-0.000 007 820 53 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 007 820 53(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
-0.000 007 820 53(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. Start with the positive version of the number:

|-0.000 007 820 53| = 0.000 007 820 53


2. 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;

3. 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)


4. Convert to binary (base 2) the fractional part: 0.000 007 820 53.

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) 0.000 007 820 53 × 2 = 0 + 0.000 015 641 06;
  • 2) 0.000 015 641 06 × 2 = 0 + 0.000 031 282 12;
  • 3) 0.000 031 282 12 × 2 = 0 + 0.000 062 564 24;
  • 4) 0.000 062 564 24 × 2 = 0 + 0.000 125 128 48;
  • 5) 0.000 125 128 48 × 2 = 0 + 0.000 250 256 96;
  • 6) 0.000 250 256 96 × 2 = 0 + 0.000 500 513 92;
  • 7) 0.000 500 513 92 × 2 = 0 + 0.001 001 027 84;
  • 8) 0.001 001 027 84 × 2 = 0 + 0.002 002 055 68;
  • 9) 0.002 002 055 68 × 2 = 0 + 0.004 004 111 36;
  • 10) 0.004 004 111 36 × 2 = 0 + 0.008 008 222 72;
  • 11) 0.008 008 222 72 × 2 = 0 + 0.016 016 445 44;
  • 12) 0.016 016 445 44 × 2 = 0 + 0.032 032 890 88;
  • 13) 0.032 032 890 88 × 2 = 0 + 0.064 065 781 76;
  • 14) 0.064 065 781 76 × 2 = 0 + 0.128 131 563 52;
  • 15) 0.128 131 563 52 × 2 = 0 + 0.256 263 127 04;
  • 16) 0.256 263 127 04 × 2 = 0 + 0.512 526 254 08;
  • 17) 0.512 526 254 08 × 2 = 1 + 0.025 052 508 16;
  • 18) 0.025 052 508 16 × 2 = 0 + 0.050 105 016 32;
  • 19) 0.050 105 016 32 × 2 = 0 + 0.100 210 032 64;
  • 20) 0.100 210 032 64 × 2 = 0 + 0.200 420 065 28;
  • 21) 0.200 420 065 28 × 2 = 0 + 0.400 840 130 56;
  • 22) 0.400 840 130 56 × 2 = 0 + 0.801 680 261 12;
  • 23) 0.801 680 261 12 × 2 = 1 + 0.603 360 522 24;
  • 24) 0.603 360 522 24 × 2 = 1 + 0.206 721 044 48;
  • 25) 0.206 721 044 48 × 2 = 0 + 0.413 442 088 96;
  • 26) 0.413 442 088 96 × 2 = 0 + 0.826 884 177 92;
  • 27) 0.826 884 177 92 × 2 = 1 + 0.653 768 355 84;
  • 28) 0.653 768 355 84 × 2 = 1 + 0.307 536 711 68;
  • 29) 0.307 536 711 68 × 2 = 0 + 0.615 073 423 36;
  • 30) 0.615 073 423 36 × 2 = 1 + 0.230 146 846 72;
  • 31) 0.230 146 846 72 × 2 = 0 + 0.460 293 693 44;
  • 32) 0.460 293 693 44 × 2 = 0 + 0.920 587 386 88;
  • 33) 0.920 587 386 88 × 2 = 1 + 0.841 174 773 76;
  • 34) 0.841 174 773 76 × 2 = 1 + 0.682 349 547 52;
  • 35) 0.682 349 547 52 × 2 = 1 + 0.364 699 095 04;
  • 36) 0.364 699 095 04 × 2 = 0 + 0.729 398 190 08;
  • 37) 0.729 398 190 08 × 2 = 1 + 0.458 796 380 16;
  • 38) 0.458 796 380 16 × 2 = 0 + 0.917 592 760 32;
  • 39) 0.917 592 760 32 × 2 = 1 + 0.835 185 520 64;
  • 40) 0.835 185 520 64 × 2 = 1 + 0.670 371 041 28;
  • 41) 0.670 371 041 28 × 2 = 1 + 0.340 742 082 56;
  • 42) 0.340 742 082 56 × 2 = 0 + 0.681 484 165 12;
  • 43) 0.681 484 165 12 × 2 = 1 + 0.362 968 330 24;
  • 44) 0.362 968 330 24 × 2 = 0 + 0.725 936 660 48;
  • 45) 0.725 936 660 48 × 2 = 1 + 0.451 873 320 96;
  • 46) 0.451 873 320 96 × 2 = 0 + 0.903 746 641 92;
  • 47) 0.903 746 641 92 × 2 = 1 + 0.807 493 283 84;
  • 48) 0.807 493 283 84 × 2 = 1 + 0.614 986 567 68;
  • 49) 0.614 986 567 68 × 2 = 1 + 0.229 973 135 36;
  • 50) 0.229 973 135 36 × 2 = 0 + 0.459 946 270 72;
  • 51) 0.459 946 270 72 × 2 = 0 + 0.919 892 541 44;
  • 52) 0.919 892 541 44 × 2 = 1 + 0.839 785 082 88;
  • 53) 0.839 785 082 88 × 2 = 1 + 0.679 570 165 76;
  • 54) 0.679 570 165 76 × 2 = 1 + 0.359 140 331 52;
  • 55) 0.359 140 331 52 × 2 = 0 + 0.718 280 663 04;
  • 56) 0.718 280 663 04 × 2 = 1 + 0.436 561 326 08;
  • 57) 0.436 561 326 08 × 2 = 0 + 0.873 122 652 16;
  • 58) 0.873 122 652 16 × 2 = 1 + 0.746 245 304 32;
  • 59) 0.746 245 304 32 × 2 = 1 + 0.492 490 608 64;
  • 60) 0.492 490 608 64 × 2 = 0 + 0.984 981 217 28;
  • 61) 0.984 981 217 28 × 2 = 1 + 0.969 962 434 56;
  • 62) 0.969 962 434 56 × 2 = 1 + 0.939 924 869 12;
  • 63) 0.939 924 869 12 × 2 = 1 + 0.879 849 738 24;
  • 64) 0.879 849 738 24 × 2 = 1 + 0.759 699 476 48;
  • 65) 0.759 699 476 48 × 2 = 1 + 0.519 398 952 96;
  • 66) 0.519 398 952 96 × 2 = 1 + 0.038 797 905 92;
  • 67) 0.038 797 905 92 × 2 = 0 + 0.077 595 811 84;
  • 68) 0.077 595 811 84 × 2 = 0 + 0.155 191 623 68;
  • 69) 0.155 191 623 68 × 2 = 0 + 0.310 383 247 36;

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).


5. 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:


0.000 007 820 53(10) =

0.0000 0000 0000 0000 1000 0011 0011 0100 1110 1011 1010 1011 1001 1101 0110 1111 1100 0(2)

6. Positive number before normalization:

0.000 007 820 53(10) =

0.0000 0000 0000 0000 1000 0011 0011 0100 1110 1011 1010 1011 1001 1101 0110 1111 1100 0(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 17 positions to the right, so that only one non zero digit remains to the left of it:


0.000 007 820 53(10) =

0.0000 0000 0000 0000 1000 0011 0011 0100 1110 1011 1010 1011 1001 1101 0110 1111 1100 0(2) =

0.0000 0000 0000 0000 1000 0011 0011 0100 1110 1011 1010 1011 1001 1101 0110 1111 1100 0(2) × 20 =

1.0000 0110 0110 1001 1101 0111 0101 0111 0011 1010 1101 1111 1000(2) × 2-17


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): -17

Mantissa (not normalized):
1.0000 0110 0110 1001 1101 0111 0101 0111 0011 1010 1101 1111 1000


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-17 + 2(11-1) - 1 =

(-17 + 1 023)(10) =

1 006(10)


10. 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;
  • 1 006 ÷ 2 = 503 + 0;
  • 503 ÷ 2 = 251 + 1;
  • 251 ÷ 2 = 125 + 1;
  • 125 ÷ 2 = 62 + 1;
  • 62 ÷ 2 = 31 + 0;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

1006(10) =

011 1110 1110(2)


12. 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. 0000 0110 0110 1001 1101 0111 0101 0111 0011 1010 1101 1111 1000 =

0000 0110 0110 1001 1101 0111 0101 0111 0011 1010 1101 1111 1000


13. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (11 bits) =
011 1110 1110

Mantissa (52 bits) =
0000 0110 0110 1001 1101 0111 0101 0111 0011 1010 1101 1111 1000


Decimal number -0.000 007 820 53 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1110 1110 - 0000 0110 0110 1001 1101 0111 0101 0111 0011 1010 1101 1111 1000

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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