0.000 046 473 646 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 046 473 646(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
0.000 046 473 646(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: 0.000 046 473 646.

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) 0.000 046 473 646 × 2 = 0 + 0.000 092 947 292;
  • 2) 0.000 092 947 292 × 2 = 0 + 0.000 185 894 584;
  • 3) 0.000 185 894 584 × 2 = 0 + 0.000 371 789 168;
  • 4) 0.000 371 789 168 × 2 = 0 + 0.000 743 578 336;
  • 5) 0.000 743 578 336 × 2 = 0 + 0.001 487 156 672;
  • 6) 0.001 487 156 672 × 2 = 0 + 0.002 974 313 344;
  • 7) 0.002 974 313 344 × 2 = 0 + 0.005 948 626 688;
  • 8) 0.005 948 626 688 × 2 = 0 + 0.011 897 253 376;
  • 9) 0.011 897 253 376 × 2 = 0 + 0.023 794 506 752;
  • 10) 0.023 794 506 752 × 2 = 0 + 0.047 589 013 504;
  • 11) 0.047 589 013 504 × 2 = 0 + 0.095 178 027 008;
  • 12) 0.095 178 027 008 × 2 = 0 + 0.190 356 054 016;
  • 13) 0.190 356 054 016 × 2 = 0 + 0.380 712 108 032;
  • 14) 0.380 712 108 032 × 2 = 0 + 0.761 424 216 064;
  • 15) 0.761 424 216 064 × 2 = 1 + 0.522 848 432 128;
  • 16) 0.522 848 432 128 × 2 = 1 + 0.045 696 864 256;
  • 17) 0.045 696 864 256 × 2 = 0 + 0.091 393 728 512;
  • 18) 0.091 393 728 512 × 2 = 0 + 0.182 787 457 024;
  • 19) 0.182 787 457 024 × 2 = 0 + 0.365 574 914 048;
  • 20) 0.365 574 914 048 × 2 = 0 + 0.731 149 828 096;
  • 21) 0.731 149 828 096 × 2 = 1 + 0.462 299 656 192;
  • 22) 0.462 299 656 192 × 2 = 0 + 0.924 599 312 384;
  • 23) 0.924 599 312 384 × 2 = 1 + 0.849 198 624 768;
  • 24) 0.849 198 624 768 × 2 = 1 + 0.698 397 249 536;
  • 25) 0.698 397 249 536 × 2 = 1 + 0.396 794 499 072;
  • 26) 0.396 794 499 072 × 2 = 0 + 0.793 588 998 144;
  • 27) 0.793 588 998 144 × 2 = 1 + 0.587 177 996 288;
  • 28) 0.587 177 996 288 × 2 = 1 + 0.174 355 992 576;
  • 29) 0.174 355 992 576 × 2 = 0 + 0.348 711 985 152;
  • 30) 0.348 711 985 152 × 2 = 0 + 0.697 423 970 304;
  • 31) 0.697 423 970 304 × 2 = 1 + 0.394 847 940 608;
  • 32) 0.394 847 940 608 × 2 = 0 + 0.789 695 881 216;
  • 33) 0.789 695 881 216 × 2 = 1 + 0.579 391 762 432;
  • 34) 0.579 391 762 432 × 2 = 1 + 0.158 783 524 864;
  • 35) 0.158 783 524 864 × 2 = 0 + 0.317 567 049 728;
  • 36) 0.317 567 049 728 × 2 = 0 + 0.635 134 099 456;
  • 37) 0.635 134 099 456 × 2 = 1 + 0.270 268 198 912;
  • 38) 0.270 268 198 912 × 2 = 0 + 0.540 536 397 824;
  • 39) 0.540 536 397 824 × 2 = 1 + 0.081 072 795 648;
  • 40) 0.081 072 795 648 × 2 = 0 + 0.162 145 591 296;
  • 41) 0.162 145 591 296 × 2 = 0 + 0.324 291 182 592;
  • 42) 0.324 291 182 592 × 2 = 0 + 0.648 582 365 184;
  • 43) 0.648 582 365 184 × 2 = 1 + 0.297 164 730 368;
  • 44) 0.297 164 730 368 × 2 = 0 + 0.594 329 460 736;
  • 45) 0.594 329 460 736 × 2 = 1 + 0.188 658 921 472;
  • 46) 0.188 658 921 472 × 2 = 0 + 0.377 317 842 944;
  • 47) 0.377 317 842 944 × 2 = 0 + 0.754 635 685 888;
  • 48) 0.754 635 685 888 × 2 = 1 + 0.509 271 371 776;
  • 49) 0.509 271 371 776 × 2 = 1 + 0.018 542 743 552;
  • 50) 0.018 542 743 552 × 2 = 0 + 0.037 085 487 104;
  • 51) 0.037 085 487 104 × 2 = 0 + 0.074 170 974 208;
  • 52) 0.074 170 974 208 × 2 = 0 + 0.148 341 948 416;
  • 53) 0.148 341 948 416 × 2 = 0 + 0.296 683 896 832;
  • 54) 0.296 683 896 832 × 2 = 0 + 0.593 367 793 664;
  • 55) 0.593 367 793 664 × 2 = 1 + 0.186 735 587 328;
  • 56) 0.186 735 587 328 × 2 = 0 + 0.373 471 174 656;
  • 57) 0.373 471 174 656 × 2 = 0 + 0.746 942 349 312;
  • 58) 0.746 942 349 312 × 2 = 1 + 0.493 884 698 624;
  • 59) 0.493 884 698 624 × 2 = 0 + 0.987 769 397 248;
  • 60) 0.987 769 397 248 × 2 = 1 + 0.975 538 794 496;
  • 61) 0.975 538 794 496 × 2 = 1 + 0.951 077 588 992;
  • 62) 0.951 077 588 992 × 2 = 1 + 0.902 155 177 984;
  • 63) 0.902 155 177 984 × 2 = 1 + 0.804 310 355 968;
  • 64) 0.804 310 355 968 × 2 = 1 + 0.608 620 711 936;
  • 65) 0.608 620 711 936 × 2 = 1 + 0.217 241 423 872;
  • 66) 0.217 241 423 872 × 2 = 0 + 0.434 482 847 744;
  • 67) 0.434 482 847 744 × 2 = 0 + 0.868 965 695 488;

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:


0.000 046 473 646(10) =

0.0000 0000 0000 0011 0000 1011 1011 0010 1100 1010 0010 1001 1000 0010 0101 1111 100(2)

5. Positive number before normalization:

0.000 046 473 646(10) =

0.0000 0000 0000 0011 0000 1011 1011 0010 1100 1010 0010 1001 1000 0010 0101 1111 100(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 15 positions to the right, so that only one non zero digit remains to the left of it:


0.000 046 473 646(10) =

0.0000 0000 0000 0011 0000 1011 1011 0010 1100 1010 0010 1001 1000 0010 0101 1111 100(2) =

0.0000 0000 0000 0011 0000 1011 1011 0010 1100 1010 0010 1001 1000 0010 0101 1111 100(2) × 20 =

1.1000 0101 1101 1001 0110 0101 0001 0100 1100 0001 0010 1111 1100(2) × 2-15


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): -15

Mantissa (not normalized):
1.1000 0101 1101 1001 0110 0101 0001 0100 1100 0001 0010 1111 1100


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-15 + 2(11-1) - 1 =

(-15 + 1 023)(10) =

1 008(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;
  • 1 008 ÷ 2 = 504 + 0;
  • 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;

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) =

1008(10) =

011 1111 0000(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. 1000 0101 1101 1001 0110 0101 0001 0100 1100 0001 0010 1111 1100 =

1000 0101 1101 1001 0110 0101 0001 0100 1100 0001 0010 1111 1100


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 1111 0000

Mantissa (52 bits) =
1000 0101 1101 1001 0110 0101 0001 0100 1100 0001 0010 1111 1100


Decimal number 0.000 046 473 646 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 0000 - 1000 0101 1101 1001 0110 0101 0001 0100 1100 0001 0010 1111 1100

» Convert decimal 0.000 046 473 679 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

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