5.143 999 999 999 999 239 719 272 736 692 801 117 897 033 691 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 5.143 999 999 999 999 239 719 272 736 692 801 117 897 033 691(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
5.143 999 999 999 999 239 719 272 736 692 801 117 897 033 691(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: 5.
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;
  • 5 ÷ 2 = 2 + 1;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

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.

5(10) =

101(2)


3. Convert to binary (base 2) the fractional part: 0.143 999 999 999 999 239 719 272 736 692 801 117 897 033 691.

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) 0.143 999 999 999 999 239 719 272 736 692 801 117 897 033 691 × 2 = 0 + 0.287 999 999 999 998 479 438 545 473 385 602 235 794 067 382;
  • 2) 0.287 999 999 999 998 479 438 545 473 385 602 235 794 067 382 × 2 = 0 + 0.575 999 999 999 996 958 877 090 946 771 204 471 588 134 764;
  • 3) 0.575 999 999 999 996 958 877 090 946 771 204 471 588 134 764 × 2 = 1 + 0.151 999 999 999 993 917 754 181 893 542 408 943 176 269 528;
  • 4) 0.151 999 999 999 993 917 754 181 893 542 408 943 176 269 528 × 2 = 0 + 0.303 999 999 999 987 835 508 363 787 084 817 886 352 539 056;
  • 5) 0.303 999 999 999 987 835 508 363 787 084 817 886 352 539 056 × 2 = 0 + 0.607 999 999 999 975 671 016 727 574 169 635 772 705 078 112;
  • 6) 0.607 999 999 999 975 671 016 727 574 169 635 772 705 078 112 × 2 = 1 + 0.215 999 999 999 951 342 033 455 148 339 271 545 410 156 224;
  • 7) 0.215 999 999 999 951 342 033 455 148 339 271 545 410 156 224 × 2 = 0 + 0.431 999 999 999 902 684 066 910 296 678 543 090 820 312 448;
  • 8) 0.431 999 999 999 902 684 066 910 296 678 543 090 820 312 448 × 2 = 0 + 0.863 999 999 999 805 368 133 820 593 357 086 181 640 624 896;
  • 9) 0.863 999 999 999 805 368 133 820 593 357 086 181 640 624 896 × 2 = 1 + 0.727 999 999 999 610 736 267 641 186 714 172 363 281 249 792;
  • 10) 0.727 999 999 999 610 736 267 641 186 714 172 363 281 249 792 × 2 = 1 + 0.455 999 999 999 221 472 535 282 373 428 344 726 562 499 584;
  • 11) 0.455 999 999 999 221 472 535 282 373 428 344 726 562 499 584 × 2 = 0 + 0.911 999 999 998 442 945 070 564 746 856 689 453 124 999 168;
  • 12) 0.911 999 999 998 442 945 070 564 746 856 689 453 124 999 168 × 2 = 1 + 0.823 999 999 996 885 890 141 129 493 713 378 906 249 998 336;
  • 13) 0.823 999 999 996 885 890 141 129 493 713 378 906 249 998 336 × 2 = 1 + 0.647 999 999 993 771 780 282 258 987 426 757 812 499 996 672;
  • 14) 0.647 999 999 993 771 780 282 258 987 426 757 812 499 996 672 × 2 = 1 + 0.295 999 999 987 543 560 564 517 974 853 515 624 999 993 344;
  • 15) 0.295 999 999 987 543 560 564 517 974 853 515 624 999 993 344 × 2 = 0 + 0.591 999 999 975 087 121 129 035 949 707 031 249 999 986 688;
  • 16) 0.591 999 999 975 087 121 129 035 949 707 031 249 999 986 688 × 2 = 1 + 0.183 999 999 950 174 242 258 071 899 414 062 499 999 973 376;
  • 17) 0.183 999 999 950 174 242 258 071 899 414 062 499 999 973 376 × 2 = 0 + 0.367 999 999 900 348 484 516 143 798 828 124 999 999 946 752;
  • 18) 0.367 999 999 900 348 484 516 143 798 828 124 999 999 946 752 × 2 = 0 + 0.735 999 999 800 696 969 032 287 597 656 249 999 999 893 504;
  • 19) 0.735 999 999 800 696 969 032 287 597 656 249 999 999 893 504 × 2 = 1 + 0.471 999 999 601 393 938 064 575 195 312 499 999 999 787 008;
  • 20) 0.471 999 999 601 393 938 064 575 195 312 499 999 999 787 008 × 2 = 0 + 0.943 999 999 202 787 876 129 150 390 624 999 999 999 574 016;
  • 21) 0.943 999 999 202 787 876 129 150 390 624 999 999 999 574 016 × 2 = 1 + 0.887 999 998 405 575 752 258 300 781 249 999 999 999 148 032;
  • 22) 0.887 999 998 405 575 752 258 300 781 249 999 999 999 148 032 × 2 = 1 + 0.775 999 996 811 151 504 516 601 562 499 999 999 998 296 064;
  • 23) 0.775 999 996 811 151 504 516 601 562 499 999 999 998 296 064 × 2 = 1 + 0.551 999 993 622 303 009 033 203 124 999 999 999 996 592 128;
  • 24) 0.551 999 993 622 303 009 033 203 124 999 999 999 996 592 128 × 2 = 1 + 0.103 999 987 244 606 018 066 406 249 999 999 999 993 184 256;
  • 25) 0.103 999 987 244 606 018 066 406 249 999 999 999 993 184 256 × 2 = 0 + 0.207 999 974 489 212 036 132 812 499 999 999 999 986 368 512;
  • 26) 0.207 999 974 489 212 036 132 812 499 999 999 999 986 368 512 × 2 = 0 + 0.415 999 948 978 424 072 265 624 999 999 999 999 972 737 024;
  • 27) 0.415 999 948 978 424 072 265 624 999 999 999 999 972 737 024 × 2 = 0 + 0.831 999 897 956 848 144 531 249 999 999 999 999 945 474 048;
  • 28) 0.831 999 897 956 848 144 531 249 999 999 999 999 945 474 048 × 2 = 1 + 0.663 999 795 913 696 289 062 499 999 999 999 999 890 948 096;
  • 29) 0.663 999 795 913 696 289 062 499 999 999 999 999 890 948 096 × 2 = 1 + 0.327 999 591 827 392 578 124 999 999 999 999 999 781 896 192;
  • 30) 0.327 999 591 827 392 578 124 999 999 999 999 999 781 896 192 × 2 = 0 + 0.655 999 183 654 785 156 249 999 999 999 999 999 563 792 384;
  • 31) 0.655 999 183 654 785 156 249 999 999 999 999 999 563 792 384 × 2 = 1 + 0.311 998 367 309 570 312 499 999 999 999 999 999 127 584 768;
  • 32) 0.311 998 367 309 570 312 499 999 999 999 999 999 127 584 768 × 2 = 0 + 0.623 996 734 619 140 624 999 999 999 999 999 998 255 169 536;
  • 33) 0.623 996 734 619 140 624 999 999 999 999 999 998 255 169 536 × 2 = 1 + 0.247 993 469 238 281 249 999 999 999 999 999 996 510 339 072;
  • 34) 0.247 993 469 238 281 249 999 999 999 999 999 996 510 339 072 × 2 = 0 + 0.495 986 938 476 562 499 999 999 999 999 999 993 020 678 144;
  • 35) 0.495 986 938 476 562 499 999 999 999 999 999 993 020 678 144 × 2 = 0 + 0.991 973 876 953 124 999 999 999 999 999 999 986 041 356 288;
  • 36) 0.991 973 876 953 124 999 999 999 999 999 999 986 041 356 288 × 2 = 1 + 0.983 947 753 906 249 999 999 999 999 999 999 972 082 712 576;
  • 37) 0.983 947 753 906 249 999 999 999 999 999 999 972 082 712 576 × 2 = 1 + 0.967 895 507 812 499 999 999 999 999 999 999 944 165 425 152;
  • 38) 0.967 895 507 812 499 999 999 999 999 999 999 944 165 425 152 × 2 = 1 + 0.935 791 015 624 999 999 999 999 999 999 999 888 330 850 304;
  • 39) 0.935 791 015 624 999 999 999 999 999 999 999 888 330 850 304 × 2 = 1 + 0.871 582 031 249 999 999 999 999 999 999 999 776 661 700 608;
  • 40) 0.871 582 031 249 999 999 999 999 999 999 999 776 661 700 608 × 2 = 1 + 0.743 164 062 499 999 999 999 999 999 999 999 553 323 401 216;
  • 41) 0.743 164 062 499 999 999 999 999 999 999 999 553 323 401 216 × 2 = 1 + 0.486 328 124 999 999 999 999 999 999 999 999 106 646 802 432;
  • 42) 0.486 328 124 999 999 999 999 999 999 999 999 106 646 802 432 × 2 = 0 + 0.972 656 249 999 999 999 999 999 999 999 998 213 293 604 864;
  • 43) 0.972 656 249 999 999 999 999 999 999 999 998 213 293 604 864 × 2 = 1 + 0.945 312 499 999 999 999 999 999 999 999 996 426 587 209 728;
  • 44) 0.945 312 499 999 999 999 999 999 999 999 996 426 587 209 728 × 2 = 1 + 0.890 624 999 999 999 999 999 999 999 999 992 853 174 419 456;
  • 45) 0.890 624 999 999 999 999 999 999 999 999 992 853 174 419 456 × 2 = 1 + 0.781 249 999 999 999 999 999 999 999 999 985 706 348 838 912;
  • 46) 0.781 249 999 999 999 999 999 999 999 999 985 706 348 838 912 × 2 = 1 + 0.562 499 999 999 999 999 999 999 999 999 971 412 697 677 824;
  • 47) 0.562 499 999 999 999 999 999 999 999 999 971 412 697 677 824 × 2 = 1 + 0.124 999 999 999 999 999 999 999 999 999 942 825 395 355 648;
  • 48) 0.124 999 999 999 999 999 999 999 999 999 942 825 395 355 648 × 2 = 0 + 0.249 999 999 999 999 999 999 999 999 999 885 650 790 711 296;
  • 49) 0.249 999 999 999 999 999 999 999 999 999 885 650 790 711 296 × 2 = 0 + 0.499 999 999 999 999 999 999 999 999 999 771 301 581 422 592;
  • 50) 0.499 999 999 999 999 999 999 999 999 999 771 301 581 422 592 × 2 = 0 + 0.999 999 999 999 999 999 999 999 999 999 542 603 162 845 184;
  • 51) 0.999 999 999 999 999 999 999 999 999 999 542 603 162 845 184 × 2 = 1 + 0.999 999 999 999 999 999 999 999 999 999 085 206 325 690 368;
  • 52) 0.999 999 999 999 999 999 999 999 999 999 085 206 325 690 368 × 2 = 1 + 0.999 999 999 999 999 999 999 999 999 998 170 412 651 380 736;
  • 53) 0.999 999 999 999 999 999 999 999 999 998 170 412 651 380 736 × 2 = 1 + 0.999 999 999 999 999 999 999 999 999 996 340 825 302 761 472;

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:


0.143 999 999 999 999 239 719 272 736 692 801 117 897 033 691(10) =

0.0010 0100 1101 1101 0010 1111 0001 1010 1001 1111 1011 1110 0011 1(2)

5. Positive number before normalization:

5.143 999 999 999 999 239 719 272 736 692 801 117 897 033 691(10) =

101.0010 0100 1101 1101 0010 1111 0001 1010 1001 1111 1011 1110 0011 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 2 positions to the left, so that only one non zero digit remains to the left of it:


5.143 999 999 999 999 239 719 272 736 692 801 117 897 033 691(10) =

101.0010 0100 1101 1101 0010 1111 0001 1010 1001 1111 1011 1110 0011 1(2) =

101.0010 0100 1101 1101 0010 1111 0001 1010 1001 1111 1011 1110 0011 1(2) × 20 =

1.0100 1001 0011 0111 0100 1011 1100 0110 1010 0111 1110 1111 1000 111(2) × 22


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): 2

Mantissa (not normalized):
1.0100 1001 0011 0111 0100 1011 1100 0110 1010 0111 1110 1111 1000 111


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

2 + 2(11-1) - 1 =

(2 + 1 023)(10) =

1 025(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;
  • 1 025 ÷ 2 = 512 + 1;
  • 512 ÷ 2 = 256 + 0;
  • 256 ÷ 2 = 128 + 0;
  • 128 ÷ 2 = 64 + 0;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 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) =

1025(10) =

100 0000 0001(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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =

1. 0100 1001 0011 0111 0100 1011 1100 0110 1010 0111 1110 1111 1000 111 =

0100 1001 0011 0111 0100 1011 1100 0110 1010 0111 1110 1111 1000


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) =
100 0000 0001

Mantissa (52 bits) =
0100 1001 0011 0111 0100 1011 1100 0110 1010 0111 1110 1111 1000


Decimal number 5.143 999 999 999 999 239 719 272 736 692 801 117 897 033 691 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0001 - 0100 1001 0011 0111 0100 1011 1100 0110 1010 0111 1110 1111 1000

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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