0.000 043 774 1 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 043 774 1(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
0.000 043 774 1(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: 0.000 043 774 1.

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) 0.000 043 774 1 × 2 = 0 + 0.000 087 548 2;
  • 2) 0.000 087 548 2 × 2 = 0 + 0.000 175 096 4;
  • 3) 0.000 175 096 4 × 2 = 0 + 0.000 350 192 8;
  • 4) 0.000 350 192 8 × 2 = 0 + 0.000 700 385 6;
  • 5) 0.000 700 385 6 × 2 = 0 + 0.001 400 771 2;
  • 6) 0.001 400 771 2 × 2 = 0 + 0.002 801 542 4;
  • 7) 0.002 801 542 4 × 2 = 0 + 0.005 603 084 8;
  • 8) 0.005 603 084 8 × 2 = 0 + 0.011 206 169 6;
  • 9) 0.011 206 169 6 × 2 = 0 + 0.022 412 339 2;
  • 10) 0.022 412 339 2 × 2 = 0 + 0.044 824 678 4;
  • 11) 0.044 824 678 4 × 2 = 0 + 0.089 649 356 8;
  • 12) 0.089 649 356 8 × 2 = 0 + 0.179 298 713 6;
  • 13) 0.179 298 713 6 × 2 = 0 + 0.358 597 427 2;
  • 14) 0.358 597 427 2 × 2 = 0 + 0.717 194 854 4;
  • 15) 0.717 194 854 4 × 2 = 1 + 0.434 389 708 8;
  • 16) 0.434 389 708 8 × 2 = 0 + 0.868 779 417 6;
  • 17) 0.868 779 417 6 × 2 = 1 + 0.737 558 835 2;
  • 18) 0.737 558 835 2 × 2 = 1 + 0.475 117 670 4;
  • 19) 0.475 117 670 4 × 2 = 0 + 0.950 235 340 8;
  • 20) 0.950 235 340 8 × 2 = 1 + 0.900 470 681 6;
  • 21) 0.900 470 681 6 × 2 = 1 + 0.800 941 363 2;
  • 22) 0.800 941 363 2 × 2 = 1 + 0.601 882 726 4;
  • 23) 0.601 882 726 4 × 2 = 1 + 0.203 765 452 8;
  • 24) 0.203 765 452 8 × 2 = 0 + 0.407 530 905 6;
  • 25) 0.407 530 905 6 × 2 = 0 + 0.815 061 811 2;
  • 26) 0.815 061 811 2 × 2 = 1 + 0.630 123 622 4;
  • 27) 0.630 123 622 4 × 2 = 1 + 0.260 247 244 8;
  • 28) 0.260 247 244 8 × 2 = 0 + 0.520 494 489 6;
  • 29) 0.520 494 489 6 × 2 = 1 + 0.040 988 979 2;
  • 30) 0.040 988 979 2 × 2 = 0 + 0.081 977 958 4;
  • 31) 0.081 977 958 4 × 2 = 0 + 0.163 955 916 8;
  • 32) 0.163 955 916 8 × 2 = 0 + 0.327 911 833 6;
  • 33) 0.327 911 833 6 × 2 = 0 + 0.655 823 667 2;
  • 34) 0.655 823 667 2 × 2 = 1 + 0.311 647 334 4;
  • 35) 0.311 647 334 4 × 2 = 0 + 0.623 294 668 8;
  • 36) 0.623 294 668 8 × 2 = 1 + 0.246 589 337 6;
  • 37) 0.246 589 337 6 × 2 = 0 + 0.493 178 675 2;
  • 38) 0.493 178 675 2 × 2 = 0 + 0.986 357 350 4;
  • 39) 0.986 357 350 4 × 2 = 1 + 0.972 714 700 8;
  • 40) 0.972 714 700 8 × 2 = 1 + 0.945 429 401 6;
  • 41) 0.945 429 401 6 × 2 = 1 + 0.890 858 803 2;
  • 42) 0.890 858 803 2 × 2 = 1 + 0.781 717 606 4;
  • 43) 0.781 717 606 4 × 2 = 1 + 0.563 435 212 8;
  • 44) 0.563 435 212 8 × 2 = 1 + 0.126 870 425 6;
  • 45) 0.126 870 425 6 × 2 = 0 + 0.253 740 851 2;
  • 46) 0.253 740 851 2 × 2 = 0 + 0.507 481 702 4;
  • 47) 0.507 481 702 4 × 2 = 1 + 0.014 963 404 8;
  • 48) 0.014 963 404 8 × 2 = 0 + 0.029 926 809 6;
  • 49) 0.029 926 809 6 × 2 = 0 + 0.059 853 619 2;
  • 50) 0.059 853 619 2 × 2 = 0 + 0.119 707 238 4;
  • 51) 0.119 707 238 4 × 2 = 0 + 0.239 414 476 8;
  • 52) 0.239 414 476 8 × 2 = 0 + 0.478 828 953 6;
  • 53) 0.478 828 953 6 × 2 = 0 + 0.957 657 907 2;
  • 54) 0.957 657 907 2 × 2 = 1 + 0.915 315 814 4;
  • 55) 0.915 315 814 4 × 2 = 1 + 0.830 631 628 8;
  • 56) 0.830 631 628 8 × 2 = 1 + 0.661 263 257 6;
  • 57) 0.661 263 257 6 × 2 = 1 + 0.322 526 515 2;
  • 58) 0.322 526 515 2 × 2 = 0 + 0.645 053 030 4;
  • 59) 0.645 053 030 4 × 2 = 1 + 0.290 106 060 8;
  • 60) 0.290 106 060 8 × 2 = 0 + 0.580 212 121 6;
  • 61) 0.580 212 121 6 × 2 = 1 + 0.160 424 243 2;
  • 62) 0.160 424 243 2 × 2 = 0 + 0.320 848 486 4;
  • 63) 0.320 848 486 4 × 2 = 0 + 0.641 696 972 8;
  • 64) 0.641 696 972 8 × 2 = 1 + 0.283 393 945 6;
  • 65) 0.283 393 945 6 × 2 = 0 + 0.566 787 891 2;
  • 66) 0.566 787 891 2 × 2 = 1 + 0.133 575 782 4;
  • 67) 0.133 575 782 4 × 2 = 0 + 0.267 151 564 8;

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:


0.000 043 774 1(10) =

0.0000 0000 0000 0010 1101 1110 0110 1000 0101 0011 1111 0010 0000 0111 1010 1001 010(2)

5. Positive number before normalization:

0.000 043 774 1(10) =

0.0000 0000 0000 0010 1101 1110 0110 1000 0101 0011 1111 0010 0000 0111 1010 1001 010(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:


0.000 043 774 1(10) =

0.0000 0000 0000 0010 1101 1110 0110 1000 0101 0011 1111 0010 0000 0111 1010 1001 010(2) =

0.0000 0000 0000 0010 1101 1110 0110 1000 0101 0011 1111 0010 0000 0111 1010 1001 010(2) × 20 =

1.0110 1111 0011 0100 0010 1001 1111 1001 0000 0011 1101 0100 1010(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 0011 0100 0010 1001 1111 1001 0000 0011 1101 0100 1010


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 0011 0100 0010 1001 1111 1001 0000 0011 1101 0100 1010 =

0110 1111 0011 0100 0010 1001 1111 1001 0000 0011 1101 0100 1010


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 0011 0100 0010 1001 1111 1001 0000 0011 1101 0100 1010


Decimal number 0.000 043 774 1 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 0000 - 0110 1111 0011 0100 0010 1001 1111 1001 0000 0011 1101 0100 1010

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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