0.000 000 000 000 003 8 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 000 000 003 8(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
0.000 000 000 000 003 8(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: 0.000 000 000 000 003 8.

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) 0.000 000 000 000 003 8 × 2 = 0 + 0.000 000 000 000 007 6;
  • 2) 0.000 000 000 000 007 6 × 2 = 0 + 0.000 000 000 000 015 2;
  • 3) 0.000 000 000 000 015 2 × 2 = 0 + 0.000 000 000 000 030 4;
  • 4) 0.000 000 000 000 030 4 × 2 = 0 + 0.000 000 000 000 060 8;
  • 5) 0.000 000 000 000 060 8 × 2 = 0 + 0.000 000 000 000 121 6;
  • 6) 0.000 000 000 000 121 6 × 2 = 0 + 0.000 000 000 000 243 2;
  • 7) 0.000 000 000 000 243 2 × 2 = 0 + 0.000 000 000 000 486 4;
  • 8) 0.000 000 000 000 486 4 × 2 = 0 + 0.000 000 000 000 972 8;
  • 9) 0.000 000 000 000 972 8 × 2 = 0 + 0.000 000 000 001 945 6;
  • 10) 0.000 000 000 001 945 6 × 2 = 0 + 0.000 000 000 003 891 2;
  • 11) 0.000 000 000 003 891 2 × 2 = 0 + 0.000 000 000 007 782 4;
  • 12) 0.000 000 000 007 782 4 × 2 = 0 + 0.000 000 000 015 564 8;
  • 13) 0.000 000 000 015 564 8 × 2 = 0 + 0.000 000 000 031 129 6;
  • 14) 0.000 000 000 031 129 6 × 2 = 0 + 0.000 000 000 062 259 2;
  • 15) 0.000 000 000 062 259 2 × 2 = 0 + 0.000 000 000 124 518 4;
  • 16) 0.000 000 000 124 518 4 × 2 = 0 + 0.000 000 000 249 036 8;
  • 17) 0.000 000 000 249 036 8 × 2 = 0 + 0.000 000 000 498 073 6;
  • 18) 0.000 000 000 498 073 6 × 2 = 0 + 0.000 000 000 996 147 2;
  • 19) 0.000 000 000 996 147 2 × 2 = 0 + 0.000 000 001 992 294 4;
  • 20) 0.000 000 001 992 294 4 × 2 = 0 + 0.000 000 003 984 588 8;
  • 21) 0.000 000 003 984 588 8 × 2 = 0 + 0.000 000 007 969 177 6;
  • 22) 0.000 000 007 969 177 6 × 2 = 0 + 0.000 000 015 938 355 2;
  • 23) 0.000 000 015 938 355 2 × 2 = 0 + 0.000 000 031 876 710 4;
  • 24) 0.000 000 031 876 710 4 × 2 = 0 + 0.000 000 063 753 420 8;
  • 25) 0.000 000 063 753 420 8 × 2 = 0 + 0.000 000 127 506 841 6;
  • 26) 0.000 000 127 506 841 6 × 2 = 0 + 0.000 000 255 013 683 2;
  • 27) 0.000 000 255 013 683 2 × 2 = 0 + 0.000 000 510 027 366 4;
  • 28) 0.000 000 510 027 366 4 × 2 = 0 + 0.000 001 020 054 732 8;
  • 29) 0.000 001 020 054 732 8 × 2 = 0 + 0.000 002 040 109 465 6;
  • 30) 0.000 002 040 109 465 6 × 2 = 0 + 0.000 004 080 218 931 2;
  • 31) 0.000 004 080 218 931 2 × 2 = 0 + 0.000 008 160 437 862 4;
  • 32) 0.000 008 160 437 862 4 × 2 = 0 + 0.000 016 320 875 724 8;
  • 33) 0.000 016 320 875 724 8 × 2 = 0 + 0.000 032 641 751 449 6;
  • 34) 0.000 032 641 751 449 6 × 2 = 0 + 0.000 065 283 502 899 2;
  • 35) 0.000 065 283 502 899 2 × 2 = 0 + 0.000 130 567 005 798 4;
  • 36) 0.000 130 567 005 798 4 × 2 = 0 + 0.000 261 134 011 596 8;
  • 37) 0.000 261 134 011 596 8 × 2 = 0 + 0.000 522 268 023 193 6;
  • 38) 0.000 522 268 023 193 6 × 2 = 0 + 0.001 044 536 046 387 2;
  • 39) 0.001 044 536 046 387 2 × 2 = 0 + 0.002 089 072 092 774 4;
  • 40) 0.002 089 072 092 774 4 × 2 = 0 + 0.004 178 144 185 548 8;
  • 41) 0.004 178 144 185 548 8 × 2 = 0 + 0.008 356 288 371 097 6;
  • 42) 0.008 356 288 371 097 6 × 2 = 0 + 0.016 712 576 742 195 2;
  • 43) 0.016 712 576 742 195 2 × 2 = 0 + 0.033 425 153 484 390 4;
  • 44) 0.033 425 153 484 390 4 × 2 = 0 + 0.066 850 306 968 780 8;
  • 45) 0.066 850 306 968 780 8 × 2 = 0 + 0.133 700 613 937 561 6;
  • 46) 0.133 700 613 937 561 6 × 2 = 0 + 0.267 401 227 875 123 2;
  • 47) 0.267 401 227 875 123 2 × 2 = 0 + 0.534 802 455 750 246 4;
  • 48) 0.534 802 455 750 246 4 × 2 = 1 + 0.069 604 911 500 492 8;
  • 49) 0.069 604 911 500 492 8 × 2 = 0 + 0.139 209 823 000 985 6;
  • 50) 0.139 209 823 000 985 6 × 2 = 0 + 0.278 419 646 001 971 2;
  • 51) 0.278 419 646 001 971 2 × 2 = 0 + 0.556 839 292 003 942 4;
  • 52) 0.556 839 292 003 942 4 × 2 = 1 + 0.113 678 584 007 884 8;
  • 53) 0.113 678 584 007 884 8 × 2 = 0 + 0.227 357 168 015 769 6;
  • 54) 0.227 357 168 015 769 6 × 2 = 0 + 0.454 714 336 031 539 2;
  • 55) 0.454 714 336 031 539 2 × 2 = 0 + 0.909 428 672 063 078 4;
  • 56) 0.909 428 672 063 078 4 × 2 = 1 + 0.818 857 344 126 156 8;
  • 57) 0.818 857 344 126 156 8 × 2 = 1 + 0.637 714 688 252 313 6;
  • 58) 0.637 714 688 252 313 6 × 2 = 1 + 0.275 429 376 504 627 2;
  • 59) 0.275 429 376 504 627 2 × 2 = 0 + 0.550 858 753 009 254 4;
  • 60) 0.550 858 753 009 254 4 × 2 = 1 + 0.101 717 506 018 508 8;
  • 61) 0.101 717 506 018 508 8 × 2 = 0 + 0.203 435 012 037 017 6;
  • 62) 0.203 435 012 037 017 6 × 2 = 0 + 0.406 870 024 074 035 2;
  • 63) 0.406 870 024 074 035 2 × 2 = 0 + 0.813 740 048 148 070 4;
  • 64) 0.813 740 048 148 070 4 × 2 = 1 + 0.627 480 096 296 140 8;
  • 65) 0.627 480 096 296 140 8 × 2 = 1 + 0.254 960 192 592 281 6;
  • 66) 0.254 960 192 592 281 6 × 2 = 0 + 0.509 920 385 184 563 2;
  • 67) 0.509 920 385 184 563 2 × 2 = 1 + 0.019 840 770 369 126 4;
  • 68) 0.019 840 770 369 126 4 × 2 = 0 + 0.039 681 540 738 252 8;
  • 69) 0.039 681 540 738 252 8 × 2 = 0 + 0.079 363 081 476 505 6;
  • 70) 0.079 363 081 476 505 6 × 2 = 0 + 0.158 726 162 953 011 2;
  • 71) 0.158 726 162 953 011 2 × 2 = 0 + 0.317 452 325 906 022 4;
  • 72) 0.317 452 325 906 022 4 × 2 = 0 + 0.634 904 651 812 044 8;
  • 73) 0.634 904 651 812 044 8 × 2 = 1 + 0.269 809 303 624 089 6;
  • 74) 0.269 809 303 624 089 6 × 2 = 0 + 0.539 618 607 248 179 2;
  • 75) 0.539 618 607 248 179 2 × 2 = 1 + 0.079 237 214 496 358 4;
  • 76) 0.079 237 214 496 358 4 × 2 = 0 + 0.158 474 428 992 716 8;
  • 77) 0.158 474 428 992 716 8 × 2 = 0 + 0.316 948 857 985 433 6;
  • 78) 0.316 948 857 985 433 6 × 2 = 0 + 0.633 897 715 970 867 2;
  • 79) 0.633 897 715 970 867 2 × 2 = 1 + 0.267 795 431 941 734 4;
  • 80) 0.267 795 431 941 734 4 × 2 = 0 + 0.535 590 863 883 468 8;
  • 81) 0.535 590 863 883 468 8 × 2 = 1 + 0.071 181 727 766 937 6;
  • 82) 0.071 181 727 766 937 6 × 2 = 0 + 0.142 363 455 533 875 2;
  • 83) 0.142 363 455 533 875 2 × 2 = 0 + 0.284 726 911 067 750 4;
  • 84) 0.284 726 911 067 750 4 × 2 = 0 + 0.569 453 822 135 500 8;
  • 85) 0.569 453 822 135 500 8 × 2 = 1 + 0.138 907 644 271 001 6;
  • 86) 0.138 907 644 271 001 6 × 2 = 0 + 0.277 815 288 542 003 2;
  • 87) 0.277 815 288 542 003 2 × 2 = 0 + 0.555 630 577 084 006 4;
  • 88) 0.555 630 577 084 006 4 × 2 = 1 + 0.111 261 154 168 012 8;
  • 89) 0.111 261 154 168 012 8 × 2 = 0 + 0.222 522 308 336 025 6;
  • 90) 0.222 522 308 336 025 6 × 2 = 0 + 0.445 044 616 672 051 2;
  • 91) 0.445 044 616 672 051 2 × 2 = 0 + 0.890 089 233 344 102 4;
  • 92) 0.890 089 233 344 102 4 × 2 = 1 + 0.780 178 466 688 204 8;
  • 93) 0.780 178 466 688 204 8 × 2 = 1 + 0.560 356 933 376 409 6;
  • 94) 0.560 356 933 376 409 6 × 2 = 1 + 0.120 713 866 752 819 2;
  • 95) 0.120 713 866 752 819 2 × 2 = 0 + 0.241 427 733 505 638 4;
  • 96) 0.241 427 733 505 638 4 × 2 = 0 + 0.482 855 467 011 276 8;
  • 97) 0.482 855 467 011 276 8 × 2 = 0 + 0.965 710 934 022 553 6;
  • 98) 0.965 710 934 022 553 6 × 2 = 1 + 0.931 421 868 045 107 2;
  • 99) 0.931 421 868 045 107 2 × 2 = 1 + 0.862 843 736 090 214 4;
  • 100) 0.862 843 736 090 214 4 × 2 = 1 + 0.725 687 472 180 428 8;

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:


0.000 000 000 000 003 8(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 0001 0001 1101 0001 1010 0000 1010 0010 1000 1001 0001 1100 0111(2)

5. Positive number before normalization:

0.000 000 000 000 003 8(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 0001 0001 1101 0001 1010 0000 1010 0010 1000 1001 0001 1100 0111(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 48 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 003 8(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 0001 0001 1101 0001 1010 0000 1010 0010 1000 1001 0001 1100 0111(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 0001 0001 1101 0001 1010 0000 1010 0010 1000 1001 0001 1100 0111(2) × 20 =

1.0001 0001 1101 0001 1010 0000 1010 0010 1000 1001 0001 1100 0111(2) × 2-48


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): -48

Mantissa (not normalized):
1.0001 0001 1101 0001 1010 0000 1010 0010 1000 1001 0001 1100 0111


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-48 + 2(11-1) - 1 =

(-48 + 1 023)(10) =

975(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;
  • 975 ÷ 2 = 487 + 1;
  • 487 ÷ 2 = 243 + 1;
  • 243 ÷ 2 = 121 + 1;
  • 121 ÷ 2 = 60 + 1;
  • 60 ÷ 2 = 30 + 0;
  • 30 ÷ 2 = 15 + 0;
  • 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) =

975(10) =

011 1100 1111(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. 0001 0001 1101 0001 1010 0000 1010 0010 1000 1001 0001 1100 0111 =

0001 0001 1101 0001 1010 0000 1010 0010 1000 1001 0001 1100 0111


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 1100 1111

Mantissa (52 bits) =
0001 0001 1101 0001 1010 0000 1010 0010 1000 1001 0001 1100 0111


Decimal number 0.000 000 000 000 003 8 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1100 1111 - 0001 0001 1101 0001 1010 0000 1010 0010 1000 1001 0001 1100 0111

» Convert decimal 0.000 000 000 000 013 4 to 64 bit double precision IEEE 754 binary floating point representation standard

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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