-0.000 084 993 381 976 765 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 084 993 381 976 765(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
-0.000 084 993 381 976 765(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:

|-0.000 084 993 381 976 765| = 0.000 084 993 381 976 765


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: 0.000 084 993 381 976 765.

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) 0.000 084 993 381 976 765 × 2 = 0 + 0.000 169 986 763 953 53;
  • 2) 0.000 169 986 763 953 53 × 2 = 0 + 0.000 339 973 527 907 06;
  • 3) 0.000 339 973 527 907 06 × 2 = 0 + 0.000 679 947 055 814 12;
  • 4) 0.000 679 947 055 814 12 × 2 = 0 + 0.001 359 894 111 628 24;
  • 5) 0.001 359 894 111 628 24 × 2 = 0 + 0.002 719 788 223 256 48;
  • 6) 0.002 719 788 223 256 48 × 2 = 0 + 0.005 439 576 446 512 96;
  • 7) 0.005 439 576 446 512 96 × 2 = 0 + 0.010 879 152 893 025 92;
  • 8) 0.010 879 152 893 025 92 × 2 = 0 + 0.021 758 305 786 051 84;
  • 9) 0.021 758 305 786 051 84 × 2 = 0 + 0.043 516 611 572 103 68;
  • 10) 0.043 516 611 572 103 68 × 2 = 0 + 0.087 033 223 144 207 36;
  • 11) 0.087 033 223 144 207 36 × 2 = 0 + 0.174 066 446 288 414 72;
  • 12) 0.174 066 446 288 414 72 × 2 = 0 + 0.348 132 892 576 829 44;
  • 13) 0.348 132 892 576 829 44 × 2 = 0 + 0.696 265 785 153 658 88;
  • 14) 0.696 265 785 153 658 88 × 2 = 1 + 0.392 531 570 307 317 76;
  • 15) 0.392 531 570 307 317 76 × 2 = 0 + 0.785 063 140 614 635 52;
  • 16) 0.785 063 140 614 635 52 × 2 = 1 + 0.570 126 281 229 271 04;
  • 17) 0.570 126 281 229 271 04 × 2 = 1 + 0.140 252 562 458 542 08;
  • 18) 0.140 252 562 458 542 08 × 2 = 0 + 0.280 505 124 917 084 16;
  • 19) 0.280 505 124 917 084 16 × 2 = 0 + 0.561 010 249 834 168 32;
  • 20) 0.561 010 249 834 168 32 × 2 = 1 + 0.122 020 499 668 336 64;
  • 21) 0.122 020 499 668 336 64 × 2 = 0 + 0.244 040 999 336 673 28;
  • 22) 0.244 040 999 336 673 28 × 2 = 0 + 0.488 081 998 673 346 56;
  • 23) 0.488 081 998 673 346 56 × 2 = 0 + 0.976 163 997 346 693 12;
  • 24) 0.976 163 997 346 693 12 × 2 = 1 + 0.952 327 994 693 386 24;
  • 25) 0.952 327 994 693 386 24 × 2 = 1 + 0.904 655 989 386 772 48;
  • 26) 0.904 655 989 386 772 48 × 2 = 1 + 0.809 311 978 773 544 96;
  • 27) 0.809 311 978 773 544 96 × 2 = 1 + 0.618 623 957 547 089 92;
  • 28) 0.618 623 957 547 089 92 × 2 = 1 + 0.237 247 915 094 179 84;
  • 29) 0.237 247 915 094 179 84 × 2 = 0 + 0.474 495 830 188 359 68;
  • 30) 0.474 495 830 188 359 68 × 2 = 0 + 0.948 991 660 376 719 36;
  • 31) 0.948 991 660 376 719 36 × 2 = 1 + 0.897 983 320 753 438 72;
  • 32) 0.897 983 320 753 438 72 × 2 = 1 + 0.795 966 641 506 877 44;
  • 33) 0.795 966 641 506 877 44 × 2 = 1 + 0.591 933 283 013 754 88;
  • 34) 0.591 933 283 013 754 88 × 2 = 1 + 0.183 866 566 027 509 76;
  • 35) 0.183 866 566 027 509 76 × 2 = 0 + 0.367 733 132 055 019 52;
  • 36) 0.367 733 132 055 019 52 × 2 = 0 + 0.735 466 264 110 039 04;
  • 37) 0.735 466 264 110 039 04 × 2 = 1 + 0.470 932 528 220 078 08;
  • 38) 0.470 932 528 220 078 08 × 2 = 0 + 0.941 865 056 440 156 16;
  • 39) 0.941 865 056 440 156 16 × 2 = 1 + 0.883 730 112 880 312 32;
  • 40) 0.883 730 112 880 312 32 × 2 = 1 + 0.767 460 225 760 624 64;
  • 41) 0.767 460 225 760 624 64 × 2 = 1 + 0.534 920 451 521 249 28;
  • 42) 0.534 920 451 521 249 28 × 2 = 1 + 0.069 840 903 042 498 56;
  • 43) 0.069 840 903 042 498 56 × 2 = 0 + 0.139 681 806 084 997 12;
  • 44) 0.139 681 806 084 997 12 × 2 = 0 + 0.279 363 612 169 994 24;
  • 45) 0.279 363 612 169 994 24 × 2 = 0 + 0.558 727 224 339 988 48;
  • 46) 0.558 727 224 339 988 48 × 2 = 1 + 0.117 454 448 679 976 96;
  • 47) 0.117 454 448 679 976 96 × 2 = 0 + 0.234 908 897 359 953 92;
  • 48) 0.234 908 897 359 953 92 × 2 = 0 + 0.469 817 794 719 907 84;
  • 49) 0.469 817 794 719 907 84 × 2 = 0 + 0.939 635 589 439 815 68;
  • 50) 0.939 635 589 439 815 68 × 2 = 1 + 0.879 271 178 879 631 36;
  • 51) 0.879 271 178 879 631 36 × 2 = 1 + 0.758 542 357 759 262 72;
  • 52) 0.758 542 357 759 262 72 × 2 = 1 + 0.517 084 715 518 525 44;
  • 53) 0.517 084 715 518 525 44 × 2 = 1 + 0.034 169 431 037 050 88;
  • 54) 0.034 169 431 037 050 88 × 2 = 0 + 0.068 338 862 074 101 76;
  • 55) 0.068 338 862 074 101 76 × 2 = 0 + 0.136 677 724 148 203 52;
  • 56) 0.136 677 724 148 203 52 × 2 = 0 + 0.273 355 448 296 407 04;
  • 57) 0.273 355 448 296 407 04 × 2 = 0 + 0.546 710 896 592 814 08;
  • 58) 0.546 710 896 592 814 08 × 2 = 1 + 0.093 421 793 185 628 16;
  • 59) 0.093 421 793 185 628 16 × 2 = 0 + 0.186 843 586 371 256 32;
  • 60) 0.186 843 586 371 256 32 × 2 = 0 + 0.373 687 172 742 512 64;
  • 61) 0.373 687 172 742 512 64 × 2 = 0 + 0.747 374 345 485 025 28;
  • 62) 0.747 374 345 485 025 28 × 2 = 1 + 0.494 748 690 970 050 56;
  • 63) 0.494 748 690 970 050 56 × 2 = 0 + 0.989 497 381 940 101 12;
  • 64) 0.989 497 381 940 101 12 × 2 = 1 + 0.978 994 763 880 202 24;
  • 65) 0.978 994 763 880 202 24 × 2 = 1 + 0.957 989 527 760 404 48;
  • 66) 0.957 989 527 760 404 48 × 2 = 1 + 0.915 979 055 520 808 96;

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:


0.000 084 993 381 976 765(10) =

0.0000 0000 0000 0101 1001 0001 1111 0011 1100 1011 1100 0100 0111 1000 0100 0101 11(2)

6. Positive number before normalization:

0.000 084 993 381 976 765(10) =

0.0000 0000 0000 0101 1001 0001 1111 0011 1100 1011 1100 0100 0111 1000 0100 0101 11(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:


0.000 084 993 381 976 765(10) =

0.0000 0000 0000 0101 1001 0001 1111 0011 1100 1011 1100 0100 0111 1000 0100 0101 11(2) =

0.0000 0000 0000 0101 1001 0001 1111 0011 1100 1011 1100 0100 0111 1000 0100 0101 11(2) × 20 =

1.0110 0100 0111 1100 1111 0010 1111 0001 0001 1110 0001 0001 0111(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 0001 0001 0111


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 0001 0001 0111 =

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


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 0001 0001 0111


Decimal number -0.000 084 993 381 976 765 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 0001 0001 0111

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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