-0.000 000 005 917 031 260 475 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 005 917 031 260 475(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
-0.000 000 005 917 031 260 475(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:

|-0.000 000 005 917 031 260 475| = 0.000 000 005 917 031 260 475


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: 0.000 000 005 917 031 260 475.

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) 0.000 000 005 917 031 260 475 × 2 = 0 + 0.000 000 011 834 062 520 95;
  • 2) 0.000 000 011 834 062 520 95 × 2 = 0 + 0.000 000 023 668 125 041 9;
  • 3) 0.000 000 023 668 125 041 9 × 2 = 0 + 0.000 000 047 336 250 083 8;
  • 4) 0.000 000 047 336 250 083 8 × 2 = 0 + 0.000 000 094 672 500 167 6;
  • 5) 0.000 000 094 672 500 167 6 × 2 = 0 + 0.000 000 189 345 000 335 2;
  • 6) 0.000 000 189 345 000 335 2 × 2 = 0 + 0.000 000 378 690 000 670 4;
  • 7) 0.000 000 378 690 000 670 4 × 2 = 0 + 0.000 000 757 380 001 340 8;
  • 8) 0.000 000 757 380 001 340 8 × 2 = 0 + 0.000 001 514 760 002 681 6;
  • 9) 0.000 001 514 760 002 681 6 × 2 = 0 + 0.000 003 029 520 005 363 2;
  • 10) 0.000 003 029 520 005 363 2 × 2 = 0 + 0.000 006 059 040 010 726 4;
  • 11) 0.000 006 059 040 010 726 4 × 2 = 0 + 0.000 012 118 080 021 452 8;
  • 12) 0.000 012 118 080 021 452 8 × 2 = 0 + 0.000 024 236 160 042 905 6;
  • 13) 0.000 024 236 160 042 905 6 × 2 = 0 + 0.000 048 472 320 085 811 2;
  • 14) 0.000 048 472 320 085 811 2 × 2 = 0 + 0.000 096 944 640 171 622 4;
  • 15) 0.000 096 944 640 171 622 4 × 2 = 0 + 0.000 193 889 280 343 244 8;
  • 16) 0.000 193 889 280 343 244 8 × 2 = 0 + 0.000 387 778 560 686 489 6;
  • 17) 0.000 387 778 560 686 489 6 × 2 = 0 + 0.000 775 557 121 372 979 2;
  • 18) 0.000 775 557 121 372 979 2 × 2 = 0 + 0.001 551 114 242 745 958 4;
  • 19) 0.001 551 114 242 745 958 4 × 2 = 0 + 0.003 102 228 485 491 916 8;
  • 20) 0.003 102 228 485 491 916 8 × 2 = 0 + 0.006 204 456 970 983 833 6;
  • 21) 0.006 204 456 970 983 833 6 × 2 = 0 + 0.012 408 913 941 967 667 2;
  • 22) 0.012 408 913 941 967 667 2 × 2 = 0 + 0.024 817 827 883 935 334 4;
  • 23) 0.024 817 827 883 935 334 4 × 2 = 0 + 0.049 635 655 767 870 668 8;
  • 24) 0.049 635 655 767 870 668 8 × 2 = 0 + 0.099 271 311 535 741 337 6;
  • 25) 0.099 271 311 535 741 337 6 × 2 = 0 + 0.198 542 623 071 482 675 2;
  • 26) 0.198 542 623 071 482 675 2 × 2 = 0 + 0.397 085 246 142 965 350 4;
  • 27) 0.397 085 246 142 965 350 4 × 2 = 0 + 0.794 170 492 285 930 700 8;
  • 28) 0.794 170 492 285 930 700 8 × 2 = 1 + 0.588 340 984 571 861 401 6;
  • 29) 0.588 340 984 571 861 401 6 × 2 = 1 + 0.176 681 969 143 722 803 2;
  • 30) 0.176 681 969 143 722 803 2 × 2 = 0 + 0.353 363 938 287 445 606 4;
  • 31) 0.353 363 938 287 445 606 4 × 2 = 0 + 0.706 727 876 574 891 212 8;
  • 32) 0.706 727 876 574 891 212 8 × 2 = 1 + 0.413 455 753 149 782 425 6;
  • 33) 0.413 455 753 149 782 425 6 × 2 = 0 + 0.826 911 506 299 564 851 2;
  • 34) 0.826 911 506 299 564 851 2 × 2 = 1 + 0.653 823 012 599 129 702 4;
  • 35) 0.653 823 012 599 129 702 4 × 2 = 1 + 0.307 646 025 198 259 404 8;
  • 36) 0.307 646 025 198 259 404 8 × 2 = 0 + 0.615 292 050 396 518 809 6;
  • 37) 0.615 292 050 396 518 809 6 × 2 = 1 + 0.230 584 100 793 037 619 2;
  • 38) 0.230 584 100 793 037 619 2 × 2 = 0 + 0.461 168 201 586 075 238 4;
  • 39) 0.461 168 201 586 075 238 4 × 2 = 0 + 0.922 336 403 172 150 476 8;
  • 40) 0.922 336 403 172 150 476 8 × 2 = 1 + 0.844 672 806 344 300 953 6;
  • 41) 0.844 672 806 344 300 953 6 × 2 = 1 + 0.689 345 612 688 601 907 2;
  • 42) 0.689 345 612 688 601 907 2 × 2 = 1 + 0.378 691 225 377 203 814 4;
  • 43) 0.378 691 225 377 203 814 4 × 2 = 0 + 0.757 382 450 754 407 628 8;
  • 44) 0.757 382 450 754 407 628 8 × 2 = 1 + 0.514 764 901 508 815 257 6;
  • 45) 0.514 764 901 508 815 257 6 × 2 = 1 + 0.029 529 803 017 630 515 2;
  • 46) 0.029 529 803 017 630 515 2 × 2 = 0 + 0.059 059 606 035 261 030 4;
  • 47) 0.059 059 606 035 261 030 4 × 2 = 0 + 0.118 119 212 070 522 060 8;
  • 48) 0.118 119 212 070 522 060 8 × 2 = 0 + 0.236 238 424 141 044 121 6;
  • 49) 0.236 238 424 141 044 121 6 × 2 = 0 + 0.472 476 848 282 088 243 2;
  • 50) 0.472 476 848 282 088 243 2 × 2 = 0 + 0.944 953 696 564 176 486 4;
  • 51) 0.944 953 696 564 176 486 4 × 2 = 1 + 0.889 907 393 128 352 972 8;
  • 52) 0.889 907 393 128 352 972 8 × 2 = 1 + 0.779 814 786 256 705 945 6;
  • 53) 0.779 814 786 256 705 945 6 × 2 = 1 + 0.559 629 572 513 411 891 2;
  • 54) 0.559 629 572 513 411 891 2 × 2 = 1 + 0.119 259 145 026 823 782 4;
  • 55) 0.119 259 145 026 823 782 4 × 2 = 0 + 0.238 518 290 053 647 564 8;
  • 56) 0.238 518 290 053 647 564 8 × 2 = 0 + 0.477 036 580 107 295 129 6;
  • 57) 0.477 036 580 107 295 129 6 × 2 = 0 + 0.954 073 160 214 590 259 2;
  • 58) 0.954 073 160 214 590 259 2 × 2 = 1 + 0.908 146 320 429 180 518 4;
  • 59) 0.908 146 320 429 180 518 4 × 2 = 1 + 0.816 292 640 858 361 036 8;
  • 60) 0.816 292 640 858 361 036 8 × 2 = 1 + 0.632 585 281 716 722 073 6;
  • 61) 0.632 585 281 716 722 073 6 × 2 = 1 + 0.265 170 563 433 444 147 2;
  • 62) 0.265 170 563 433 444 147 2 × 2 = 0 + 0.530 341 126 866 888 294 4;
  • 63) 0.530 341 126 866 888 294 4 × 2 = 1 + 0.060 682 253 733 776 588 8;
  • 64) 0.060 682 253 733 776 588 8 × 2 = 0 + 0.121 364 507 467 553 177 6;
  • 65) 0.121 364 507 467 553 177 6 × 2 = 0 + 0.242 729 014 935 106 355 2;
  • 66) 0.242 729 014 935 106 355 2 × 2 = 0 + 0.485 458 029 870 212 710 4;
  • 67) 0.485 458 029 870 212 710 4 × 2 = 0 + 0.970 916 059 740 425 420 8;
  • 68) 0.970 916 059 740 425 420 8 × 2 = 1 + 0.941 832 119 480 850 841 6;
  • 69) 0.941 832 119 480 850 841 6 × 2 = 1 + 0.883 664 238 961 701 683 2;
  • 70) 0.883 664 238 961 701 683 2 × 2 = 1 + 0.767 328 477 923 403 366 4;
  • 71) 0.767 328 477 923 403 366 4 × 2 = 1 + 0.534 656 955 846 806 732 8;
  • 72) 0.534 656 955 846 806 732 8 × 2 = 1 + 0.069 313 911 693 613 465 6;
  • 73) 0.069 313 911 693 613 465 6 × 2 = 0 + 0.138 627 823 387 226 931 2;
  • 74) 0.138 627 823 387 226 931 2 × 2 = 0 + 0.277 255 646 774 453 862 4;
  • 75) 0.277 255 646 774 453 862 4 × 2 = 0 + 0.554 511 293 548 907 724 8;
  • 76) 0.554 511 293 548 907 724 8 × 2 = 1 + 0.109 022 587 097 815 449 6;
  • 77) 0.109 022 587 097 815 449 6 × 2 = 0 + 0.218 045 174 195 630 899 2;
  • 78) 0.218 045 174 195 630 899 2 × 2 = 0 + 0.436 090 348 391 261 798 4;
  • 79) 0.436 090 348 391 261 798 4 × 2 = 0 + 0.872 180 696 782 523 596 8;
  • 80) 0.872 180 696 782 523 596 8 × 2 = 1 + 0.744 361 393 565 047 193 6;

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:


0.000 000 005 917 031 260 475(10) =

0.0000 0000 0000 0000 0000 0000 0001 1001 0110 1001 1101 1000 0011 1100 0111 1010 0001 1111 0001 0001(2)

6. Positive number before normalization:

0.000 000 005 917 031 260 475(10) =

0.0000 0000 0000 0000 0000 0000 0001 1001 0110 1001 1101 1000 0011 1100 0111 1010 0001 1111 0001 0001(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 28 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 005 917 031 260 475(10) =

0.0000 0000 0000 0000 0000 0000 0001 1001 0110 1001 1101 1000 0011 1100 0111 1010 0001 1111 0001 0001(2) =

0.0000 0000 0000 0000 0000 0000 0001 1001 0110 1001 1101 1000 0011 1100 0111 1010 0001 1111 0001 0001(2) × 20 =

1.1001 0110 1001 1101 1000 0011 1100 0111 1010 0001 1111 0001 0001(2) × 2-28


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): -28

Mantissa (not normalized):
1.1001 0110 1001 1101 1000 0011 1100 0111 1010 0001 1111 0001 0001


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-28 + 2(11-1) - 1 =

(-28 + 1 023)(10) =

995(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;
  • 995 ÷ 2 = 497 + 1;
  • 497 ÷ 2 = 248 + 1;
  • 248 ÷ 2 = 124 + 0;
  • 124 ÷ 2 = 62 + 0;
  • 62 ÷ 2 = 31 + 0;
  • 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) =

995(10) =

011 1110 0011(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. 1001 0110 1001 1101 1000 0011 1100 0111 1010 0001 1111 0001 0001 =

1001 0110 1001 1101 1000 0011 1100 0111 1010 0001 1111 0001 0001


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 1110 0011

Mantissa (52 bits) =
1001 0110 1001 1101 1000 0011 1100 0111 1010 0001 1111 0001 0001


Decimal number -0.000 000 005 917 031 260 475 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1110 0011 - 1001 0110 1001 1101 1000 0011 1100 0111 1010 0001 1111 0001 0001

Share

» Convert decimal -0.000 000 005 917 031 260 533 to 64 bit double precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 09, 2026 [September]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: September - by our visitors

» Improve your social skills: The Art of Strategic Ambiguity. Gemini Notebook YouTube Video of 6.59 minutes

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