-0.000 084 993 381 976 776 51 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 084 993 381 976 776 51(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
-0.000 084 993 381 976 776 51(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:

|-0.000 084 993 381 976 776 51| = 0.000 084 993 381 976 776 51


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: 0.000 084 993 381 976 776 51.

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) 0.000 084 993 381 976 776 51 × 2 = 0 + 0.000 169 986 763 953 553 02;
  • 2) 0.000 169 986 763 953 553 02 × 2 = 0 + 0.000 339 973 527 907 106 04;
  • 3) 0.000 339 973 527 907 106 04 × 2 = 0 + 0.000 679 947 055 814 212 08;
  • 4) 0.000 679 947 055 814 212 08 × 2 = 0 + 0.001 359 894 111 628 424 16;
  • 5) 0.001 359 894 111 628 424 16 × 2 = 0 + 0.002 719 788 223 256 848 32;
  • 6) 0.002 719 788 223 256 848 32 × 2 = 0 + 0.005 439 576 446 513 696 64;
  • 7) 0.005 439 576 446 513 696 64 × 2 = 0 + 0.010 879 152 893 027 393 28;
  • 8) 0.010 879 152 893 027 393 28 × 2 = 0 + 0.021 758 305 786 054 786 56;
  • 9) 0.021 758 305 786 054 786 56 × 2 = 0 + 0.043 516 611 572 109 573 12;
  • 10) 0.043 516 611 572 109 573 12 × 2 = 0 + 0.087 033 223 144 219 146 24;
  • 11) 0.087 033 223 144 219 146 24 × 2 = 0 + 0.174 066 446 288 438 292 48;
  • 12) 0.174 066 446 288 438 292 48 × 2 = 0 + 0.348 132 892 576 876 584 96;
  • 13) 0.348 132 892 576 876 584 96 × 2 = 0 + 0.696 265 785 153 753 169 92;
  • 14) 0.696 265 785 153 753 169 92 × 2 = 1 + 0.392 531 570 307 506 339 84;
  • 15) 0.392 531 570 307 506 339 84 × 2 = 0 + 0.785 063 140 615 012 679 68;
  • 16) 0.785 063 140 615 012 679 68 × 2 = 1 + 0.570 126 281 230 025 359 36;
  • 17) 0.570 126 281 230 025 359 36 × 2 = 1 + 0.140 252 562 460 050 718 72;
  • 18) 0.140 252 562 460 050 718 72 × 2 = 0 + 0.280 505 124 920 101 437 44;
  • 19) 0.280 505 124 920 101 437 44 × 2 = 0 + 0.561 010 249 840 202 874 88;
  • 20) 0.561 010 249 840 202 874 88 × 2 = 1 + 0.122 020 499 680 405 749 76;
  • 21) 0.122 020 499 680 405 749 76 × 2 = 0 + 0.244 040 999 360 811 499 52;
  • 22) 0.244 040 999 360 811 499 52 × 2 = 0 + 0.488 081 998 721 622 999 04;
  • 23) 0.488 081 998 721 622 999 04 × 2 = 0 + 0.976 163 997 443 245 998 08;
  • 24) 0.976 163 997 443 245 998 08 × 2 = 1 + 0.952 327 994 886 491 996 16;
  • 25) 0.952 327 994 886 491 996 16 × 2 = 1 + 0.904 655 989 772 983 992 32;
  • 26) 0.904 655 989 772 983 992 32 × 2 = 1 + 0.809 311 979 545 967 984 64;
  • 27) 0.809 311 979 545 967 984 64 × 2 = 1 + 0.618 623 959 091 935 969 28;
  • 28) 0.618 623 959 091 935 969 28 × 2 = 1 + 0.237 247 918 183 871 938 56;
  • 29) 0.237 247 918 183 871 938 56 × 2 = 0 + 0.474 495 836 367 743 877 12;
  • 30) 0.474 495 836 367 743 877 12 × 2 = 0 + 0.948 991 672 735 487 754 24;
  • 31) 0.948 991 672 735 487 754 24 × 2 = 1 + 0.897 983 345 470 975 508 48;
  • 32) 0.897 983 345 470 975 508 48 × 2 = 1 + 0.795 966 690 941 951 016 96;
  • 33) 0.795 966 690 941 951 016 96 × 2 = 1 + 0.591 933 381 883 902 033 92;
  • 34) 0.591 933 381 883 902 033 92 × 2 = 1 + 0.183 866 763 767 804 067 84;
  • 35) 0.183 866 763 767 804 067 84 × 2 = 0 + 0.367 733 527 535 608 135 68;
  • 36) 0.367 733 527 535 608 135 68 × 2 = 0 + 0.735 467 055 071 216 271 36;
  • 37) 0.735 467 055 071 216 271 36 × 2 = 1 + 0.470 934 110 142 432 542 72;
  • 38) 0.470 934 110 142 432 542 72 × 2 = 0 + 0.941 868 220 284 865 085 44;
  • 39) 0.941 868 220 284 865 085 44 × 2 = 1 + 0.883 736 440 569 730 170 88;
  • 40) 0.883 736 440 569 730 170 88 × 2 = 1 + 0.767 472 881 139 460 341 76;
  • 41) 0.767 472 881 139 460 341 76 × 2 = 1 + 0.534 945 762 278 920 683 52;
  • 42) 0.534 945 762 278 920 683 52 × 2 = 1 + 0.069 891 524 557 841 367 04;
  • 43) 0.069 891 524 557 841 367 04 × 2 = 0 + 0.139 783 049 115 682 734 08;
  • 44) 0.139 783 049 115 682 734 08 × 2 = 0 + 0.279 566 098 231 365 468 16;
  • 45) 0.279 566 098 231 365 468 16 × 2 = 0 + 0.559 132 196 462 730 936 32;
  • 46) 0.559 132 196 462 730 936 32 × 2 = 1 + 0.118 264 392 925 461 872 64;
  • 47) 0.118 264 392 925 461 872 64 × 2 = 0 + 0.236 528 785 850 923 745 28;
  • 48) 0.236 528 785 850 923 745 28 × 2 = 0 + 0.473 057 571 701 847 490 56;
  • 49) 0.473 057 571 701 847 490 56 × 2 = 0 + 0.946 115 143 403 694 981 12;
  • 50) 0.946 115 143 403 694 981 12 × 2 = 1 + 0.892 230 286 807 389 962 24;
  • 51) 0.892 230 286 807 389 962 24 × 2 = 1 + 0.784 460 573 614 779 924 48;
  • 52) 0.784 460 573 614 779 924 48 × 2 = 1 + 0.568 921 147 229 559 848 96;
  • 53) 0.568 921 147 229 559 848 96 × 2 = 1 + 0.137 842 294 459 119 697 92;
  • 54) 0.137 842 294 459 119 697 92 × 2 = 0 + 0.275 684 588 918 239 395 84;
  • 55) 0.275 684 588 918 239 395 84 × 2 = 0 + 0.551 369 177 836 478 791 68;
  • 56) 0.551 369 177 836 478 791 68 × 2 = 1 + 0.102 738 355 672 957 583 36;
  • 57) 0.102 738 355 672 957 583 36 × 2 = 0 + 0.205 476 711 345 915 166 72;
  • 58) 0.205 476 711 345 915 166 72 × 2 = 0 + 0.410 953 422 691 830 333 44;
  • 59) 0.410 953 422 691 830 333 44 × 2 = 0 + 0.821 906 845 383 660 666 88;
  • 60) 0.821 906 845 383 660 666 88 × 2 = 1 + 0.643 813 690 767 321 333 76;
  • 61) 0.643 813 690 767 321 333 76 × 2 = 1 + 0.287 627 381 534 642 667 52;
  • 62) 0.287 627 381 534 642 667 52 × 2 = 0 + 0.575 254 763 069 285 335 04;
  • 63) 0.575 254 763 069 285 335 04 × 2 = 1 + 0.150 509 526 138 570 670 08;
  • 64) 0.150 509 526 138 570 670 08 × 2 = 0 + 0.301 019 052 277 141 340 16;
  • 65) 0.301 019 052 277 141 340 16 × 2 = 0 + 0.602 038 104 554 282 680 32;
  • 66) 0.602 038 104 554 282 680 32 × 2 = 1 + 0.204 076 209 108 565 360 64;

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:


0.000 084 993 381 976 776 51(10) =

0.0000 0000 0000 0101 1001 0001 1111 0011 1100 1011 1100 0100 0111 1001 0001 1010 01(2)

6. Positive number before normalization:

0.000 084 993 381 976 776 51(10) =

0.0000 0000 0000 0101 1001 0001 1111 0011 1100 1011 1100 0100 0111 1001 0001 1010 01(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 14 positions to the right, so that only one non zero digit remains to the left of it:


0.000 084 993 381 976 776 51(10) =

0.0000 0000 0000 0101 1001 0001 1111 0011 1100 1011 1100 0100 0111 1001 0001 1010 01(2) =

0.0000 0000 0000 0101 1001 0001 1111 0011 1100 1011 1100 0100 0111 1001 0001 1010 01(2) × 20 =

1.0110 0100 0111 1100 1111 0010 1111 0001 0001 1110 0100 0110 1001(2) × 2-14


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): -14

Mantissa (not normalized):
1.0110 0100 0111 1100 1111 0010 1111 0001 0001 1110 0100 0110 1001


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-14 + 2(11-1) - 1 =

(-14 + 1 023)(10) =

1 009(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;
  • 1 009 ÷ 2 = 504 + 1;
  • 504 ÷ 2 = 252 + 0;
  • 252 ÷ 2 = 126 + 0;
  • 126 ÷ 2 = 63 + 0;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 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) =

1009(10) =

011 1111 0001(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. 0110 0100 0111 1100 1111 0010 1111 0001 0001 1110 0100 0110 1001 =

0110 0100 0111 1100 1111 0010 1111 0001 0001 1110 0100 0110 1001


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 1111 0001

Mantissa (52 bits) =
0110 0100 0111 1100 1111 0010 1111 0001 0001 1110 0100 0110 1001


Decimal number -0.000 084 993 381 976 776 51 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0001 - 0110 0100 0111 1100 1111 0010 1111 0001 0001 1110 0100 0110 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