Decimal to 64 Bit IEEE 754 Binary: Convert Number -0.000 105 923 4 to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From Base Ten Decimal System

Number -0.000 105 923 4(10) converted and written in 64 bit double precision IEEE 754 binary floating point representation standard (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. Start with the positive version of the number:

|-0.000 105 923 4| = 0.000 105 923 4

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: 0.000 105 923 4.

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) 0.000 105 923 4 × 2 = 0 + 0.000 211 846 8;
  • 2) 0.000 211 846 8 × 2 = 0 + 0.000 423 693 6;
  • 3) 0.000 423 693 6 × 2 = 0 + 0.000 847 387 2;
  • 4) 0.000 847 387 2 × 2 = 0 + 0.001 694 774 4;
  • 5) 0.001 694 774 4 × 2 = 0 + 0.003 389 548 8;
  • 6) 0.003 389 548 8 × 2 = 0 + 0.006 779 097 6;
  • 7) 0.006 779 097 6 × 2 = 0 + 0.013 558 195 2;
  • 8) 0.013 558 195 2 × 2 = 0 + 0.027 116 390 4;
  • 9) 0.027 116 390 4 × 2 = 0 + 0.054 232 780 8;
  • 10) 0.054 232 780 8 × 2 = 0 + 0.108 465 561 6;
  • 11) 0.108 465 561 6 × 2 = 0 + 0.216 931 123 2;
  • 12) 0.216 931 123 2 × 2 = 0 + 0.433 862 246 4;
  • 13) 0.433 862 246 4 × 2 = 0 + 0.867 724 492 8;
  • 14) 0.867 724 492 8 × 2 = 1 + 0.735 448 985 6;
  • 15) 0.735 448 985 6 × 2 = 1 + 0.470 897 971 2;
  • 16) 0.470 897 971 2 × 2 = 0 + 0.941 795 942 4;
  • 17) 0.941 795 942 4 × 2 = 1 + 0.883 591 884 8;
  • 18) 0.883 591 884 8 × 2 = 1 + 0.767 183 769 6;
  • 19) 0.767 183 769 6 × 2 = 1 + 0.534 367 539 2;
  • 20) 0.534 367 539 2 × 2 = 1 + 0.068 735 078 4;
  • 21) 0.068 735 078 4 × 2 = 0 + 0.137 470 156 8;
  • 22) 0.137 470 156 8 × 2 = 0 + 0.274 940 313 6;
  • 23) 0.274 940 313 6 × 2 = 0 + 0.549 880 627 2;
  • 24) 0.549 880 627 2 × 2 = 1 + 0.099 761 254 4;
  • 25) 0.099 761 254 4 × 2 = 0 + 0.199 522 508 8;
  • 26) 0.199 522 508 8 × 2 = 0 + 0.399 045 017 6;
  • 27) 0.399 045 017 6 × 2 = 0 + 0.798 090 035 2;
  • 28) 0.798 090 035 2 × 2 = 1 + 0.596 180 070 4;
  • 29) 0.596 180 070 4 × 2 = 1 + 0.192 360 140 8;
  • 30) 0.192 360 140 8 × 2 = 0 + 0.384 720 281 6;
  • 31) 0.384 720 281 6 × 2 = 0 + 0.769 440 563 2;
  • 32) 0.769 440 563 2 × 2 = 1 + 0.538 881 126 4;
  • 33) 0.538 881 126 4 × 2 = 1 + 0.077 762 252 8;
  • 34) 0.077 762 252 8 × 2 = 0 + 0.155 524 505 6;
  • 35) 0.155 524 505 6 × 2 = 0 + 0.311 049 011 2;
  • 36) 0.311 049 011 2 × 2 = 0 + 0.622 098 022 4;
  • 37) 0.622 098 022 4 × 2 = 1 + 0.244 196 044 8;
  • 38) 0.244 196 044 8 × 2 = 0 + 0.488 392 089 6;
  • 39) 0.488 392 089 6 × 2 = 0 + 0.976 784 179 2;
  • 40) 0.976 784 179 2 × 2 = 1 + 0.953 568 358 4;
  • 41) 0.953 568 358 4 × 2 = 1 + 0.907 136 716 8;
  • 42) 0.907 136 716 8 × 2 = 1 + 0.814 273 433 6;
  • 43) 0.814 273 433 6 × 2 = 1 + 0.628 546 867 2;
  • 44) 0.628 546 867 2 × 2 = 1 + 0.257 093 734 4;
  • 45) 0.257 093 734 4 × 2 = 0 + 0.514 187 468 8;
  • 46) 0.514 187 468 8 × 2 = 1 + 0.028 374 937 6;
  • 47) 0.028 374 937 6 × 2 = 0 + 0.056 749 875 2;
  • 48) 0.056 749 875 2 × 2 = 0 + 0.113 499 750 4;
  • 49) 0.113 499 750 4 × 2 = 0 + 0.226 999 500 8;
  • 50) 0.226 999 500 8 × 2 = 0 + 0.453 999 001 6;
  • 51) 0.453 999 001 6 × 2 = 0 + 0.907 998 003 2;
  • 52) 0.907 998 003 2 × 2 = 1 + 0.815 996 006 4;
  • 53) 0.815 996 006 4 × 2 = 1 + 0.631 992 012 8;
  • 54) 0.631 992 012 8 × 2 = 1 + 0.263 984 025 6;
  • 55) 0.263 984 025 6 × 2 = 0 + 0.527 968 051 2;
  • 56) 0.527 968 051 2 × 2 = 1 + 0.055 936 102 4;
  • 57) 0.055 936 102 4 × 2 = 0 + 0.111 872 204 8;
  • 58) 0.111 872 204 8 × 2 = 0 + 0.223 744 409 6;
  • 59) 0.223 744 409 6 × 2 = 0 + 0.447 488 819 2;
  • 60) 0.447 488 819 2 × 2 = 0 + 0.894 977 638 4;
  • 61) 0.894 977 638 4 × 2 = 1 + 0.789 955 276 8;
  • 62) 0.789 955 276 8 × 2 = 1 + 0.579 910 553 6;
  • 63) 0.579 910 553 6 × 2 = 1 + 0.159 821 107 2;
  • 64) 0.159 821 107 2 × 2 = 0 + 0.319 642 214 4;
  • 65) 0.319 642 214 4 × 2 = 0 + 0.639 284 428 8;
  • 66) 0.639 284 428 8 × 2 = 1 + 0.278 568 857 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).


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:


0.000 105 923 4(10) =


0.0000 0000 0000 0110 1111 0001 0001 1001 1000 1001 1111 0100 0001 1101 0000 1110 01(2)


6. Positive number before normalization:

0.000 105 923 4(10) =


0.0000 0000 0000 0110 1111 0001 0001 1001 1000 1001 1111 0100 0001 1101 0000 1110 01(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 14 positions to the right, so that only one non zero digit remains to the left of it:


0.000 105 923 4(10) =


0.0000 0000 0000 0110 1111 0001 0001 1001 1000 1001 1111 0100 0001 1101 0000 1110 01(2) =


0.0000 0000 0000 0110 1111 0001 0001 1001 1000 1001 1111 0100 0001 1101 0000 1110 01(2) × 20 =


1.1011 1100 0100 0110 0110 0010 0111 1101 0000 0111 0100 0011 1001(2) × 2-14


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): -14


Mantissa (not normalized):
1.1011 1100 0100 0110 0110 0010 0111 1101 0000 0111 0100 0011 1001


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =


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


-14 + 2(11-1) - 1 =


(-14 + 1 023)(10) =


1 009(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;
  • 1 009 ÷ 2 = 504 + 1;
  • 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;

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) =


1009(10) =


011 1111 0001(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. 1011 1100 0100 0110 0110 0010 0111 1101 0000 0111 0100 0011 1001 =


1011 1100 0100 0110 0110 0010 0111 1101 0000 0111 0100 0011 1001


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 1111 0001


Mantissa (52 bits) =
1011 1100 0100 0110 0110 0010 0111 1101 0000 0111 0100 0011 1001


The base ten decimal number -0.000 105 923 4 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
1 - 011 1111 0001 - 1011 1100 0100 0110 0110 0010 0111 1101 0000 0111 0100 0011 1001

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