Decimal to 64 Bit IEEE 754 Binary: Convert Number 0.000 000 347 323 to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From Base Ten Decimal System

Number 0.000 000 347 323(10) converted and written in 64 bit double precision IEEE 754 binary floating point representation standard (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: 0.000 000 347 323.

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) 0.000 000 347 323 × 2 = 0 + 0.000 000 694 646;
  • 2) 0.000 000 694 646 × 2 = 0 + 0.000 001 389 292;
  • 3) 0.000 001 389 292 × 2 = 0 + 0.000 002 778 584;
  • 4) 0.000 002 778 584 × 2 = 0 + 0.000 005 557 168;
  • 5) 0.000 005 557 168 × 2 = 0 + 0.000 011 114 336;
  • 6) 0.000 011 114 336 × 2 = 0 + 0.000 022 228 672;
  • 7) 0.000 022 228 672 × 2 = 0 + 0.000 044 457 344;
  • 8) 0.000 044 457 344 × 2 = 0 + 0.000 088 914 688;
  • 9) 0.000 088 914 688 × 2 = 0 + 0.000 177 829 376;
  • 10) 0.000 177 829 376 × 2 = 0 + 0.000 355 658 752;
  • 11) 0.000 355 658 752 × 2 = 0 + 0.000 711 317 504;
  • 12) 0.000 711 317 504 × 2 = 0 + 0.001 422 635 008;
  • 13) 0.001 422 635 008 × 2 = 0 + 0.002 845 270 016;
  • 14) 0.002 845 270 016 × 2 = 0 + 0.005 690 540 032;
  • 15) 0.005 690 540 032 × 2 = 0 + 0.011 381 080 064;
  • 16) 0.011 381 080 064 × 2 = 0 + 0.022 762 160 128;
  • 17) 0.022 762 160 128 × 2 = 0 + 0.045 524 320 256;
  • 18) 0.045 524 320 256 × 2 = 0 + 0.091 048 640 512;
  • 19) 0.091 048 640 512 × 2 = 0 + 0.182 097 281 024;
  • 20) 0.182 097 281 024 × 2 = 0 + 0.364 194 562 048;
  • 21) 0.364 194 562 048 × 2 = 0 + 0.728 389 124 096;
  • 22) 0.728 389 124 096 × 2 = 1 + 0.456 778 248 192;
  • 23) 0.456 778 248 192 × 2 = 0 + 0.913 556 496 384;
  • 24) 0.913 556 496 384 × 2 = 1 + 0.827 112 992 768;
  • 25) 0.827 112 992 768 × 2 = 1 + 0.654 225 985 536;
  • 26) 0.654 225 985 536 × 2 = 1 + 0.308 451 971 072;
  • 27) 0.308 451 971 072 × 2 = 0 + 0.616 903 942 144;
  • 28) 0.616 903 942 144 × 2 = 1 + 0.233 807 884 288;
  • 29) 0.233 807 884 288 × 2 = 0 + 0.467 615 768 576;
  • 30) 0.467 615 768 576 × 2 = 0 + 0.935 231 537 152;
  • 31) 0.935 231 537 152 × 2 = 1 + 0.870 463 074 304;
  • 32) 0.870 463 074 304 × 2 = 1 + 0.740 926 148 608;
  • 33) 0.740 926 148 608 × 2 = 1 + 0.481 852 297 216;
  • 34) 0.481 852 297 216 × 2 = 0 + 0.963 704 594 432;
  • 35) 0.963 704 594 432 × 2 = 1 + 0.927 409 188 864;
  • 36) 0.927 409 188 864 × 2 = 1 + 0.854 818 377 728;
  • 37) 0.854 818 377 728 × 2 = 1 + 0.709 636 755 456;
  • 38) 0.709 636 755 456 × 2 = 1 + 0.419 273 510 912;
  • 39) 0.419 273 510 912 × 2 = 0 + 0.838 547 021 824;
  • 40) 0.838 547 021 824 × 2 = 1 + 0.677 094 043 648;
  • 41) 0.677 094 043 648 × 2 = 1 + 0.354 188 087 296;
  • 42) 0.354 188 087 296 × 2 = 0 + 0.708 376 174 592;
  • 43) 0.708 376 174 592 × 2 = 1 + 0.416 752 349 184;
  • 44) 0.416 752 349 184 × 2 = 0 + 0.833 504 698 368;
  • 45) 0.833 504 698 368 × 2 = 1 + 0.667 009 396 736;
  • 46) 0.667 009 396 736 × 2 = 1 + 0.334 018 793 472;
  • 47) 0.334 018 793 472 × 2 = 0 + 0.668 037 586 944;
  • 48) 0.668 037 586 944 × 2 = 1 + 0.336 075 173 888;
  • 49) 0.336 075 173 888 × 2 = 0 + 0.672 150 347 776;
  • 50) 0.672 150 347 776 × 2 = 1 + 0.344 300 695 552;
  • 51) 0.344 300 695 552 × 2 = 0 + 0.688 601 391 104;
  • 52) 0.688 601 391 104 × 2 = 1 + 0.377 202 782 208;
  • 53) 0.377 202 782 208 × 2 = 0 + 0.754 405 564 416;
  • 54) 0.754 405 564 416 × 2 = 1 + 0.508 811 128 832;
  • 55) 0.508 811 128 832 × 2 = 1 + 0.017 622 257 664;
  • 56) 0.017 622 257 664 × 2 = 0 + 0.035 244 515 328;
  • 57) 0.035 244 515 328 × 2 = 0 + 0.070 489 030 656;
  • 58) 0.070 489 030 656 × 2 = 0 + 0.140 978 061 312;
  • 59) 0.140 978 061 312 × 2 = 0 + 0.281 956 122 624;
  • 60) 0.281 956 122 624 × 2 = 0 + 0.563 912 245 248;
  • 61) 0.563 912 245 248 × 2 = 1 + 0.127 824 490 496;
  • 62) 0.127 824 490 496 × 2 = 0 + 0.255 648 980 992;
  • 63) 0.255 648 980 992 × 2 = 0 + 0.511 297 961 984;
  • 64) 0.511 297 961 984 × 2 = 1 + 0.022 595 923 968;
  • 65) 0.022 595 923 968 × 2 = 0 + 0.045 191 847 936;
  • 66) 0.045 191 847 936 × 2 = 0 + 0.090 383 695 872;
  • 67) 0.090 383 695 872 × 2 = 0 + 0.180 767 391 744;
  • 68) 0.180 767 391 744 × 2 = 0 + 0.361 534 783 488;
  • 69) 0.361 534 783 488 × 2 = 0 + 0.723 069 566 976;
  • 70) 0.723 069 566 976 × 2 = 1 + 0.446 139 133 952;
  • 71) 0.446 139 133 952 × 2 = 0 + 0.892 278 267 904;
  • 72) 0.892 278 267 904 × 2 = 1 + 0.784 556 535 808;
  • 73) 0.784 556 535 808 × 2 = 1 + 0.569 113 071 616;
  • 74) 0.569 113 071 616 × 2 = 1 + 0.138 226 143 232;

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:


0.000 000 347 323(10) =

0.0000 0000 0000 0000 0000 0101 1101 0011 1011 1101 1010 1101 0101 0110 0000 1001 0000 0101 11(2)


5. Positive number before normalization:

0.000 000 347 323(10) =

0.0000 0000 0000 0000 0000 0101 1101 0011 1011 1101 1010 1101 0101 0110 0000 1001 0000 0101 11(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 22 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 347 323(10) =

0.0000 0000 0000 0000 0000 0101 1101 0011 1011 1101 1010 1101 0101 0110 0000 1001 0000 0101 11(2) =

0.0000 0000 0000 0000 0000 0101 1101 0011 1011 1101 1010 1101 0101 0110 0000 1001 0000 0101 11(2) × 20 =

1.0111 0100 1110 1111 0110 1011 0101 0101 1000 0010 0100 0001 0111(2) × 2-22


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): -22

Mantissa (not normalized):
1.0111 0100 1110 1111 0110 1011 0101 0101 1000 0010 0100 0001 0111


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-22 + 2(11-1) - 1 =

(-22 + 1 023)(10) =

1 001(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;
  • 1 001 ÷ 2 = 500 + 1;
  • 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) =

1001(10) =

011 1110 1001(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 0100 1110 1111 0110 1011 0101 0101 1000 0010 0100 0001 0111 =

0111 0100 1110 1111 0110 1011 0101 0101 1000 0010 0100 0001 0111


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 1001

Mantissa (52 bits) =
0111 0100 1110 1111 0110 1011 0101 0101 1000 0010 0100 0001 0111


The base ten decimal number 0.000 000 347 323 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
0 - 011 1110 1001 - 0111 0100 1110 1111 0110 1011 0101 0101 1000 0010 0100 0001 0111

» Decimal number in base ten 0.000 000 347 393 converted and written as 64 bit double precision IEEE 754 binary floating point representation standard

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

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