0.000 046 473 679 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 046 473 679(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
0.000 046 473 679(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: 0.000 046 473 679.

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) 0.000 046 473 679 × 2 = 0 + 0.000 092 947 358;
  • 2) 0.000 092 947 358 × 2 = 0 + 0.000 185 894 716;
  • 3) 0.000 185 894 716 × 2 = 0 + 0.000 371 789 432;
  • 4) 0.000 371 789 432 × 2 = 0 + 0.000 743 578 864;
  • 5) 0.000 743 578 864 × 2 = 0 + 0.001 487 157 728;
  • 6) 0.001 487 157 728 × 2 = 0 + 0.002 974 315 456;
  • 7) 0.002 974 315 456 × 2 = 0 + 0.005 948 630 912;
  • 8) 0.005 948 630 912 × 2 = 0 + 0.011 897 261 824;
  • 9) 0.011 897 261 824 × 2 = 0 + 0.023 794 523 648;
  • 10) 0.023 794 523 648 × 2 = 0 + 0.047 589 047 296;
  • 11) 0.047 589 047 296 × 2 = 0 + 0.095 178 094 592;
  • 12) 0.095 178 094 592 × 2 = 0 + 0.190 356 189 184;
  • 13) 0.190 356 189 184 × 2 = 0 + 0.380 712 378 368;
  • 14) 0.380 712 378 368 × 2 = 0 + 0.761 424 756 736;
  • 15) 0.761 424 756 736 × 2 = 1 + 0.522 849 513 472;
  • 16) 0.522 849 513 472 × 2 = 1 + 0.045 699 026 944;
  • 17) 0.045 699 026 944 × 2 = 0 + 0.091 398 053 888;
  • 18) 0.091 398 053 888 × 2 = 0 + 0.182 796 107 776;
  • 19) 0.182 796 107 776 × 2 = 0 + 0.365 592 215 552;
  • 20) 0.365 592 215 552 × 2 = 0 + 0.731 184 431 104;
  • 21) 0.731 184 431 104 × 2 = 1 + 0.462 368 862 208;
  • 22) 0.462 368 862 208 × 2 = 0 + 0.924 737 724 416;
  • 23) 0.924 737 724 416 × 2 = 1 + 0.849 475 448 832;
  • 24) 0.849 475 448 832 × 2 = 1 + 0.698 950 897 664;
  • 25) 0.698 950 897 664 × 2 = 1 + 0.397 901 795 328;
  • 26) 0.397 901 795 328 × 2 = 0 + 0.795 803 590 656;
  • 27) 0.795 803 590 656 × 2 = 1 + 0.591 607 181 312;
  • 28) 0.591 607 181 312 × 2 = 1 + 0.183 214 362 624;
  • 29) 0.183 214 362 624 × 2 = 0 + 0.366 428 725 248;
  • 30) 0.366 428 725 248 × 2 = 0 + 0.732 857 450 496;
  • 31) 0.732 857 450 496 × 2 = 1 + 0.465 714 900 992;
  • 32) 0.465 714 900 992 × 2 = 0 + 0.931 429 801 984;
  • 33) 0.931 429 801 984 × 2 = 1 + 0.862 859 603 968;
  • 34) 0.862 859 603 968 × 2 = 1 + 0.725 719 207 936;
  • 35) 0.725 719 207 936 × 2 = 1 + 0.451 438 415 872;
  • 36) 0.451 438 415 872 × 2 = 0 + 0.902 876 831 744;
  • 37) 0.902 876 831 744 × 2 = 1 + 0.805 753 663 488;
  • 38) 0.805 753 663 488 × 2 = 1 + 0.611 507 326 976;
  • 39) 0.611 507 326 976 × 2 = 1 + 0.223 014 653 952;
  • 40) 0.223 014 653 952 × 2 = 0 + 0.446 029 307 904;
  • 41) 0.446 029 307 904 × 2 = 0 + 0.892 058 615 808;
  • 42) 0.892 058 615 808 × 2 = 1 + 0.784 117 231 616;
  • 43) 0.784 117 231 616 × 2 = 1 + 0.568 234 463 232;
  • 44) 0.568 234 463 232 × 2 = 1 + 0.136 468 926 464;
  • 45) 0.136 468 926 464 × 2 = 0 + 0.272 937 852 928;
  • 46) 0.272 937 852 928 × 2 = 0 + 0.545 875 705 856;
  • 47) 0.545 875 705 856 × 2 = 1 + 0.091 751 411 712;
  • 48) 0.091 751 411 712 × 2 = 0 + 0.183 502 823 424;
  • 49) 0.183 502 823 424 × 2 = 0 + 0.367 005 646 848;
  • 50) 0.367 005 646 848 × 2 = 0 + 0.734 011 293 696;
  • 51) 0.734 011 293 696 × 2 = 1 + 0.468 022 587 392;
  • 52) 0.468 022 587 392 × 2 = 0 + 0.936 045 174 784;
  • 53) 0.936 045 174 784 × 2 = 1 + 0.872 090 349 568;
  • 54) 0.872 090 349 568 × 2 = 1 + 0.744 180 699 136;
  • 55) 0.744 180 699 136 × 2 = 1 + 0.488 361 398 272;
  • 56) 0.488 361 398 272 × 2 = 0 + 0.976 722 796 544;
  • 57) 0.976 722 796 544 × 2 = 1 + 0.953 445 593 088;
  • 58) 0.953 445 593 088 × 2 = 1 + 0.906 891 186 176;
  • 59) 0.906 891 186 176 × 2 = 1 + 0.813 782 372 352;
  • 60) 0.813 782 372 352 × 2 = 1 + 0.627 564 744 704;
  • 61) 0.627 564 744 704 × 2 = 1 + 0.255 129 489 408;
  • 62) 0.255 129 489 408 × 2 = 0 + 0.510 258 978 816;
  • 63) 0.510 258 978 816 × 2 = 1 + 0.020 517 957 632;
  • 64) 0.020 517 957 632 × 2 = 0 + 0.041 035 915 264;
  • 65) 0.041 035 915 264 × 2 = 0 + 0.082 071 830 528;
  • 66) 0.082 071 830 528 × 2 = 0 + 0.164 143 661 056;
  • 67) 0.164 143 661 056 × 2 = 0 + 0.328 287 322 112;

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:


0.000 046 473 679(10) =

0.0000 0000 0000 0011 0000 1011 1011 0010 1110 1110 0111 0010 0010 1110 1111 1010 000(2)

5. Positive number before normalization:

0.000 046 473 679(10) =

0.0000 0000 0000 0011 0000 1011 1011 0010 1110 1110 0111 0010 0010 1110 1111 1010 000(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:


0.000 046 473 679(10) =

0.0000 0000 0000 0011 0000 1011 1011 0010 1110 1110 0111 0010 0010 1110 1111 1010 000(2) =

0.0000 0000 0000 0011 0000 1011 1011 0010 1110 1110 0111 0010 0010 1110 1111 1010 000(2) × 20 =

1.1000 0101 1101 1001 0111 0111 0011 1001 0001 0111 0111 1101 0000(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 0111 0111 0011 1001 0001 0111 0111 1101 0000


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 0111 0111 0011 1001 0001 0111 0111 1101 0000 =

1000 0101 1101 1001 0111 0111 0011 1001 0001 0111 0111 1101 0000


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 0111 0111 0011 1001 0001 0111 0111 1101 0000


Decimal number 0.000 046 473 679 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 0000 - 1000 0101 1101 1001 0111 0111 0011 1001 0001 0111 0111 1101 0000

» Convert decimal 0.000 046 473 606 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