0.000 043 759 7 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 043 759 7(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
0.000 043 759 7(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: 0.000 043 759 7.

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) 0.000 043 759 7 × 2 = 0 + 0.000 087 519 4;
  • 2) 0.000 087 519 4 × 2 = 0 + 0.000 175 038 8;
  • 3) 0.000 175 038 8 × 2 = 0 + 0.000 350 077 6;
  • 4) 0.000 350 077 6 × 2 = 0 + 0.000 700 155 2;
  • 5) 0.000 700 155 2 × 2 = 0 + 0.001 400 310 4;
  • 6) 0.001 400 310 4 × 2 = 0 + 0.002 800 620 8;
  • 7) 0.002 800 620 8 × 2 = 0 + 0.005 601 241 6;
  • 8) 0.005 601 241 6 × 2 = 0 + 0.011 202 483 2;
  • 9) 0.011 202 483 2 × 2 = 0 + 0.022 404 966 4;
  • 10) 0.022 404 966 4 × 2 = 0 + 0.044 809 932 8;
  • 11) 0.044 809 932 8 × 2 = 0 + 0.089 619 865 6;
  • 12) 0.089 619 865 6 × 2 = 0 + 0.179 239 731 2;
  • 13) 0.179 239 731 2 × 2 = 0 + 0.358 479 462 4;
  • 14) 0.358 479 462 4 × 2 = 0 + 0.716 958 924 8;
  • 15) 0.716 958 924 8 × 2 = 1 + 0.433 917 849 6;
  • 16) 0.433 917 849 6 × 2 = 0 + 0.867 835 699 2;
  • 17) 0.867 835 699 2 × 2 = 1 + 0.735 671 398 4;
  • 18) 0.735 671 398 4 × 2 = 1 + 0.471 342 796 8;
  • 19) 0.471 342 796 8 × 2 = 0 + 0.942 685 593 6;
  • 20) 0.942 685 593 6 × 2 = 1 + 0.885 371 187 2;
  • 21) 0.885 371 187 2 × 2 = 1 + 0.770 742 374 4;
  • 22) 0.770 742 374 4 × 2 = 1 + 0.541 484 748 8;
  • 23) 0.541 484 748 8 × 2 = 1 + 0.082 969 497 6;
  • 24) 0.082 969 497 6 × 2 = 0 + 0.165 938 995 2;
  • 25) 0.165 938 995 2 × 2 = 0 + 0.331 877 990 4;
  • 26) 0.331 877 990 4 × 2 = 0 + 0.663 755 980 8;
  • 27) 0.663 755 980 8 × 2 = 1 + 0.327 511 961 6;
  • 28) 0.327 511 961 6 × 2 = 0 + 0.655 023 923 2;
  • 29) 0.655 023 923 2 × 2 = 1 + 0.310 047 846 4;
  • 30) 0.310 047 846 4 × 2 = 0 + 0.620 095 692 8;
  • 31) 0.620 095 692 8 × 2 = 1 + 0.240 191 385 6;
  • 32) 0.240 191 385 6 × 2 = 0 + 0.480 382 771 2;
  • 33) 0.480 382 771 2 × 2 = 0 + 0.960 765 542 4;
  • 34) 0.960 765 542 4 × 2 = 1 + 0.921 531 084 8;
  • 35) 0.921 531 084 8 × 2 = 1 + 0.843 062 169 6;
  • 36) 0.843 062 169 6 × 2 = 1 + 0.686 124 339 2;
  • 37) 0.686 124 339 2 × 2 = 1 + 0.372 248 678 4;
  • 38) 0.372 248 678 4 × 2 = 0 + 0.744 497 356 8;
  • 39) 0.744 497 356 8 × 2 = 1 + 0.488 994 713 6;
  • 40) 0.488 994 713 6 × 2 = 0 + 0.977 989 427 2;
  • 41) 0.977 989 427 2 × 2 = 1 + 0.955 978 854 4;
  • 42) 0.955 978 854 4 × 2 = 1 + 0.911 957 708 8;
  • 43) 0.911 957 708 8 × 2 = 1 + 0.823 915 417 6;
  • 44) 0.823 915 417 6 × 2 = 1 + 0.647 830 835 2;
  • 45) 0.647 830 835 2 × 2 = 1 + 0.295 661 670 4;
  • 46) 0.295 661 670 4 × 2 = 0 + 0.591 323 340 8;
  • 47) 0.591 323 340 8 × 2 = 1 + 0.182 646 681 6;
  • 48) 0.182 646 681 6 × 2 = 0 + 0.365 293 363 2;
  • 49) 0.365 293 363 2 × 2 = 0 + 0.730 586 726 4;
  • 50) 0.730 586 726 4 × 2 = 1 + 0.461 173 452 8;
  • 51) 0.461 173 452 8 × 2 = 0 + 0.922 346 905 6;
  • 52) 0.922 346 905 6 × 2 = 1 + 0.844 693 811 2;
  • 53) 0.844 693 811 2 × 2 = 1 + 0.689 387 622 4;
  • 54) 0.689 387 622 4 × 2 = 1 + 0.378 775 244 8;
  • 55) 0.378 775 244 8 × 2 = 0 + 0.757 550 489 6;
  • 56) 0.757 550 489 6 × 2 = 1 + 0.515 100 979 2;
  • 57) 0.515 100 979 2 × 2 = 1 + 0.030 201 958 4;
  • 58) 0.030 201 958 4 × 2 = 0 + 0.060 403 916 8;
  • 59) 0.060 403 916 8 × 2 = 0 + 0.120 807 833 6;
  • 60) 0.120 807 833 6 × 2 = 0 + 0.241 615 667 2;
  • 61) 0.241 615 667 2 × 2 = 0 + 0.483 231 334 4;
  • 62) 0.483 231 334 4 × 2 = 0 + 0.966 462 668 8;
  • 63) 0.966 462 668 8 × 2 = 1 + 0.932 925 337 6;
  • 64) 0.932 925 337 6 × 2 = 1 + 0.865 850 675 2;
  • 65) 0.865 850 675 2 × 2 = 1 + 0.731 701 350 4;
  • 66) 0.731 701 350 4 × 2 = 1 + 0.463 402 700 8;
  • 67) 0.463 402 700 8 × 2 = 0 + 0.926 805 401 6;

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:


0.000 043 759 7(10) =

0.0000 0000 0000 0010 1101 1110 0010 1010 0111 1010 1111 1010 0101 1101 1000 0011 110(2)

5. Positive number before normalization:

0.000 043 759 7(10) =

0.0000 0000 0000 0010 1101 1110 0010 1010 0111 1010 1111 1010 0101 1101 1000 0011 110(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:


0.000 043 759 7(10) =

0.0000 0000 0000 0010 1101 1110 0010 1010 0111 1010 1111 1010 0101 1101 1000 0011 110(2) =

0.0000 0000 0000 0010 1101 1110 0010 1010 0111 1010 1111 1010 0101 1101 1000 0011 110(2) × 20 =

1.0110 1111 0001 0101 0011 1101 0111 1101 0010 1110 1100 0001 1110(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):
1.0110 1111 0001 0101 0011 1101 0111 1101 0010 1110 1100 0001 1110


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. 0110 1111 0001 0101 0011 1101 0111 1101 0010 1110 1100 0001 1110 =

0110 1111 0001 0101 0011 1101 0111 1101 0010 1110 1100 0001 1110


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) =
0110 1111 0001 0101 0011 1101 0111 1101 0010 1110 1100 0001 1110


Decimal number 0.000 043 759 7 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 0000 - 0110 1111 0001 0101 0011 1101 0111 1101 0010 1110 1100 0001 1110

» Convert decimal 0.000 043 758 2 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 10, 2026 [October]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: October - 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