-0.000 084 993 381 976 777 51 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 084 993 381 976 777 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
-0.000 084 993 381 976 777 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:

|-0.000 084 993 381 976 777 51| = 0.000 084 993 381 976 777 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: 0.000 084 993 381 976 777 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) 0.000 084 993 381 976 777 51 × 2 = 0 + 0.000 169 986 763 953 555 02;
  • 2) 0.000 169 986 763 953 555 02 × 2 = 0 + 0.000 339 973 527 907 110 04;
  • 3) 0.000 339 973 527 907 110 04 × 2 = 0 + 0.000 679 947 055 814 220 08;
  • 4) 0.000 679 947 055 814 220 08 × 2 = 0 + 0.001 359 894 111 628 440 16;
  • 5) 0.001 359 894 111 628 440 16 × 2 = 0 + 0.002 719 788 223 256 880 32;
  • 6) 0.002 719 788 223 256 880 32 × 2 = 0 + 0.005 439 576 446 513 760 64;
  • 7) 0.005 439 576 446 513 760 64 × 2 = 0 + 0.010 879 152 893 027 521 28;
  • 8) 0.010 879 152 893 027 521 28 × 2 = 0 + 0.021 758 305 786 055 042 56;
  • 9) 0.021 758 305 786 055 042 56 × 2 = 0 + 0.043 516 611 572 110 085 12;
  • 10) 0.043 516 611 572 110 085 12 × 2 = 0 + 0.087 033 223 144 220 170 24;
  • 11) 0.087 033 223 144 220 170 24 × 2 = 0 + 0.174 066 446 288 440 340 48;
  • 12) 0.174 066 446 288 440 340 48 × 2 = 0 + 0.348 132 892 576 880 680 96;
  • 13) 0.348 132 892 576 880 680 96 × 2 = 0 + 0.696 265 785 153 761 361 92;
  • 14) 0.696 265 785 153 761 361 92 × 2 = 1 + 0.392 531 570 307 522 723 84;
  • 15) 0.392 531 570 307 522 723 84 × 2 = 0 + 0.785 063 140 615 045 447 68;
  • 16) 0.785 063 140 615 045 447 68 × 2 = 1 + 0.570 126 281 230 090 895 36;
  • 17) 0.570 126 281 230 090 895 36 × 2 = 1 + 0.140 252 562 460 181 790 72;
  • 18) 0.140 252 562 460 181 790 72 × 2 = 0 + 0.280 505 124 920 363 581 44;
  • 19) 0.280 505 124 920 363 581 44 × 2 = 0 + 0.561 010 249 840 727 162 88;
  • 20) 0.561 010 249 840 727 162 88 × 2 = 1 + 0.122 020 499 681 454 325 76;
  • 21) 0.122 020 499 681 454 325 76 × 2 = 0 + 0.244 040 999 362 908 651 52;
  • 22) 0.244 040 999 362 908 651 52 × 2 = 0 + 0.488 081 998 725 817 303 04;
  • 23) 0.488 081 998 725 817 303 04 × 2 = 0 + 0.976 163 997 451 634 606 08;
  • 24) 0.976 163 997 451 634 606 08 × 2 = 1 + 0.952 327 994 903 269 212 16;
  • 25) 0.952 327 994 903 269 212 16 × 2 = 1 + 0.904 655 989 806 538 424 32;
  • 26) 0.904 655 989 806 538 424 32 × 2 = 1 + 0.809 311 979 613 076 848 64;
  • 27) 0.809 311 979 613 076 848 64 × 2 = 1 + 0.618 623 959 226 153 697 28;
  • 28) 0.618 623 959 226 153 697 28 × 2 = 1 + 0.237 247 918 452 307 394 56;
  • 29) 0.237 247 918 452 307 394 56 × 2 = 0 + 0.474 495 836 904 614 789 12;
  • 30) 0.474 495 836 904 614 789 12 × 2 = 0 + 0.948 991 673 809 229 578 24;
  • 31) 0.948 991 673 809 229 578 24 × 2 = 1 + 0.897 983 347 618 459 156 48;
  • 32) 0.897 983 347 618 459 156 48 × 2 = 1 + 0.795 966 695 236 918 312 96;
  • 33) 0.795 966 695 236 918 312 96 × 2 = 1 + 0.591 933 390 473 836 625 92;
  • 34) 0.591 933 390 473 836 625 92 × 2 = 1 + 0.183 866 780 947 673 251 84;
  • 35) 0.183 866 780 947 673 251 84 × 2 = 0 + 0.367 733 561 895 346 503 68;
  • 36) 0.367 733 561 895 346 503 68 × 2 = 0 + 0.735 467 123 790 693 007 36;
  • 37) 0.735 467 123 790 693 007 36 × 2 = 1 + 0.470 934 247 581 386 014 72;
  • 38) 0.470 934 247 581 386 014 72 × 2 = 0 + 0.941 868 495 162 772 029 44;
  • 39) 0.941 868 495 162 772 029 44 × 2 = 1 + 0.883 736 990 325 544 058 88;
  • 40) 0.883 736 990 325 544 058 88 × 2 = 1 + 0.767 473 980 651 088 117 76;
  • 41) 0.767 473 980 651 088 117 76 × 2 = 1 + 0.534 947 961 302 176 235 52;
  • 42) 0.534 947 961 302 176 235 52 × 2 = 1 + 0.069 895 922 604 352 471 04;
  • 43) 0.069 895 922 604 352 471 04 × 2 = 0 + 0.139 791 845 208 704 942 08;
  • 44) 0.139 791 845 208 704 942 08 × 2 = 0 + 0.279 583 690 417 409 884 16;
  • 45) 0.279 583 690 417 409 884 16 × 2 = 0 + 0.559 167 380 834 819 768 32;
  • 46) 0.559 167 380 834 819 768 32 × 2 = 1 + 0.118 334 761 669 639 536 64;
  • 47) 0.118 334 761 669 639 536 64 × 2 = 0 + 0.236 669 523 339 279 073 28;
  • 48) 0.236 669 523 339 279 073 28 × 2 = 0 + 0.473 339 046 678 558 146 56;
  • 49) 0.473 339 046 678 558 146 56 × 2 = 0 + 0.946 678 093 357 116 293 12;
  • 50) 0.946 678 093 357 116 293 12 × 2 = 1 + 0.893 356 186 714 232 586 24;
  • 51) 0.893 356 186 714 232 586 24 × 2 = 1 + 0.786 712 373 428 465 172 48;
  • 52) 0.786 712 373 428 465 172 48 × 2 = 1 + 0.573 424 746 856 930 344 96;
  • 53) 0.573 424 746 856 930 344 96 × 2 = 1 + 0.146 849 493 713 860 689 92;
  • 54) 0.146 849 493 713 860 689 92 × 2 = 0 + 0.293 698 987 427 721 379 84;
  • 55) 0.293 698 987 427 721 379 84 × 2 = 0 + 0.587 397 974 855 442 759 68;
  • 56) 0.587 397 974 855 442 759 68 × 2 = 1 + 0.174 795 949 710 885 519 36;
  • 57) 0.174 795 949 710 885 519 36 × 2 = 0 + 0.349 591 899 421 771 038 72;
  • 58) 0.349 591 899 421 771 038 72 × 2 = 0 + 0.699 183 798 843 542 077 44;
  • 59) 0.699 183 798 843 542 077 44 × 2 = 1 + 0.398 367 597 687 084 154 88;
  • 60) 0.398 367 597 687 084 154 88 × 2 = 0 + 0.796 735 195 374 168 309 76;
  • 61) 0.796 735 195 374 168 309 76 × 2 = 1 + 0.593 470 390 748 336 619 52;
  • 62) 0.593 470 390 748 336 619 52 × 2 = 1 + 0.186 940 781 496 673 239 04;
  • 63) 0.186 940 781 496 673 239 04 × 2 = 0 + 0.373 881 562 993 346 478 08;
  • 64) 0.373 881 562 993 346 478 08 × 2 = 0 + 0.747 763 125 986 692 956 16;
  • 65) 0.747 763 125 986 692 956 16 × 2 = 1 + 0.495 526 251 973 385 912 32;
  • 66) 0.495 526 251 973 385 912 32 × 2 = 0 + 0.991 052 503 946 771 824 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:


0.000 084 993 381 976 777 51(10) =

0.0000 0000 0000 0101 1001 0001 1111 0011 1100 1011 1100 0100 0111 1001 0010 1100 10(2)

6. Positive number before normalization:

0.000 084 993 381 976 777 51(10) =

0.0000 0000 0000 0101 1001 0001 1111 0011 1100 1011 1100 0100 0111 1001 0010 1100 10(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:


0.000 084 993 381 976 777 51(10) =

0.0000 0000 0000 0101 1001 0001 1111 0011 1100 1011 1100 0100 0111 1001 0010 1100 10(2) =

0.0000 0000 0000 0101 1001 0001 1111 0011 1100 1011 1100 0100 0111 1001 0010 1100 10(2) × 20 =

1.0110 0100 0111 1100 1111 0010 1111 0001 0001 1110 0100 1011 0010(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 1011 0010


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 1011 0010 =

0110 0100 0111 1100 1111 0010 1111 0001 0001 1110 0100 1011 0010


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 1011 0010


Decimal number -0.000 084 993 381 976 777 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 1011 0010

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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