0.000 043 772 2 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 043 772 2(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
0.000 043 772 2(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: 0.000 043 772 2.

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) 0.000 043 772 2 × 2 = 0 + 0.000 087 544 4;
  • 2) 0.000 087 544 4 × 2 = 0 + 0.000 175 088 8;
  • 3) 0.000 175 088 8 × 2 = 0 + 0.000 350 177 6;
  • 4) 0.000 350 177 6 × 2 = 0 + 0.000 700 355 2;
  • 5) 0.000 700 355 2 × 2 = 0 + 0.001 400 710 4;
  • 6) 0.001 400 710 4 × 2 = 0 + 0.002 801 420 8;
  • 7) 0.002 801 420 8 × 2 = 0 + 0.005 602 841 6;
  • 8) 0.005 602 841 6 × 2 = 0 + 0.011 205 683 2;
  • 9) 0.011 205 683 2 × 2 = 0 + 0.022 411 366 4;
  • 10) 0.022 411 366 4 × 2 = 0 + 0.044 822 732 8;
  • 11) 0.044 822 732 8 × 2 = 0 + 0.089 645 465 6;
  • 12) 0.089 645 465 6 × 2 = 0 + 0.179 290 931 2;
  • 13) 0.179 290 931 2 × 2 = 0 + 0.358 581 862 4;
  • 14) 0.358 581 862 4 × 2 = 0 + 0.717 163 724 8;
  • 15) 0.717 163 724 8 × 2 = 1 + 0.434 327 449 6;
  • 16) 0.434 327 449 6 × 2 = 0 + 0.868 654 899 2;
  • 17) 0.868 654 899 2 × 2 = 1 + 0.737 309 798 4;
  • 18) 0.737 309 798 4 × 2 = 1 + 0.474 619 596 8;
  • 19) 0.474 619 596 8 × 2 = 0 + 0.949 239 193 6;
  • 20) 0.949 239 193 6 × 2 = 1 + 0.898 478 387 2;
  • 21) 0.898 478 387 2 × 2 = 1 + 0.796 956 774 4;
  • 22) 0.796 956 774 4 × 2 = 1 + 0.593 913 548 8;
  • 23) 0.593 913 548 8 × 2 = 1 + 0.187 827 097 6;
  • 24) 0.187 827 097 6 × 2 = 0 + 0.375 654 195 2;
  • 25) 0.375 654 195 2 × 2 = 0 + 0.751 308 390 4;
  • 26) 0.751 308 390 4 × 2 = 1 + 0.502 616 780 8;
  • 27) 0.502 616 780 8 × 2 = 1 + 0.005 233 561 6;
  • 28) 0.005 233 561 6 × 2 = 0 + 0.010 467 123 2;
  • 29) 0.010 467 123 2 × 2 = 0 + 0.020 934 246 4;
  • 30) 0.020 934 246 4 × 2 = 0 + 0.041 868 492 8;
  • 31) 0.041 868 492 8 × 2 = 0 + 0.083 736 985 6;
  • 32) 0.083 736 985 6 × 2 = 0 + 0.167 473 971 2;
  • 33) 0.167 473 971 2 × 2 = 0 + 0.334 947 942 4;
  • 34) 0.334 947 942 4 × 2 = 0 + 0.669 895 884 8;
  • 35) 0.669 895 884 8 × 2 = 1 + 0.339 791 769 6;
  • 36) 0.339 791 769 6 × 2 = 0 + 0.679 583 539 2;
  • 37) 0.679 583 539 2 × 2 = 1 + 0.359 167 078 4;
  • 38) 0.359 167 078 4 × 2 = 0 + 0.718 334 156 8;
  • 39) 0.718 334 156 8 × 2 = 1 + 0.436 668 313 6;
  • 40) 0.436 668 313 6 × 2 = 0 + 0.873 336 627 2;
  • 41) 0.873 336 627 2 × 2 = 1 + 0.746 673 254 4;
  • 42) 0.746 673 254 4 × 2 = 1 + 0.493 346 508 8;
  • 43) 0.493 346 508 8 × 2 = 0 + 0.986 693 017 6;
  • 44) 0.986 693 017 6 × 2 = 1 + 0.973 386 035 2;
  • 45) 0.973 386 035 2 × 2 = 1 + 0.946 772 070 4;
  • 46) 0.946 772 070 4 × 2 = 1 + 0.893 544 140 8;
  • 47) 0.893 544 140 8 × 2 = 1 + 0.787 088 281 6;
  • 48) 0.787 088 281 6 × 2 = 1 + 0.574 176 563 2;
  • 49) 0.574 176 563 2 × 2 = 1 + 0.148 353 126 4;
  • 50) 0.148 353 126 4 × 2 = 0 + 0.296 706 252 8;
  • 51) 0.296 706 252 8 × 2 = 0 + 0.593 412 505 6;
  • 52) 0.593 412 505 6 × 2 = 1 + 0.186 825 011 2;
  • 53) 0.186 825 011 2 × 2 = 0 + 0.373 650 022 4;
  • 54) 0.373 650 022 4 × 2 = 0 + 0.747 300 044 8;
  • 55) 0.747 300 044 8 × 2 = 1 + 0.494 600 089 6;
  • 56) 0.494 600 089 6 × 2 = 0 + 0.989 200 179 2;
  • 57) 0.989 200 179 2 × 2 = 1 + 0.978 400 358 4;
  • 58) 0.978 400 358 4 × 2 = 1 + 0.956 800 716 8;
  • 59) 0.956 800 716 8 × 2 = 1 + 0.913 601 433 6;
  • 60) 0.913 601 433 6 × 2 = 1 + 0.827 202 867 2;
  • 61) 0.827 202 867 2 × 2 = 1 + 0.654 405 734 4;
  • 62) 0.654 405 734 4 × 2 = 1 + 0.308 811 468 8;
  • 63) 0.308 811 468 8 × 2 = 0 + 0.617 622 937 6;
  • 64) 0.617 622 937 6 × 2 = 1 + 0.235 245 875 2;
  • 65) 0.235 245 875 2 × 2 = 0 + 0.470 491 750 4;
  • 66) 0.470 491 750 4 × 2 = 0 + 0.940 983 500 8;
  • 67) 0.940 983 500 8 × 2 = 1 + 0.881 967 001 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:


0.000 043 772 2(10) =

0.0000 0000 0000 0010 1101 1110 0110 0000 0010 1010 1101 1111 1001 0010 1111 1101 001(2)

5. Positive number before normalization:

0.000 043 772 2(10) =

0.0000 0000 0000 0010 1101 1110 0110 0000 0010 1010 1101 1111 1001 0010 1111 1101 001(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:


0.000 043 772 2(10) =

0.0000 0000 0000 0010 1101 1110 0110 0000 0010 1010 1101 1111 1001 0010 1111 1101 001(2) =

0.0000 0000 0000 0010 1101 1110 0110 0000 0010 1010 1101 1111 1001 0010 1111 1101 001(2) × 20 =

1.0110 1111 0011 0000 0001 0101 0110 1111 1100 1001 0111 1110 1001(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 0000 0001 0101 0110 1111 1100 1001 0111 1110 1001


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 0000 0001 0101 0110 1111 1100 1001 0111 1110 1001 =

0110 1111 0011 0000 0001 0101 0110 1111 1100 1001 0111 1110 1001


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 0000 0001 0101 0110 1111 1100 1001 0111 1110 1001


Decimal number 0.000 043 772 2 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 0000 - 0110 1111 0011 0000 0001 0101 0110 1111 1100 1001 0111 1110 1001

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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