0.000 000 021 979 552 668 138 406 845 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 021 979 552 668 138 406 845(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
0.000 000 021 979 552 668 138 406 845(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. 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;

2. 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)


3. Convert to binary (base 2) the fractional part: 0.000 000 021 979 552 668 138 406 845.

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) 0.000 000 021 979 552 668 138 406 845 × 2 = 0 + 0.000 000 043 959 105 336 276 813 69;
  • 2) 0.000 000 043 959 105 336 276 813 69 × 2 = 0 + 0.000 000 087 918 210 672 553 627 38;
  • 3) 0.000 000 087 918 210 672 553 627 38 × 2 = 0 + 0.000 000 175 836 421 345 107 254 76;
  • 4) 0.000 000 175 836 421 345 107 254 76 × 2 = 0 + 0.000 000 351 672 842 690 214 509 52;
  • 5) 0.000 000 351 672 842 690 214 509 52 × 2 = 0 + 0.000 000 703 345 685 380 429 019 04;
  • 6) 0.000 000 703 345 685 380 429 019 04 × 2 = 0 + 0.000 001 406 691 370 760 858 038 08;
  • 7) 0.000 001 406 691 370 760 858 038 08 × 2 = 0 + 0.000 002 813 382 741 521 716 076 16;
  • 8) 0.000 002 813 382 741 521 716 076 16 × 2 = 0 + 0.000 005 626 765 483 043 432 152 32;
  • 9) 0.000 005 626 765 483 043 432 152 32 × 2 = 0 + 0.000 011 253 530 966 086 864 304 64;
  • 10) 0.000 011 253 530 966 086 864 304 64 × 2 = 0 + 0.000 022 507 061 932 173 728 609 28;
  • 11) 0.000 022 507 061 932 173 728 609 28 × 2 = 0 + 0.000 045 014 123 864 347 457 218 56;
  • 12) 0.000 045 014 123 864 347 457 218 56 × 2 = 0 + 0.000 090 028 247 728 694 914 437 12;
  • 13) 0.000 090 028 247 728 694 914 437 12 × 2 = 0 + 0.000 180 056 495 457 389 828 874 24;
  • 14) 0.000 180 056 495 457 389 828 874 24 × 2 = 0 + 0.000 360 112 990 914 779 657 748 48;
  • 15) 0.000 360 112 990 914 779 657 748 48 × 2 = 0 + 0.000 720 225 981 829 559 315 496 96;
  • 16) 0.000 720 225 981 829 559 315 496 96 × 2 = 0 + 0.001 440 451 963 659 118 630 993 92;
  • 17) 0.001 440 451 963 659 118 630 993 92 × 2 = 0 + 0.002 880 903 927 318 237 261 987 84;
  • 18) 0.002 880 903 927 318 237 261 987 84 × 2 = 0 + 0.005 761 807 854 636 474 523 975 68;
  • 19) 0.005 761 807 854 636 474 523 975 68 × 2 = 0 + 0.011 523 615 709 272 949 047 951 36;
  • 20) 0.011 523 615 709 272 949 047 951 36 × 2 = 0 + 0.023 047 231 418 545 898 095 902 72;
  • 21) 0.023 047 231 418 545 898 095 902 72 × 2 = 0 + 0.046 094 462 837 091 796 191 805 44;
  • 22) 0.046 094 462 837 091 796 191 805 44 × 2 = 0 + 0.092 188 925 674 183 592 383 610 88;
  • 23) 0.092 188 925 674 183 592 383 610 88 × 2 = 0 + 0.184 377 851 348 367 184 767 221 76;
  • 24) 0.184 377 851 348 367 184 767 221 76 × 2 = 0 + 0.368 755 702 696 734 369 534 443 52;
  • 25) 0.368 755 702 696 734 369 534 443 52 × 2 = 0 + 0.737 511 405 393 468 739 068 887 04;
  • 26) 0.737 511 405 393 468 739 068 887 04 × 2 = 1 + 0.475 022 810 786 937 478 137 774 08;
  • 27) 0.475 022 810 786 937 478 137 774 08 × 2 = 0 + 0.950 045 621 573 874 956 275 548 16;
  • 28) 0.950 045 621 573 874 956 275 548 16 × 2 = 1 + 0.900 091 243 147 749 912 551 096 32;
  • 29) 0.900 091 243 147 749 912 551 096 32 × 2 = 1 + 0.800 182 486 295 499 825 102 192 64;
  • 30) 0.800 182 486 295 499 825 102 192 64 × 2 = 1 + 0.600 364 972 590 999 650 204 385 28;
  • 31) 0.600 364 972 590 999 650 204 385 28 × 2 = 1 + 0.200 729 945 181 999 300 408 770 56;
  • 32) 0.200 729 945 181 999 300 408 770 56 × 2 = 0 + 0.401 459 890 363 998 600 817 541 12;
  • 33) 0.401 459 890 363 998 600 817 541 12 × 2 = 0 + 0.802 919 780 727 997 201 635 082 24;
  • 34) 0.802 919 780 727 997 201 635 082 24 × 2 = 1 + 0.605 839 561 455 994 403 270 164 48;
  • 35) 0.605 839 561 455 994 403 270 164 48 × 2 = 1 + 0.211 679 122 911 988 806 540 328 96;
  • 36) 0.211 679 122 911 988 806 540 328 96 × 2 = 0 + 0.423 358 245 823 977 613 080 657 92;
  • 37) 0.423 358 245 823 977 613 080 657 92 × 2 = 0 + 0.846 716 491 647 955 226 161 315 84;
  • 38) 0.846 716 491 647 955 226 161 315 84 × 2 = 1 + 0.693 432 983 295 910 452 322 631 68;
  • 39) 0.693 432 983 295 910 452 322 631 68 × 2 = 1 + 0.386 865 966 591 820 904 645 263 36;
  • 40) 0.386 865 966 591 820 904 645 263 36 × 2 = 0 + 0.773 731 933 183 641 809 290 526 72;
  • 41) 0.773 731 933 183 641 809 290 526 72 × 2 = 1 + 0.547 463 866 367 283 618 581 053 44;
  • 42) 0.547 463 866 367 283 618 581 053 44 × 2 = 1 + 0.094 927 732 734 567 237 162 106 88;
  • 43) 0.094 927 732 734 567 237 162 106 88 × 2 = 0 + 0.189 855 465 469 134 474 324 213 76;
  • 44) 0.189 855 465 469 134 474 324 213 76 × 2 = 0 + 0.379 710 930 938 268 948 648 427 52;
  • 45) 0.379 710 930 938 268 948 648 427 52 × 2 = 0 + 0.759 421 861 876 537 897 296 855 04;
  • 46) 0.759 421 861 876 537 897 296 855 04 × 2 = 1 + 0.518 843 723 753 075 794 593 710 08;
  • 47) 0.518 843 723 753 075 794 593 710 08 × 2 = 1 + 0.037 687 447 506 151 589 187 420 16;
  • 48) 0.037 687 447 506 151 589 187 420 16 × 2 = 0 + 0.075 374 895 012 303 178 374 840 32;
  • 49) 0.075 374 895 012 303 178 374 840 32 × 2 = 0 + 0.150 749 790 024 606 356 749 680 64;
  • 50) 0.150 749 790 024 606 356 749 680 64 × 2 = 0 + 0.301 499 580 049 212 713 499 361 28;
  • 51) 0.301 499 580 049 212 713 499 361 28 × 2 = 0 + 0.602 999 160 098 425 426 998 722 56;
  • 52) 0.602 999 160 098 425 426 998 722 56 × 2 = 1 + 0.205 998 320 196 850 853 997 445 12;
  • 53) 0.205 998 320 196 850 853 997 445 12 × 2 = 0 + 0.411 996 640 393 701 707 994 890 24;
  • 54) 0.411 996 640 393 701 707 994 890 24 × 2 = 0 + 0.823 993 280 787 403 415 989 780 48;
  • 55) 0.823 993 280 787 403 415 989 780 48 × 2 = 1 + 0.647 986 561 574 806 831 979 560 96;
  • 56) 0.647 986 561 574 806 831 979 560 96 × 2 = 1 + 0.295 973 123 149 613 663 959 121 92;
  • 57) 0.295 973 123 149 613 663 959 121 92 × 2 = 0 + 0.591 946 246 299 227 327 918 243 84;
  • 58) 0.591 946 246 299 227 327 918 243 84 × 2 = 1 + 0.183 892 492 598 454 655 836 487 68;
  • 59) 0.183 892 492 598 454 655 836 487 68 × 2 = 0 + 0.367 784 985 196 909 311 672 975 36;
  • 60) 0.367 784 985 196 909 311 672 975 36 × 2 = 0 + 0.735 569 970 393 818 623 345 950 72;
  • 61) 0.735 569 970 393 818 623 345 950 72 × 2 = 1 + 0.471 139 940 787 637 246 691 901 44;
  • 62) 0.471 139 940 787 637 246 691 901 44 × 2 = 0 + 0.942 279 881 575 274 493 383 802 88;
  • 63) 0.942 279 881 575 274 493 383 802 88 × 2 = 1 + 0.884 559 763 150 548 986 767 605 76;
  • 64) 0.884 559 763 150 548 986 767 605 76 × 2 = 1 + 0.769 119 526 301 097 973 535 211 52;
  • 65) 0.769 119 526 301 097 973 535 211 52 × 2 = 1 + 0.538 239 052 602 195 947 070 423 04;
  • 66) 0.538 239 052 602 195 947 070 423 04 × 2 = 1 + 0.076 478 105 204 391 894 140 846 08;
  • 67) 0.076 478 105 204 391 894 140 846 08 × 2 = 0 + 0.152 956 210 408 783 788 281 692 16;
  • 68) 0.152 956 210 408 783 788 281 692 16 × 2 = 0 + 0.305 912 420 817 567 576 563 384 32;
  • 69) 0.305 912 420 817 567 576 563 384 32 × 2 = 0 + 0.611 824 841 635 135 153 126 768 64;
  • 70) 0.611 824 841 635 135 153 126 768 64 × 2 = 1 + 0.223 649 683 270 270 306 253 537 28;
  • 71) 0.223 649 683 270 270 306 253 537 28 × 2 = 0 + 0.447 299 366 540 540 612 507 074 56;
  • 72) 0.447 299 366 540 540 612 507 074 56 × 2 = 0 + 0.894 598 733 081 081 225 014 149 12;
  • 73) 0.894 598 733 081 081 225 014 149 12 × 2 = 1 + 0.789 197 466 162 162 450 028 298 24;
  • 74) 0.789 197 466 162 162 450 028 298 24 × 2 = 1 + 0.578 394 932 324 324 900 056 596 48;
  • 75) 0.578 394 932 324 324 900 056 596 48 × 2 = 1 + 0.156 789 864 648 649 800 113 192 96;
  • 76) 0.156 789 864 648 649 800 113 192 96 × 2 = 0 + 0.313 579 729 297 299 600 226 385 92;
  • 77) 0.313 579 729 297 299 600 226 385 92 × 2 = 0 + 0.627 159 458 594 599 200 452 771 84;
  • 78) 0.627 159 458 594 599 200 452 771 84 × 2 = 1 + 0.254 318 917 189 198 400 905 543 68;

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).


4. 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:


0.000 000 021 979 552 668 138 406 845(10) =

0.0000 0000 0000 0000 0000 0000 0101 1110 0110 0110 1100 0110 0001 0011 0100 1011 1100 0100 1110 01(2)

5. Positive number before normalization:

0.000 000 021 979 552 668 138 406 845(10) =

0.0000 0000 0000 0000 0000 0000 0101 1110 0110 0110 1100 0110 0001 0011 0100 1011 1100 0100 1110 01(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 26 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 021 979 552 668 138 406 845(10) =

0.0000 0000 0000 0000 0000 0000 0101 1110 0110 0110 1100 0110 0001 0011 0100 1011 1100 0100 1110 01(2) =

0.0000 0000 0000 0000 0000 0000 0101 1110 0110 0110 1100 0110 0001 0011 0100 1011 1100 0100 1110 01(2) × 20 =

1.0111 1001 1001 1011 0001 1000 0100 1101 0010 1111 0001 0011 1001(2) × 2-26


7. 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 0 (a positive number)

Exponent (unadjusted): -26

Mantissa (not normalized):
1.0111 1001 1001 1011 0001 1000 0100 1101 0010 1111 0001 0011 1001


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-26 + 2(11-1) - 1 =

(-26 + 1 023)(10) =

997(10)


9. 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;
  • 997 ÷ 2 = 498 + 1;
  • 498 ÷ 2 = 249 + 0;
  • 249 ÷ 2 = 124 + 1;
  • 124 ÷ 2 = 62 + 0;
  • 62 ÷ 2 = 31 + 0;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

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

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


Exponent (adjusted) =

997(10) =

011 1110 0101(2)


11. 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. 0111 1001 1001 1011 0001 1000 0100 1101 0010 1111 0001 0011 1001 =

0111 1001 1001 1011 0001 1000 0100 1101 0010 1111 0001 0011 1001


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

Sign (1 bit) =
0 (a positive number)

Exponent (11 bits) =
011 1110 0101

Mantissa (52 bits) =
0111 1001 1001 1011 0001 1000 0100 1101 0010 1111 0001 0011 1001


Decimal number 0.000 000 021 979 552 668 138 406 845 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1110 0101 - 0111 1001 1001 1011 0001 1000 0100 1101 0010 1111 0001 0011 1001

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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