0.000 000 003 710 851 743 744;
  • 36) 0.000 000 003 710 851 743 744 × 2 = 0 + 0.000 000 007 421 703 487 488;
  • 37) 0.000 000 007 421 703 487 488 × 2 = 0 + 0.000 000 014 843 406 974 976;
  • 38) 0.000 000 014 843 406 974 976 × 2 = 0 + 0.000 000 029 686 813 949 952;
  • 39) 0.000 000 029 686 813 949 952 × 2 = 0 + 0.000 000 059 373 627 899 904;
  • 40) 0.000 000 059 373 627 899 904 × 2 = 0 + 0.000 000 118 747 255 799 808;
  • 41) 0.000 000 118 747 255 799 808 × 2 = 0 + 0.000 000 237 494 511 599 616;
  • 42) 0.000 000 237 494 511 599 616 × 2 = 0 + 0.000 000 474 989 023 199 232;
  • 43) 0.000 000 474 989 023 199 232 × 2 = 0 + 0.000 000 949 978 046 398 464;
  • 44) 0.000 000 949 978 046 398 464 × 2 = 0 + 0.000 001 899 956 092 796 928;
  • 45) 0.000 001 899 956 092 796 928 × 2 = 0 + 0.000 003 799 912 185 593 856;
  • 46) 0.000 003 799 912 185 593 856 × 2 = 0 + 0.000 007 599 824 371 187 712;
  • 47) 0.000 007 599 824 371 187 712 × 2 = 0 + 0.000 015 199 648 742 375 424;
  • 48) 0.000 015 199 648 742 375 424 × 2 = 0 + 0.000 030 399 297 484 750 848;
  • 49) 0.000 030 399 297 484 750 848 × 2 = 0 + 0.000 060 798 594 969 501 696;
  • 50) 0.000 060 798 594 969 501 696 × 2 = 0 + 0.000 121 597 189 939 003 392;
  • 51) 0.000 121 597 189 939 003 392 × 2 = 0 + 0.000 243 194 379 878 006 784;
  • 52) 0.000 243 194 379 878 006 784 × 2 = 0 + 0.000 486 388 759 756 013 568;
  • 53) 0.000 486 388 759 756 013 568 × 2 = 0 + 0.000 972 777 519 512 027 136;
  • 54) 0.000 972 777 519 512 027 136 × 2 = 0 + 0.001 945 555 039 024 054 272;
  • 55) 0.001 945 555 039 024 054 272 × 2 = 0 + 0.003 891 110 078 048 108 544;
  • 56) 0.003 891 110 078 048 108 544 × 2 = 0 + 0.007 782 220 156 096 217 088;
  • 57) 0.007 782 220 156 096 217 088 × 2 = 0 + 0.015 564 440 312 192 434 176;
  • 58) 0.015 564 440 312 192 434 176 × 2 = 0 + 0.031 128 880 624 384 868 352;
  • 59) 0.031 128 880 624 384 868 352 × 2 = 0 + 0.062 257 761 248 769 736 704;
  • 60) 0.062 257 761 248 769 736 704 × 2 = 0 + 0.124 515 522 497 539 473 408;
  • 61) 0.124 515 522 497 539 473 408 × 2 = 0 + 0.249 031 044 995 078 946 816;
  • 62) 0.249 031 044 995 078 946 816 × 2 = 0 + 0.498 062 089 990 157 893 632;
  • 63) 0.498 062 089 990 157 893 632 × 2 = 0 + 0.996 124 179 980 315 787 264;
  • 64) 0.996 124 179 980 315 787 264 × 2 = 1 + 0.992 248 359 960 631 574 528;
  • 65) 0.992 248 359 960 631 574 528 × 2 = 1 + 0.984 496 719 921 263 149 056;
  • 66) 0.984 496 719 921 263 149 056 × 2 = 1 + 0.968 993 439 842 526 298 112;
  • 67) 0.968 993 439 842 526 298 112 × 2 = 1 + 0.937 986 879 685 052 596 224;
  • 68) 0.937 986 879 685 052 596 224 × 2 = 1 + 0.875 973 759 370 105 192 448;
  • 69) 0.875 973 759 370 105 192 448 × 2 = 1 + 0.751 947 518 740 210 384 896;
  • 70) 0.751 947 518 740 210 384 896 × 2 = 1 + 0.503 895 037 480 420 769 792;
  • 71) 0.503 895 037 480 420 769 792 × 2 = 1 + 0.007 790 074 960 841 539 584;
  • 72) 0.007 790 074 960 841 539 584 × 2 = 0 + 0.015 580 149 921 683 079 168;
  • 73) 0.015 580 149 921 683 079 168 × 2 = 0 + 0.031 160 299 843 366 158 336;
  • 74) 0.031 160 299 843 366 158 336 × 2 = 0 + 0.062 320 599 686 732 316 672;
  • 75) 0.062 320 599 686 732 316 672 × 2 = 0 + 0.124 641 199 373 464 633 344;
  • 76) 0.124 641 199 373 464 633 344 × 2 = 0 + 0.249 282 398 746 929 266 688;
  • 77) 0.249 282 398 746 929 266 688 × 2 = 0 + 0.498 564 797 493 858 533 376;
  • 78) 0.498 564 797 493 858 533 376 × 2 = 0 + 0.997 129 594 987 717 066 752;
  • 79) 0.997 129 594 987 717 066 752 × 2 = 1 + 0.994 259 189 975 434 133 504;
  • 80) 0.994 259 189 975 434 133 504 × 2 = 1 + 0.988 518 379 950 868 267 008;
  • 81) 0.988 518 379 950 868 267 008 × 2 = 1 + 0.977 036 759 901 736 534 016;
  • 82) 0.977 036 759 901 736 534 016 × 2 = 1 + 0.954 073 519 803 473 068 032;
  • 83) 0.954 073 519 803 473 068 032 × 2 = 1 + 0.908 147 039 606 946 136 064;
  • 84) 0.908 147 039 606 946 136 064 × 2 = 1 + 0.816 294 079 213 892 272 128;
  • 85) 0.816 294 079 213 892 272 128 × 2 = 1 + 0.632 588 158 427 784 544 256;
  • 86) 0.632 588 158 427 784 544 256 × 2 = 1 + 0.265 176 316 855 569 088 512;
  • 87) 0.265 176 316 855 569 088 512 × 2 = 0 + 0.530 352 633 711 138 177 024;
  • 88) 0.530 352 633 711 138 177 024 × 2 = 1 + 0.060 705 267 422 276 354 048;
  • 89) 0.060 705 267 422 276 354 048 × 2 = 0 + 0.121 410 534 844 552 708 096;
  • 90) 0.121 410 534 844 552 708 096 × 2 = 0 + 0.242 821 069 689 105 416 192;
  • 91) 0.242 821 069 689 105 416 192 × 2 = 0 + 0.485 642 139 378 210 832 384;
  • 92) 0.485 642 139 378 210 832 384 × 2 = 0 + 0.971 284 278 756 421 664 768;
  • 93) 0.971 284 278 756 421 664 768 × 2 = 1 + 0.942 568 557 512 843 329 536;
  • 94) 0.942 568 557 512 843 329 536 × 2 = 1 + 0.885 137 115 025 686 659 072;
  • 95) 0.885 137 115 025 686 659 072 × 2 = 1 + 0.770 274 230 051 373 318 144;
  • 96) 0.770 274 230 051 373 318 144 × 2 = 1 + 0.540 548 460 102 746 636 288;
  • 97) 0.540 548 460 102 746 636 288 × 2 = 1 + 0.081 096 920 205 493 272 576;
  • 98) 0.081 096 920 205 493 272 576 × 2 = 0 + 0.162 193 840 410 986 545 152;
  • 99) 0.162 193 840 410 986 545 152 × 2 = 0 + 0.324 387 680 821 973 090 304;
  • 100) 0.324 387 680 821 973 090 304 × 2 = 0 + 0.648 775 361 643 946 180 608;
  • 101) 0.648 775 361 643 946 180 608 × 2 = 1 + 0.297 550 723 287 892 361 216;
  • 102) 0.297 550 723 287 892 361 216 × 2 = 0 + 0.595 101 446 575 784 722 432;
  • 103) 0.595 101 446 575 784 722 432 × 2 = 1 + 0.190 202 893 151 569 444 864;
  • 104) 0.190 202 893 151 569 444 864 × 2 = 0 + 0.380 405 786 303 138 889 728;
  • 105) 0.380 405 786 303 138 889 728 × 2 = 0 + 0.760 811 572 606 277 779 456;
  • 106) 0.760 811 572 606 277 779 456 × 2 = 1 + 0.521 623 145 212 555 558 912;
  • 107) 0.521 623 145 212 555 558 912 × 2 = 1 + 0.043 246 290 425 111 117 824;
  • 108) 0.043 246 290 425 111 117 824 × 2 = 0 + 0.086 492 580 850 222 235 648;
  • 109) 0.086 492 580 850 222 235 648 × 2 = 0 + 0.172 985 161 700 444 471 296;
  • 110) 0.172 985 161 700 444 471 296 × 2 = 0 + 0.345 970 323 400 888 942 592;
  • 111) 0.345 970 323 400 888 942 592 × 2 = 0 + 0.691 940 646 801 777 885 184;
  • 112) 0.691 940 646 801 777 885 184 × 2 = 1 + 0.383 881 293 603 555 770 368;
  • 113) 0.383 881 293 603 555 770 368 × 2 = 0 + 0.767 762 587 207 111 540 736;
  • 114) 0.767 762 587 207 111 540 736 × 2 = 1 + 0.535 525 174 414 223 081 472;
  • 115) 0.535 525 174 414 223 081 472 × 2 = 1 + 0.071 050 348 828 446 162 944;
  • 116) 0.071 050 348 828 446 162 944 × 2 = 0 + 0.142 100 697 656 892 325 888;

We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (Losing precision - the converted number we get in the end will be just a very good approximation of the initial one).


5. Construct the base 2 representation of the fractional part of the number.

Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:


0.000 000 000 000 000 000 108(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 1111 1110 0000 0011 1111 1101 0000 1111 1000 1010 0110 0001 0110(2)

6. Positive number before normalization:

0.000 000 000 000 000 000 108(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 1111 1110 0000 0011 1111 1101 0000 1111 1000 1010 0110 0001 0110(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 64 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 000 000 108(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 1111 1110 0000 0011 1111 1101 0000 1111 1000 1010 0110 0001 0110(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 1111 1110 0000 0011 1111 1101 0000 1111 1000 1010 0110 0001 0110(2) × 20 =

1.1111 1110 0000 0011 1111 1101 0000 1111 1000 1010 0110 0001 0110(2) × 2-64


8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

Sign 1 (a negative number)

Exponent (unadjusted): -64

Mantissa (not normalized):
1.1111 1110 0000 0011 1111 1101 0000 1111 1000 1010 0110 0001 0110


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-64 + 2(11-1) - 1 =

(-64 + 1 023)(10) =

959(10)


10. Convert the adjusted exponent from the decimal (base 10) to 11 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 959 ÷ 2 = 479 + 1;
  • 479 ÷ 2 = 239 + 1;
  • 239 ÷ 2 = 119 + 1;
  • 119 ÷ 2 = 59 + 1;
  • 59 ÷ 2 = 29 + 1;
  • 29 ÷ 2 = 14 + 1;
  • 14 ÷ 2 = 7 + 0;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

959(10) =

011 1011 1111(2)


12. Normalize the mantissa.

a) Remove the leading (the leftmost) bit, since it's allways 1, and the decimal point, if the case.

b) Adjust its length to 52 bits, only if necessary (not the case here).


Mantissa (normalized) =

1. 1111 1110 0000 0011 1111 1101 0000 1111 1000 1010 0110 0001 0110 =

1111 1110 0000 0011 1111 1101 0000 1111 1000 1010 0110 0001 0110


13. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (11 bits) =
011 1011 1111

Mantissa (52 bits) =
1111 1110 0000 0011 1111 1101 0000 1111 1000 1010 0110 0001 0110


Decimal number -0.000 000 000 000 000 000 108 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1011 1111 - 1111 1110 0000 0011 1111 1101 0000 1111 1000 1010 0110 0001 0110

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How to convert numbers from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 64 bit double precision IEEE 754 binary floating point:

  • 1. If the number to be converted is negative, start with its the positive version.
  • 2. First convert the integer part. Divide repeatedly by 2 the positive representation of the integer number that is to be converted to binary, until we get a quotient that is equal to zero, keeping track of each remainder.
  • 3. Construct the base 2 representation of the positive integer part of the number, by taking all the remainders from the previous operations, starting from the bottom of the list constructed above. Thus, the last remainder of the divisions becomes the first symbol (the leftmost) of the base two number, while the first remainder becomes the last symbol (the rightmost).
  • 4. Then convert the fractional part. Multiply the number repeatedly by 2, until we get a fractional part that is equal to zero, keeping track of each integer part of the results.
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the multiplying operations, starting from the top of the list constructed above (they should appear in the binary representation, from left to right, in the order they have been calculated).
  • 6. Normalize the binary representation of the number, shifting the decimal mark (the decimal point) "n" positions either to the left, or to the right, so that only one non zero digit remains to the left of the decimal mark.
  • 7. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal mark, if the case) and adjust its length to 52 bits, either by removing the excess bits from the right (losing precision...) or by adding extra bits set on '0' to the right.
  • 9. Sign (it takes 1 bit) is either 1 for a negative or 0 for a positive number.

Example: convert the negative number -31.640 215 from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-31.640 215| = 31.640 215

  • 2. First convert the integer part, 31. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 31 ÷ 2 = 15 + 1;
    • 15 ÷ 2 = 7 + 1;
    • 7 ÷ 2 = 3 + 1;
    • 3 ÷ 2 = 1 + 1;
    • 1 ÷ 2 = 0 + 1;
    • We have encountered a quotient that is ZERO => FULL STOP
  • 3. Construct the base 2 representation of the integer part of the number by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above:

    31(10) = 1 1111(2)

  • 4. Then, convert the fractional part, 0.640 215. Multiply repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:
    • #) multiplying = integer + fractional part;
    • 1) 0.640 215 × 2 = 1 + 0.280 43;
    • 2) 0.280 43 × 2 = 0 + 0.560 86;
    • 3) 0.560 86 × 2 = 1 + 0.121 72;
    • 4) 0.121 72 × 2 = 0 + 0.243 44;
    • 5) 0.243 44 × 2 = 0 + 0.486 88;
    • 6) 0.486 88 × 2 = 0 + 0.973 76;
    • 7) 0.973 76 × 2 = 1 + 0.947 52;
    • 8) 0.947 52 × 2 = 1 + 0.895 04;
    • 9) 0.895 04 × 2 = 1 + 0.790 08;
    • 10) 0.790 08 × 2 = 1 + 0.580 16;
    • 11) 0.580 16 × 2 = 1 + 0.160 32;
    • 12) 0.160 32 × 2 = 0 + 0.320 64;
    • 13) 0.320 64 × 2 = 0 + 0.641 28;
    • 14) 0.641 28 × 2 = 1 + 0.282 56;
    • 15) 0.282 56 × 2 = 0 + 0.565 12;
    • 16) 0.565 12 × 2 = 1 + 0.130 24;
    • 17) 0.130 24 × 2 = 0 + 0.260 48;
    • 18) 0.260 48 × 2 = 0 + 0.520 96;
    • 19) 0.520 96 × 2 = 1 + 0.041 92;
    • 20) 0.041 92 × 2 = 0 + 0.083 84;
    • 21) 0.083 84 × 2 = 0 + 0.167 68;
    • 22) 0.167 68 × 2 = 0 + 0.335 36;
    • 23) 0.335 36 × 2 = 0 + 0.670 72;
    • 24) 0.670 72 × 2 = 1 + 0.341 44;
    • 25) 0.341 44 × 2 = 0 + 0.682 88;
    • 26) 0.682 88 × 2 = 1 + 0.365 76;
    • 27) 0.365 76 × 2 = 0 + 0.731 52;
    • 28) 0.731 52 × 2 = 1 + 0.463 04;
    • 29) 0.463 04 × 2 = 0 + 0.926 08;
    • 30) 0.926 08 × 2 = 1 + 0.852 16;
    • 31) 0.852 16 × 2 = 1 + 0.704 32;
    • 32) 0.704 32 × 2 = 1 + 0.408 64;
    • 33) 0.408 64 × 2 = 0 + 0.817 28;
    • 34) 0.817 28 × 2 = 1 + 0.634 56;
    • 35) 0.634 56 × 2 = 1 + 0.269 12;
    • 36) 0.269 12 × 2 = 0 + 0.538 24;
    • 37) 0.538 24 × 2 = 1 + 0.076 48;
    • 38) 0.076 48 × 2 = 0 + 0.152 96;
    • 39) 0.152 96 × 2 = 0 + 0.305 92;
    • 40) 0.305 92 × 2 = 0 + 0.611 84;
    • 41) 0.611 84 × 2 = 1 + 0.223 68;
    • 42) 0.223 68 × 2 = 0 + 0.447 36;
    • 43) 0.447 36 × 2 = 0 + 0.894 72;
    • 44) 0.894 72 × 2 = 1 + 0.789 44;
    • 45) 0.789 44 × 2 = 1 + 0.578 88;
    • 46) 0.578 88 × 2 = 1 + 0.157 76;
    • 47) 0.157 76 × 2 = 0 + 0.315 52;
    • 48) 0.315 52 × 2 = 0 + 0.631 04;
    • 49) 0.631 04 × 2 = 1 + 0.262 08;
    • 50) 0.262 08 × 2 = 0 + 0.524 16;
    • 51) 0.524 16 × 2 = 1 + 0.048 32;
    • 52) 0.048 32 × 2 = 0 + 0.096 64;
    • 53) 0.096 64 × 2 = 0 + 0.193 28;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 52) and at least one integer part that was different from zero => FULL STOP (losing precision...).
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above:

    0.640 215(10) = 0.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 6. Summarizing - the positive number before normalization:

    31.640 215(10) = 1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 7. Normalize the binary representation of the number, shifting the decimal mark 4 positions to the left so that only one non-zero digit stays to the left of the decimal mark:

    31.640 215(10) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 20 =
    1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

  • 9. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as shown above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1 = (4 + 1023)(10) = 1027(10) =
    100 0000 0011(2)

  • 10. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign) and adjust its length to 52 bits, by removing the excess bits, from the right (losing precision...):

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

    Mantissa (normalized): 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 100 0000 0011

    Mantissa (52 bits) = 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Number -31.640 215, converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point =
    1 - 100 0000 0011 - 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100