0.000 000 178 571 431 8 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 178 571 431 8(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
0.000 000 178 571 431 8(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. 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;

2. 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)


3. Convert to binary (base 2) the fractional part: 0.000 000 178 571 431 8.

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) 0.000 000 178 571 431 8 × 2 = 0 + 0.000 000 357 142 863 6;
  • 2) 0.000 000 357 142 863 6 × 2 = 0 + 0.000 000 714 285 727 2;
  • 3) 0.000 000 714 285 727 2 × 2 = 0 + 0.000 001 428 571 454 4;
  • 4) 0.000 001 428 571 454 4 × 2 = 0 + 0.000 002 857 142 908 8;
  • 5) 0.000 002 857 142 908 8 × 2 = 0 + 0.000 005 714 285 817 6;
  • 6) 0.000 005 714 285 817 6 × 2 = 0 + 0.000 011 428 571 635 2;
  • 7) 0.000 011 428 571 635 2 × 2 = 0 + 0.000 022 857 143 270 4;
  • 8) 0.000 022 857 143 270 4 × 2 = 0 + 0.000 045 714 286 540 8;
  • 9) 0.000 045 714 286 540 8 × 2 = 0 + 0.000 091 428 573 081 6;
  • 10) 0.000 091 428 573 081 6 × 2 = 0 + 0.000 182 857 146 163 2;
  • 11) 0.000 182 857 146 163 2 × 2 = 0 + 0.000 365 714 292 326 4;
  • 12) 0.000 365 714 292 326 4 × 2 = 0 + 0.000 731 428 584 652 8;
  • 13) 0.000 731 428 584 652 8 × 2 = 0 + 0.001 462 857 169 305 6;
  • 14) 0.001 462 857 169 305 6 × 2 = 0 + 0.002 925 714 338 611 2;
  • 15) 0.002 925 714 338 611 2 × 2 = 0 + 0.005 851 428 677 222 4;
  • 16) 0.005 851 428 677 222 4 × 2 = 0 + 0.011 702 857 354 444 8;
  • 17) 0.011 702 857 354 444 8 × 2 = 0 + 0.023 405 714 708 889 6;
  • 18) 0.023 405 714 708 889 6 × 2 = 0 + 0.046 811 429 417 779 2;
  • 19) 0.046 811 429 417 779 2 × 2 = 0 + 0.093 622 858 835 558 4;
  • 20) 0.093 622 858 835 558 4 × 2 = 0 + 0.187 245 717 671 116 8;
  • 21) 0.187 245 717 671 116 8 × 2 = 0 + 0.374 491 435 342 233 6;
  • 22) 0.374 491 435 342 233 6 × 2 = 0 + 0.748 982 870 684 467 2;
  • 23) 0.748 982 870 684 467 2 × 2 = 1 + 0.497 965 741 368 934 4;
  • 24) 0.497 965 741 368 934 4 × 2 = 0 + 0.995 931 482 737 868 8;
  • 25) 0.995 931 482 737 868 8 × 2 = 1 + 0.991 862 965 475 737 6;
  • 26) 0.991 862 965 475 737 6 × 2 = 1 + 0.983 725 930 951 475 2;
  • 27) 0.983 725 930 951 475 2 × 2 = 1 + 0.967 451 861 902 950 4;
  • 28) 0.967 451 861 902 950 4 × 2 = 1 + 0.934 903 723 805 900 8;
  • 29) 0.934 903 723 805 900 8 × 2 = 1 + 0.869 807 447 611 801 6;
  • 30) 0.869 807 447 611 801 6 × 2 = 1 + 0.739 614 895 223 603 2;
  • 31) 0.739 614 895 223 603 2 × 2 = 1 + 0.479 229 790 447 206 4;
  • 32) 0.479 229 790 447 206 4 × 2 = 0 + 0.958 459 580 894 412 8;
  • 33) 0.958 459 580 894 412 8 × 2 = 1 + 0.916 919 161 788 825 6;
  • 34) 0.916 919 161 788 825 6 × 2 = 1 + 0.833 838 323 577 651 2;
  • 35) 0.833 838 323 577 651 2 × 2 = 1 + 0.667 676 647 155 302 4;
  • 36) 0.667 676 647 155 302 4 × 2 = 1 + 0.335 353 294 310 604 8;
  • 37) 0.335 353 294 310 604 8 × 2 = 0 + 0.670 706 588 621 209 6;
  • 38) 0.670 706 588 621 209 6 × 2 = 1 + 0.341 413 177 242 419 2;
  • 39) 0.341 413 177 242 419 2 × 2 = 0 + 0.682 826 354 484 838 4;
  • 40) 0.682 826 354 484 838 4 × 2 = 1 + 0.365 652 708 969 676 8;
  • 41) 0.365 652 708 969 676 8 × 2 = 0 + 0.731 305 417 939 353 6;
  • 42) 0.731 305 417 939 353 6 × 2 = 1 + 0.462 610 835 878 707 2;
  • 43) 0.462 610 835 878 707 2 × 2 = 0 + 0.925 221 671 757 414 4;
  • 44) 0.925 221 671 757 414 4 × 2 = 1 + 0.850 443 343 514 828 8;
  • 45) 0.850 443 343 514 828 8 × 2 = 1 + 0.700 886 687 029 657 6;
  • 46) 0.700 886 687 029 657 6 × 2 = 1 + 0.401 773 374 059 315 2;
  • 47) 0.401 773 374 059 315 2 × 2 = 0 + 0.803 546 748 118 630 4;
  • 48) 0.803 546 748 118 630 4 × 2 = 1 + 0.607 093 496 237 260 8;
  • 49) 0.607 093 496 237 260 8 × 2 = 1 + 0.214 186 992 474 521 6;
  • 50) 0.214 186 992 474 521 6 × 2 = 0 + 0.428 373 984 949 043 2;
  • 51) 0.428 373 984 949 043 2 × 2 = 0 + 0.856 747 969 898 086 4;
  • 52) 0.856 747 969 898 086 4 × 2 = 1 + 0.713 495 939 796 172 8;
  • 53) 0.713 495 939 796 172 8 × 2 = 1 + 0.426 991 879 592 345 6;
  • 54) 0.426 991 879 592 345 6 × 2 = 0 + 0.853 983 759 184 691 2;
  • 55) 0.853 983 759 184 691 2 × 2 = 1 + 0.707 967 518 369 382 4;
  • 56) 0.707 967 518 369 382 4 × 2 = 1 + 0.415 935 036 738 764 8;
  • 57) 0.415 935 036 738 764 8 × 2 = 0 + 0.831 870 073 477 529 6;
  • 58) 0.831 870 073 477 529 6 × 2 = 1 + 0.663 740 146 955 059 2;
  • 59) 0.663 740 146 955 059 2 × 2 = 1 + 0.327 480 293 910 118 4;
  • 60) 0.327 480 293 910 118 4 × 2 = 0 + 0.654 960 587 820 236 8;
  • 61) 0.654 960 587 820 236 8 × 2 = 1 + 0.309 921 175 640 473 6;
  • 62) 0.309 921 175 640 473 6 × 2 = 0 + 0.619 842 351 280 947 2;
  • 63) 0.619 842 351 280 947 2 × 2 = 1 + 0.239 684 702 561 894 4;
  • 64) 0.239 684 702 561 894 4 × 2 = 0 + 0.479 369 405 123 788 8;
  • 65) 0.479 369 405 123 788 8 × 2 = 0 + 0.958 738 810 247 577 6;
  • 66) 0.958 738 810 247 577 6 × 2 = 1 + 0.917 477 620 495 155 2;
  • 67) 0.917 477 620 495 155 2 × 2 = 1 + 0.834 955 240 990 310 4;
  • 68) 0.834 955 240 990 310 4 × 2 = 1 + 0.669 910 481 980 620 8;
  • 69) 0.669 910 481 980 620 8 × 2 = 1 + 0.339 820 963 961 241 6;
  • 70) 0.339 820 963 961 241 6 × 2 = 0 + 0.679 641 927 922 483 2;
  • 71) 0.679 641 927 922 483 2 × 2 = 1 + 0.359 283 855 844 966 4;
  • 72) 0.359 283 855 844 966 4 × 2 = 0 + 0.718 567 711 689 932 8;
  • 73) 0.718 567 711 689 932 8 × 2 = 1 + 0.437 135 423 379 865 6;
  • 74) 0.437 135 423 379 865 6 × 2 = 0 + 0.874 270 846 759 731 2;
  • 75) 0.874 270 846 759 731 2 × 2 = 1 + 0.748 541 693 519 462 4;

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).


4. 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:


0.000 000 178 571 431 8(10) =

0.0000 0000 0000 0000 0000 0010 1111 1110 1111 0101 0101 1101 1001 1011 0110 1010 0111 1010 101(2)

5. Positive number before normalization:

0.000 000 178 571 431 8(10) =

0.0000 0000 0000 0000 0000 0010 1111 1110 1111 0101 0101 1101 1001 1011 0110 1010 0111 1010 101(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 23 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 178 571 431 8(10) =

0.0000 0000 0000 0000 0000 0010 1111 1110 1111 0101 0101 1101 1001 1011 0110 1010 0111 1010 101(2) =

0.0000 0000 0000 0000 0000 0010 1111 1110 1111 0101 0101 1101 1001 1011 0110 1010 0111 1010 101(2) × 20 =

1.0111 1111 0111 1010 1010 1110 1100 1101 1011 0101 0011 1101 0101(2) × 2-23


7. 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 0 (a positive number)

Exponent (unadjusted): -23

Mantissa (not normalized):
1.0111 1111 0111 1010 1010 1110 1100 1101 1011 0101 0011 1101 0101


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-23 + 2(11-1) - 1 =

(-23 + 1 023)(10) =

1 000(10)


9. 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;
  • 1 000 ÷ 2 = 500 + 0;
  • 500 ÷ 2 = 250 + 0;
  • 250 ÷ 2 = 125 + 0;
  • 125 ÷ 2 = 62 + 1;
  • 62 ÷ 2 = 31 + 0;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

1000(10) =

011 1110 1000(2)


11. 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. 0111 1111 0111 1010 1010 1110 1100 1101 1011 0101 0011 1101 0101 =

0111 1111 0111 1010 1010 1110 1100 1101 1011 0101 0011 1101 0101


12. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (11 bits) =
011 1110 1000

Mantissa (52 bits) =
0111 1111 0111 1010 1010 1110 1100 1101 1011 0101 0011 1101 0101


Decimal number 0.000 000 178 571 431 8 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1110 1000 - 0111 1111 0111 1010 1010 1110 1100 1101 1011 0101 0011 1101 0101

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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