-0.000 084 993 381 976 744 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 084 993 381 976 744 3(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
-0.000 084 993 381 976 744 3(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:

|-0.000 084 993 381 976 744 3| = 0.000 084 993 381 976 744 3


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: 0.000 084 993 381 976 744 3.

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) 0.000 084 993 381 976 744 3 × 2 = 0 + 0.000 169 986 763 953 488 6;
  • 2) 0.000 169 986 763 953 488 6 × 2 = 0 + 0.000 339 973 527 906 977 2;
  • 3) 0.000 339 973 527 906 977 2 × 2 = 0 + 0.000 679 947 055 813 954 4;
  • 4) 0.000 679 947 055 813 954 4 × 2 = 0 + 0.001 359 894 111 627 908 8;
  • 5) 0.001 359 894 111 627 908 8 × 2 = 0 + 0.002 719 788 223 255 817 6;
  • 6) 0.002 719 788 223 255 817 6 × 2 = 0 + 0.005 439 576 446 511 635 2;
  • 7) 0.005 439 576 446 511 635 2 × 2 = 0 + 0.010 879 152 893 023 270 4;
  • 8) 0.010 879 152 893 023 270 4 × 2 = 0 + 0.021 758 305 786 046 540 8;
  • 9) 0.021 758 305 786 046 540 8 × 2 = 0 + 0.043 516 611 572 093 081 6;
  • 10) 0.043 516 611 572 093 081 6 × 2 = 0 + 0.087 033 223 144 186 163 2;
  • 11) 0.087 033 223 144 186 163 2 × 2 = 0 + 0.174 066 446 288 372 326 4;
  • 12) 0.174 066 446 288 372 326 4 × 2 = 0 + 0.348 132 892 576 744 652 8;
  • 13) 0.348 132 892 576 744 652 8 × 2 = 0 + 0.696 265 785 153 489 305 6;
  • 14) 0.696 265 785 153 489 305 6 × 2 = 1 + 0.392 531 570 306 978 611 2;
  • 15) 0.392 531 570 306 978 611 2 × 2 = 0 + 0.785 063 140 613 957 222 4;
  • 16) 0.785 063 140 613 957 222 4 × 2 = 1 + 0.570 126 281 227 914 444 8;
  • 17) 0.570 126 281 227 914 444 8 × 2 = 1 + 0.140 252 562 455 828 889 6;
  • 18) 0.140 252 562 455 828 889 6 × 2 = 0 + 0.280 505 124 911 657 779 2;
  • 19) 0.280 505 124 911 657 779 2 × 2 = 0 + 0.561 010 249 823 315 558 4;
  • 20) 0.561 010 249 823 315 558 4 × 2 = 1 + 0.122 020 499 646 631 116 8;
  • 21) 0.122 020 499 646 631 116 8 × 2 = 0 + 0.244 040 999 293 262 233 6;
  • 22) 0.244 040 999 293 262 233 6 × 2 = 0 + 0.488 081 998 586 524 467 2;
  • 23) 0.488 081 998 586 524 467 2 × 2 = 0 + 0.976 163 997 173 048 934 4;
  • 24) 0.976 163 997 173 048 934 4 × 2 = 1 + 0.952 327 994 346 097 868 8;
  • 25) 0.952 327 994 346 097 868 8 × 2 = 1 + 0.904 655 988 692 195 737 6;
  • 26) 0.904 655 988 692 195 737 6 × 2 = 1 + 0.809 311 977 384 391 475 2;
  • 27) 0.809 311 977 384 391 475 2 × 2 = 1 + 0.618 623 954 768 782 950 4;
  • 28) 0.618 623 954 768 782 950 4 × 2 = 1 + 0.237 247 909 537 565 900 8;
  • 29) 0.237 247 909 537 565 900 8 × 2 = 0 + 0.474 495 819 075 131 801 6;
  • 30) 0.474 495 819 075 131 801 6 × 2 = 0 + 0.948 991 638 150 263 603 2;
  • 31) 0.948 991 638 150 263 603 2 × 2 = 1 + 0.897 983 276 300 527 206 4;
  • 32) 0.897 983 276 300 527 206 4 × 2 = 1 + 0.795 966 552 601 054 412 8;
  • 33) 0.795 966 552 601 054 412 8 × 2 = 1 + 0.591 933 105 202 108 825 6;
  • 34) 0.591 933 105 202 108 825 6 × 2 = 1 + 0.183 866 210 404 217 651 2;
  • 35) 0.183 866 210 404 217 651 2 × 2 = 0 + 0.367 732 420 808 435 302 4;
  • 36) 0.367 732 420 808 435 302 4 × 2 = 0 + 0.735 464 841 616 870 604 8;
  • 37) 0.735 464 841 616 870 604 8 × 2 = 1 + 0.470 929 683 233 741 209 6;
  • 38) 0.470 929 683 233 741 209 6 × 2 = 0 + 0.941 859 366 467 482 419 2;
  • 39) 0.941 859 366 467 482 419 2 × 2 = 1 + 0.883 718 732 934 964 838 4;
  • 40) 0.883 718 732 934 964 838 4 × 2 = 1 + 0.767 437 465 869 929 676 8;
  • 41) 0.767 437 465 869 929 676 8 × 2 = 1 + 0.534 874 931 739 859 353 6;
  • 42) 0.534 874 931 739 859 353 6 × 2 = 1 + 0.069 749 863 479 718 707 2;
  • 43) 0.069 749 863 479 718 707 2 × 2 = 0 + 0.139 499 726 959 437 414 4;
  • 44) 0.139 499 726 959 437 414 4 × 2 = 0 + 0.278 999 453 918 874 828 8;
  • 45) 0.278 999 453 918 874 828 8 × 2 = 0 + 0.557 998 907 837 749 657 6;
  • 46) 0.557 998 907 837 749 657 6 × 2 = 1 + 0.115 997 815 675 499 315 2;
  • 47) 0.115 997 815 675 499 315 2 × 2 = 0 + 0.231 995 631 350 998 630 4;
  • 48) 0.231 995 631 350 998 630 4 × 2 = 0 + 0.463 991 262 701 997 260 8;
  • 49) 0.463 991 262 701 997 260 8 × 2 = 0 + 0.927 982 525 403 994 521 6;
  • 50) 0.927 982 525 403 994 521 6 × 2 = 1 + 0.855 965 050 807 989 043 2;
  • 51) 0.855 965 050 807 989 043 2 × 2 = 1 + 0.711 930 101 615 978 086 4;
  • 52) 0.711 930 101 615 978 086 4 × 2 = 1 + 0.423 860 203 231 956 172 8;
  • 53) 0.423 860 203 231 956 172 8 × 2 = 0 + 0.847 720 406 463 912 345 6;
  • 54) 0.847 720 406 463 912 345 6 × 2 = 1 + 0.695 440 812 927 824 691 2;
  • 55) 0.695 440 812 927 824 691 2 × 2 = 1 + 0.390 881 625 855 649 382 4;
  • 56) 0.390 881 625 855 649 382 4 × 2 = 0 + 0.781 763 251 711 298 764 8;
  • 57) 0.781 763 251 711 298 764 8 × 2 = 1 + 0.563 526 503 422 597 529 6;
  • 58) 0.563 526 503 422 597 529 6 × 2 = 1 + 0.127 053 006 845 195 059 2;
  • 59) 0.127 053 006 845 195 059 2 × 2 = 0 + 0.254 106 013 690 390 118 4;
  • 60) 0.254 106 013 690 390 118 4 × 2 = 0 + 0.508 212 027 380 780 236 8;
  • 61) 0.508 212 027 380 780 236 8 × 2 = 1 + 0.016 424 054 761 560 473 6;
  • 62) 0.016 424 054 761 560 473 6 × 2 = 0 + 0.032 848 109 523 120 947 2;
  • 63) 0.032 848 109 523 120 947 2 × 2 = 0 + 0.065 696 219 046 241 894 4;
  • 64) 0.065 696 219 046 241 894 4 × 2 = 0 + 0.131 392 438 092 483 788 8;
  • 65) 0.131 392 438 092 483 788 8 × 2 = 0 + 0.262 784 876 184 967 577 6;
  • 66) 0.262 784 876 184 967 577 6 × 2 = 0 + 0.525 569 752 369 935 155 2;

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:


0.000 084 993 381 976 744 3(10) =

0.0000 0000 0000 0101 1001 0001 1111 0011 1100 1011 1100 0100 0111 0110 1100 1000 00(2)

6. Positive number before normalization:

0.000 084 993 381 976 744 3(10) =

0.0000 0000 0000 0101 1001 0001 1111 0011 1100 1011 1100 0100 0111 0110 1100 1000 00(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:


0.000 084 993 381 976 744 3(10) =

0.0000 0000 0000 0101 1001 0001 1111 0011 1100 1011 1100 0100 0111 0110 1100 1000 00(2) =

0.0000 0000 0000 0101 1001 0001 1111 0011 1100 1011 1100 0100 0111 0110 1100 1000 00(2) × 20 =

1.0110 0100 0111 1100 1111 0010 1111 0001 0001 1101 1011 0010 0000(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 1101 1011 0010 0000


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 1101 1011 0010 0000 =

0110 0100 0111 1100 1111 0010 1111 0001 0001 1101 1011 0010 0000


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 1101 1011 0010 0000


Decimal number -0.000 084 993 381 976 744 3 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 1101 1011 0010 0000

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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