0.000 000 000 000 022 04 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 000 000 022 04(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
0.000 000 000 000 022 04(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: 0.000 000 000 000 022 04.

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) 0.000 000 000 000 022 04 × 2 = 0 + 0.000 000 000 000 044 08;
  • 2) 0.000 000 000 000 044 08 × 2 = 0 + 0.000 000 000 000 088 16;
  • 3) 0.000 000 000 000 088 16 × 2 = 0 + 0.000 000 000 000 176 32;
  • 4) 0.000 000 000 000 176 32 × 2 = 0 + 0.000 000 000 000 352 64;
  • 5) 0.000 000 000 000 352 64 × 2 = 0 + 0.000 000 000 000 705 28;
  • 6) 0.000 000 000 000 705 28 × 2 = 0 + 0.000 000 000 001 410 56;
  • 7) 0.000 000 000 001 410 56 × 2 = 0 + 0.000 000 000 002 821 12;
  • 8) 0.000 000 000 002 821 12 × 2 = 0 + 0.000 000 000 005 642 24;
  • 9) 0.000 000 000 005 642 24 × 2 = 0 + 0.000 000 000 011 284 48;
  • 10) 0.000 000 000 011 284 48 × 2 = 0 + 0.000 000 000 022 568 96;
  • 11) 0.000 000 000 022 568 96 × 2 = 0 + 0.000 000 000 045 137 92;
  • 12) 0.000 000 000 045 137 92 × 2 = 0 + 0.000 000 000 090 275 84;
  • 13) 0.000 000 000 090 275 84 × 2 = 0 + 0.000 000 000 180 551 68;
  • 14) 0.000 000 000 180 551 68 × 2 = 0 + 0.000 000 000 361 103 36;
  • 15) 0.000 000 000 361 103 36 × 2 = 0 + 0.000 000 000 722 206 72;
  • 16) 0.000 000 000 722 206 72 × 2 = 0 + 0.000 000 001 444 413 44;
  • 17) 0.000 000 001 444 413 44 × 2 = 0 + 0.000 000 002 888 826 88;
  • 18) 0.000 000 002 888 826 88 × 2 = 0 + 0.000 000 005 777 653 76;
  • 19) 0.000 000 005 777 653 76 × 2 = 0 + 0.000 000 011 555 307 52;
  • 20) 0.000 000 011 555 307 52 × 2 = 0 + 0.000 000 023 110 615 04;
  • 21) 0.000 000 023 110 615 04 × 2 = 0 + 0.000 000 046 221 230 08;
  • 22) 0.000 000 046 221 230 08 × 2 = 0 + 0.000 000 092 442 460 16;
  • 23) 0.000 000 092 442 460 16 × 2 = 0 + 0.000 000 184 884 920 32;
  • 24) 0.000 000 184 884 920 32 × 2 = 0 + 0.000 000 369 769 840 64;
  • 25) 0.000 000 369 769 840 64 × 2 = 0 + 0.000 000 739 539 681 28;
  • 26) 0.000 000 739 539 681 28 × 2 = 0 + 0.000 001 479 079 362 56;
  • 27) 0.000 001 479 079 362 56 × 2 = 0 + 0.000 002 958 158 725 12;
  • 28) 0.000 002 958 158 725 12 × 2 = 0 + 0.000 005 916 317 450 24;
  • 29) 0.000 005 916 317 450 24 × 2 = 0 + 0.000 011 832 634 900 48;
  • 30) 0.000 011 832 634 900 48 × 2 = 0 + 0.000 023 665 269 800 96;
  • 31) 0.000 023 665 269 800 96 × 2 = 0 + 0.000 047 330 539 601 92;
  • 32) 0.000 047 330 539 601 92 × 2 = 0 + 0.000 094 661 079 203 84;
  • 33) 0.000 094 661 079 203 84 × 2 = 0 + 0.000 189 322 158 407 68;
  • 34) 0.000 189 322 158 407 68 × 2 = 0 + 0.000 378 644 316 815 36;
  • 35) 0.000 378 644 316 815 36 × 2 = 0 + 0.000 757 288 633 630 72;
  • 36) 0.000 757 288 633 630 72 × 2 = 0 + 0.001 514 577 267 261 44;
  • 37) 0.001 514 577 267 261 44 × 2 = 0 + 0.003 029 154 534 522 88;
  • 38) 0.003 029 154 534 522 88 × 2 = 0 + 0.006 058 309 069 045 76;
  • 39) 0.006 058 309 069 045 76 × 2 = 0 + 0.012 116 618 138 091 52;
  • 40) 0.012 116 618 138 091 52 × 2 = 0 + 0.024 233 236 276 183 04;
  • 41) 0.024 233 236 276 183 04 × 2 = 0 + 0.048 466 472 552 366 08;
  • 42) 0.048 466 472 552 366 08 × 2 = 0 + 0.096 932 945 104 732 16;
  • 43) 0.096 932 945 104 732 16 × 2 = 0 + 0.193 865 890 209 464 32;
  • 44) 0.193 865 890 209 464 32 × 2 = 0 + 0.387 731 780 418 928 64;
  • 45) 0.387 731 780 418 928 64 × 2 = 0 + 0.775 463 560 837 857 28;
  • 46) 0.775 463 560 837 857 28 × 2 = 1 + 0.550 927 121 675 714 56;
  • 47) 0.550 927 121 675 714 56 × 2 = 1 + 0.101 854 243 351 429 12;
  • 48) 0.101 854 243 351 429 12 × 2 = 0 + 0.203 708 486 702 858 24;
  • 49) 0.203 708 486 702 858 24 × 2 = 0 + 0.407 416 973 405 716 48;
  • 50) 0.407 416 973 405 716 48 × 2 = 0 + 0.814 833 946 811 432 96;
  • 51) 0.814 833 946 811 432 96 × 2 = 1 + 0.629 667 893 622 865 92;
  • 52) 0.629 667 893 622 865 92 × 2 = 1 + 0.259 335 787 245 731 84;
  • 53) 0.259 335 787 245 731 84 × 2 = 0 + 0.518 671 574 491 463 68;
  • 54) 0.518 671 574 491 463 68 × 2 = 1 + 0.037 343 148 982 927 36;
  • 55) 0.037 343 148 982 927 36 × 2 = 0 + 0.074 686 297 965 854 72;
  • 56) 0.074 686 297 965 854 72 × 2 = 0 + 0.149 372 595 931 709 44;
  • 57) 0.149 372 595 931 709 44 × 2 = 0 + 0.298 745 191 863 418 88;
  • 58) 0.298 745 191 863 418 88 × 2 = 0 + 0.597 490 383 726 837 76;
  • 59) 0.597 490 383 726 837 76 × 2 = 1 + 0.194 980 767 453 675 52;
  • 60) 0.194 980 767 453 675 52 × 2 = 0 + 0.389 961 534 907 351 04;
  • 61) 0.389 961 534 907 351 04 × 2 = 0 + 0.779 923 069 814 702 08;
  • 62) 0.779 923 069 814 702 08 × 2 = 1 + 0.559 846 139 629 404 16;
  • 63) 0.559 846 139 629 404 16 × 2 = 1 + 0.119 692 279 258 808 32;
  • 64) 0.119 692 279 258 808 32 × 2 = 0 + 0.239 384 558 517 616 64;
  • 65) 0.239 384 558 517 616 64 × 2 = 0 + 0.478 769 117 035 233 28;
  • 66) 0.478 769 117 035 233 28 × 2 = 0 + 0.957 538 234 070 466 56;
  • 67) 0.957 538 234 070 466 56 × 2 = 1 + 0.915 076 468 140 933 12;
  • 68) 0.915 076 468 140 933 12 × 2 = 1 + 0.830 152 936 281 866 24;
  • 69) 0.830 152 936 281 866 24 × 2 = 1 + 0.660 305 872 563 732 48;
  • 70) 0.660 305 872 563 732 48 × 2 = 1 + 0.320 611 745 127 464 96;
  • 71) 0.320 611 745 127 464 96 × 2 = 0 + 0.641 223 490 254 929 92;
  • 72) 0.641 223 490 254 929 92 × 2 = 1 + 0.282 446 980 509 859 84;
  • 73) 0.282 446 980 509 859 84 × 2 = 0 + 0.564 893 961 019 719 68;
  • 74) 0.564 893 961 019 719 68 × 2 = 1 + 0.129 787 922 039 439 36;
  • 75) 0.129 787 922 039 439 36 × 2 = 0 + 0.259 575 844 078 878 72;
  • 76) 0.259 575 844 078 878 72 × 2 = 0 + 0.519 151 688 157 757 44;
  • 77) 0.519 151 688 157 757 44 × 2 = 1 + 0.038 303 376 315 514 88;
  • 78) 0.038 303 376 315 514 88 × 2 = 0 + 0.076 606 752 631 029 76;
  • 79) 0.076 606 752 631 029 76 × 2 = 0 + 0.153 213 505 262 059 52;
  • 80) 0.153 213 505 262 059 52 × 2 = 0 + 0.306 427 010 524 119 04;
  • 81) 0.306 427 010 524 119 04 × 2 = 0 + 0.612 854 021 048 238 08;
  • 82) 0.612 854 021 048 238 08 × 2 = 1 + 0.225 708 042 096 476 16;
  • 83) 0.225 708 042 096 476 16 × 2 = 0 + 0.451 416 084 192 952 32;
  • 84) 0.451 416 084 192 952 32 × 2 = 0 + 0.902 832 168 385 904 64;
  • 85) 0.902 832 168 385 904 64 × 2 = 1 + 0.805 664 336 771 809 28;
  • 86) 0.805 664 336 771 809 28 × 2 = 1 + 0.611 328 673 543 618 56;
  • 87) 0.611 328 673 543 618 56 × 2 = 1 + 0.222 657 347 087 237 12;
  • 88) 0.222 657 347 087 237 12 × 2 = 0 + 0.445 314 694 174 474 24;
  • 89) 0.445 314 694 174 474 24 × 2 = 0 + 0.890 629 388 348 948 48;
  • 90) 0.890 629 388 348 948 48 × 2 = 1 + 0.781 258 776 697 896 96;
  • 91) 0.781 258 776 697 896 96 × 2 = 1 + 0.562 517 553 395 793 92;
  • 92) 0.562 517 553 395 793 92 × 2 = 1 + 0.125 035 106 791 587 84;
  • 93) 0.125 035 106 791 587 84 × 2 = 0 + 0.250 070 213 583 175 68;
  • 94) 0.250 070 213 583 175 68 × 2 = 0 + 0.500 140 427 166 351 36;
  • 95) 0.500 140 427 166 351 36 × 2 = 1 + 0.000 280 854 332 702 72;
  • 96) 0.000 280 854 332 702 72 × 2 = 0 + 0.000 561 708 665 405 44;
  • 97) 0.000 561 708 665 405 44 × 2 = 0 + 0.001 123 417 330 810 88;
  • 98) 0.001 123 417 330 810 88 × 2 = 0 + 0.002 246 834 661 621 76;

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:


0.000 000 000 000 022 04(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0110 0011 0100 0010 0110 0011 1101 0100 1000 0100 1110 0111 0010 00(2)

5. Positive number before normalization:

0.000 000 000 000 022 04(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0110 0011 0100 0010 0110 0011 1101 0100 1000 0100 1110 0111 0010 00(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 46 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 022 04(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0110 0011 0100 0010 0110 0011 1101 0100 1000 0100 1110 0111 0010 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0110 0011 0100 0010 0110 0011 1101 0100 1000 0100 1110 0111 0010 00(2) × 20 =

1.1000 1101 0000 1001 1000 1111 0101 0010 0001 0011 1001 1100 1000(2) × 2-46


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): -46

Mantissa (not normalized):
1.1000 1101 0000 1001 1000 1111 0101 0010 0001 0011 1001 1100 1000


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-46 + 2(11-1) - 1 =

(-46 + 1 023)(10) =

977(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;
  • 977 ÷ 2 = 488 + 1;
  • 488 ÷ 2 = 244 + 0;
  • 244 ÷ 2 = 122 + 0;
  • 122 ÷ 2 = 61 + 0;
  • 61 ÷ 2 = 30 + 1;
  • 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) =

977(10) =

011 1101 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, only if necessary (not the case here).


Mantissa (normalized) =

1. 1000 1101 0000 1001 1000 1111 0101 0010 0001 0011 1001 1100 1000 =

1000 1101 0000 1001 1000 1111 0101 0010 0001 0011 1001 1100 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) =
011 1101 0001

Mantissa (52 bits) =
1000 1101 0000 1001 1000 1111 0101 0010 0001 0011 1001 1100 1000


Decimal number 0.000 000 000 000 022 04 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1101 0001 - 1000 1101 0000 1001 1000 1111 0101 0010 0001 0011 1001 1100 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