64bit IEEE 754: Decimal ↗ Double Precision Floating Point Binary: -0.000 000 066 823 603 327 5 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number -0.000 000 066 823 603 327 5(10) converted and written in 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:

|-0.000 000 066 823 603 327 5| = 0.000 000 066 823 603 327 5

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: 0.000 000 066 823 603 327 5.

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) 0.000 000 066 823 603 327 5 × 2 = 0 + 0.000 000 133 647 206 655;
  • 2) 0.000 000 133 647 206 655 × 2 = 0 + 0.000 000 267 294 413 31;
  • 3) 0.000 000 267 294 413 31 × 2 = 0 + 0.000 000 534 588 826 62;
  • 4) 0.000 000 534 588 826 62 × 2 = 0 + 0.000 001 069 177 653 24;
  • 5) 0.000 001 069 177 653 24 × 2 = 0 + 0.000 002 138 355 306 48;
  • 6) 0.000 002 138 355 306 48 × 2 = 0 + 0.000 004 276 710 612 96;
  • 7) 0.000 004 276 710 612 96 × 2 = 0 + 0.000 008 553 421 225 92;
  • 8) 0.000 008 553 421 225 92 × 2 = 0 + 0.000 017 106 842 451 84;
  • 9) 0.000 017 106 842 451 84 × 2 = 0 + 0.000 034 213 684 903 68;
  • 10) 0.000 034 213 684 903 68 × 2 = 0 + 0.000 068 427 369 807 36;
  • 11) 0.000 068 427 369 807 36 × 2 = 0 + 0.000 136 854 739 614 72;
  • 12) 0.000 136 854 739 614 72 × 2 = 0 + 0.000 273 709 479 229 44;
  • 13) 0.000 273 709 479 229 44 × 2 = 0 + 0.000 547 418 958 458 88;
  • 14) 0.000 547 418 958 458 88 × 2 = 0 + 0.001 094 837 916 917 76;
  • 15) 0.001 094 837 916 917 76 × 2 = 0 + 0.002 189 675 833 835 52;
  • 16) 0.002 189 675 833 835 52 × 2 = 0 + 0.004 379 351 667 671 04;
  • 17) 0.004 379 351 667 671 04 × 2 = 0 + 0.008 758 703 335 342 08;
  • 18) 0.008 758 703 335 342 08 × 2 = 0 + 0.017 517 406 670 684 16;
  • 19) 0.017 517 406 670 684 16 × 2 = 0 + 0.035 034 813 341 368 32;
  • 20) 0.035 034 813 341 368 32 × 2 = 0 + 0.070 069 626 682 736 64;
  • 21) 0.070 069 626 682 736 64 × 2 = 0 + 0.140 139 253 365 473 28;
  • 22) 0.140 139 253 365 473 28 × 2 = 0 + 0.280 278 506 730 946 56;
  • 23) 0.280 278 506 730 946 56 × 2 = 0 + 0.560 557 013 461 893 12;
  • 24) 0.560 557 013 461 893 12 × 2 = 1 + 0.121 114 026 923 786 24;
  • 25) 0.121 114 026 923 786 24 × 2 = 0 + 0.242 228 053 847 572 48;
  • 26) 0.242 228 053 847 572 48 × 2 = 0 + 0.484 456 107 695 144 96;
  • 27) 0.484 456 107 695 144 96 × 2 = 0 + 0.968 912 215 390 289 92;
  • 28) 0.968 912 215 390 289 92 × 2 = 1 + 0.937 824 430 780 579 84;
  • 29) 0.937 824 430 780 579 84 × 2 = 1 + 0.875 648 861 561 159 68;
  • 30) 0.875 648 861 561 159 68 × 2 = 1 + 0.751 297 723 122 319 36;
  • 31) 0.751 297 723 122 319 36 × 2 = 1 + 0.502 595 446 244 638 72;
  • 32) 0.502 595 446 244 638 72 × 2 = 1 + 0.005 190 892 489 277 44;
  • 33) 0.005 190 892 489 277 44 × 2 = 0 + 0.010 381 784 978 554 88;
  • 34) 0.010 381 784 978 554 88 × 2 = 0 + 0.020 763 569 957 109 76;
  • 35) 0.020 763 569 957 109 76 × 2 = 0 + 0.041 527 139 914 219 52;
  • 36) 0.041 527 139 914 219 52 × 2 = 0 + 0.083 054 279 828 439 04;
  • 37) 0.083 054 279 828 439 04 × 2 = 0 + 0.166 108 559 656 878 08;
  • 38) 0.166 108 559 656 878 08 × 2 = 0 + 0.332 217 119 313 756 16;
  • 39) 0.332 217 119 313 756 16 × 2 = 0 + 0.664 434 238 627 512 32;
  • 40) 0.664 434 238 627 512 32 × 2 = 1 + 0.328 868 477 255 024 64;
  • 41) 0.328 868 477 255 024 64 × 2 = 0 + 0.657 736 954 510 049 28;
  • 42) 0.657 736 954 510 049 28 × 2 = 1 + 0.315 473 909 020 098 56;
  • 43) 0.315 473 909 020 098 56 × 2 = 0 + 0.630 947 818 040 197 12;
  • 44) 0.630 947 818 040 197 12 × 2 = 1 + 0.261 895 636 080 394 24;
  • 45) 0.261 895 636 080 394 24 × 2 = 0 + 0.523 791 272 160 788 48;
  • 46) 0.523 791 272 160 788 48 × 2 = 1 + 0.047 582 544 321 576 96;
  • 47) 0.047 582 544 321 576 96 × 2 = 0 + 0.095 165 088 643 153 92;
  • 48) 0.095 165 088 643 153 92 × 2 = 0 + 0.190 330 177 286 307 84;
  • 49) 0.190 330 177 286 307 84 × 2 = 0 + 0.380 660 354 572 615 68;
  • 50) 0.380 660 354 572 615 68 × 2 = 0 + 0.761 320 709 145 231 36;
  • 51) 0.761 320 709 145 231 36 × 2 = 1 + 0.522 641 418 290 462 72;
  • 52) 0.522 641 418 290 462 72 × 2 = 1 + 0.045 282 836 580 925 44;
  • 53) 0.045 282 836 580 925 44 × 2 = 0 + 0.090 565 673 161 850 88;
  • 54) 0.090 565 673 161 850 88 × 2 = 0 + 0.181 131 346 323 701 76;
  • 55) 0.181 131 346 323 701 76 × 2 = 0 + 0.362 262 692 647 403 52;
  • 56) 0.362 262 692 647 403 52 × 2 = 0 + 0.724 525 385 294 807 04;
  • 57) 0.724 525 385 294 807 04 × 2 = 1 + 0.449 050 770 589 614 08;
  • 58) 0.449 050 770 589 614 08 × 2 = 0 + 0.898 101 541 179 228 16;
  • 59) 0.898 101 541 179 228 16 × 2 = 1 + 0.796 203 082 358 456 32;
  • 60) 0.796 203 082 358 456 32 × 2 = 1 + 0.592 406 164 716 912 64;
  • 61) 0.592 406 164 716 912 64 × 2 = 1 + 0.184 812 329 433 825 28;
  • 62) 0.184 812 329 433 825 28 × 2 = 0 + 0.369 624 658 867 650 56;
  • 63) 0.369 624 658 867 650 56 × 2 = 0 + 0.739 249 317 735 301 12;
  • 64) 0.739 249 317 735 301 12 × 2 = 1 + 0.478 498 635 470 602 24;
  • 65) 0.478 498 635 470 602 24 × 2 = 0 + 0.956 997 270 941 204 48;
  • 66) 0.956 997 270 941 204 48 × 2 = 1 + 0.913 994 541 882 408 96;
  • 67) 0.913 994 541 882 408 96 × 2 = 1 + 0.827 989 083 764 817 92;
  • 68) 0.827 989 083 764 817 92 × 2 = 1 + 0.655 978 167 529 635 84;
  • 69) 0.655 978 167 529 635 84 × 2 = 1 + 0.311 956 335 059 271 68;
  • 70) 0.311 956 335 059 271 68 × 2 = 0 + 0.623 912 670 118 543 36;
  • 71) 0.623 912 670 118 543 36 × 2 = 1 + 0.247 825 340 237 086 72;
  • 72) 0.247 825 340 237 086 72 × 2 = 0 + 0.495 650 680 474 173 44;
  • 73) 0.495 650 680 474 173 44 × 2 = 0 + 0.991 301 360 948 346 88;
  • 74) 0.991 301 360 948 346 88 × 2 = 1 + 0.982 602 721 896 693 76;
  • 75) 0.982 602 721 896 693 76 × 2 = 1 + 0.965 205 443 793 387 52;
  • 76) 0.965 205 443 793 387 52 × 2 = 1 + 0.930 410 887 586 775 04;

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...)


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:


0.000 000 066 823 603 327 5(10) =

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


6. Positive number before normalization:

0.000 000 066 823 603 327 5(10) =

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

7. Normalize the binary representation of the number.

Shift the decimal mark 24 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 066 823 603 327 5(10) =

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

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

1.0001 1111 0000 0001 0101 0100 0011 0000 1011 1001 0111 1010 0111(2) × 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): -24

Mantissa (not normalized):
1.0001 1111 0000 0001 0101 0100 0011 0000 1011 1001 0111 1010 0111


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-24 + 2(11-1) - 1 =

(-24 + 1 023)(10) =

999(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;
  • 999 ÷ 2 = 499 + 1;
  • 499 ÷ 2 = 249 + 1;
  • 249 ÷ 2 = 124 + 1;
  • 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) =

999(10) =

011 1110 0111(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. 0001 1111 0000 0001 0101 0100 0011 0000 1011 1001 0111 1010 0111 =

0001 1111 0000 0001 0101 0100 0011 0000 1011 1001 0111 1010 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 1110 0111

Mantissa (52 bits) =
0001 1111 0000 0001 0101 0100 0011 0000 1011 1001 0111 1010 0111


The base ten decimal number -0.000 000 066 823 603 327 5 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
1 - 011 1110 0111 - 0001 1111 0000 0001 0101 0100 0011 0000 1011 1001 0111 1010 0111

More operations with decimal numbers converted to 64 bit double precision IEEE 754 binary floating point representation:

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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