Convert Decimal -0.000 105 918 2 to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 105 918 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
-0.000 105 918 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. Start with the positive version of the number:

|-0.000 105 918 2| = 0.000 105 918 2


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: 0.000 105 918 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) 0.000 105 918 2 × 2 = 0 + 0.000 211 836 4;
  • 2) 0.000 211 836 4 × 2 = 0 + 0.000 423 672 8;
  • 3) 0.000 423 672 8 × 2 = 0 + 0.000 847 345 6;
  • 4) 0.000 847 345 6 × 2 = 0 + 0.001 694 691 2;
  • 5) 0.001 694 691 2 × 2 = 0 + 0.003 389 382 4;
  • 6) 0.003 389 382 4 × 2 = 0 + 0.006 778 764 8;
  • 7) 0.006 778 764 8 × 2 = 0 + 0.013 557 529 6;
  • 8) 0.013 557 529 6 × 2 = 0 + 0.027 115 059 2;
  • 9) 0.027 115 059 2 × 2 = 0 + 0.054 230 118 4;
  • 10) 0.054 230 118 4 × 2 = 0 + 0.108 460 236 8;
  • 11) 0.108 460 236 8 × 2 = 0 + 0.216 920 473 6;
  • 12) 0.216 920 473 6 × 2 = 0 + 0.433 840 947 2;
  • 13) 0.433 840 947 2 × 2 = 0 + 0.867 681 894 4;
  • 14) 0.867 681 894 4 × 2 = 1 + 0.735 363 788 8;
  • 15) 0.735 363 788 8 × 2 = 1 + 0.470 727 577 6;
  • 16) 0.470 727 577 6 × 2 = 0 + 0.941 455 155 2;
  • 17) 0.941 455 155 2 × 2 = 1 + 0.882 910 310 4;
  • 18) 0.882 910 310 4 × 2 = 1 + 0.765 820 620 8;
  • 19) 0.765 820 620 8 × 2 = 1 + 0.531 641 241 6;
  • 20) 0.531 641 241 6 × 2 = 1 + 0.063 282 483 2;
  • 21) 0.063 282 483 2 × 2 = 0 + 0.126 564 966 4;
  • 22) 0.126 564 966 4 × 2 = 0 + 0.253 129 932 8;
  • 23) 0.253 129 932 8 × 2 = 0 + 0.506 259 865 6;
  • 24) 0.506 259 865 6 × 2 = 1 + 0.012 519 731 2;
  • 25) 0.012 519 731 2 × 2 = 0 + 0.025 039 462 4;
  • 26) 0.025 039 462 4 × 2 = 0 + 0.050 078 924 8;
  • 27) 0.050 078 924 8 × 2 = 0 + 0.100 157 849 6;
  • 28) 0.100 157 849 6 × 2 = 0 + 0.200 315 699 2;
  • 29) 0.200 315 699 2 × 2 = 0 + 0.400 631 398 4;
  • 30) 0.400 631 398 4 × 2 = 0 + 0.801 262 796 8;
  • 31) 0.801 262 796 8 × 2 = 1 + 0.602 525 593 6;
  • 32) 0.602 525 593 6 × 2 = 1 + 0.205 051 187 2;
  • 33) 0.205 051 187 2 × 2 = 0 + 0.410 102 374 4;
  • 34) 0.410 102 374 4 × 2 = 0 + 0.820 204 748 8;
  • 35) 0.820 204 748 8 × 2 = 1 + 0.640 409 497 6;
  • 36) 0.640 409 497 6 × 2 = 1 + 0.280 818 995 2;
  • 37) 0.280 818 995 2 × 2 = 0 + 0.561 637 990 4;
  • 38) 0.561 637 990 4 × 2 = 1 + 0.123 275 980 8;
  • 39) 0.123 275 980 8 × 2 = 0 + 0.246 551 961 6;
  • 40) 0.246 551 961 6 × 2 = 0 + 0.493 103 923 2;
  • 41) 0.493 103 923 2 × 2 = 0 + 0.986 207 846 4;
  • 42) 0.986 207 846 4 × 2 = 1 + 0.972 415 692 8;
  • 43) 0.972 415 692 8 × 2 = 1 + 0.944 831 385 6;
  • 44) 0.944 831 385 6 × 2 = 1 + 0.889 662 771 2;
  • 45) 0.889 662 771 2 × 2 = 1 + 0.779 325 542 4;
  • 46) 0.779 325 542 4 × 2 = 1 + 0.558 651 084 8;
  • 47) 0.558 651 084 8 × 2 = 1 + 0.117 302 169 6;
  • 48) 0.117 302 169 6 × 2 = 0 + 0.234 604 339 2;
  • 49) 0.234 604 339 2 × 2 = 0 + 0.469 208 678 4;
  • 50) 0.469 208 678 4 × 2 = 0 + 0.938 417 356 8;
  • 51) 0.938 417 356 8 × 2 = 1 + 0.876 834 713 6;
  • 52) 0.876 834 713 6 × 2 = 1 + 0.753 669 427 2;
  • 53) 0.753 669 427 2 × 2 = 1 + 0.507 338 854 4;
  • 54) 0.507 338 854 4 × 2 = 1 + 0.014 677 708 8;
  • 55) 0.014 677 708 8 × 2 = 0 + 0.029 355 417 6;
  • 56) 0.029 355 417 6 × 2 = 0 + 0.058 710 835 2;
  • 57) 0.058 710 835 2 × 2 = 0 + 0.117 421 670 4;
  • 58) 0.117 421 670 4 × 2 = 0 + 0.234 843 340 8;
  • 59) 0.234 843 340 8 × 2 = 0 + 0.469 686 681 6;
  • 60) 0.469 686 681 6 × 2 = 0 + 0.939 373 363 2;
  • 61) 0.939 373 363 2 × 2 = 1 + 0.878 746 726 4;
  • 62) 0.878 746 726 4 × 2 = 1 + 0.757 493 452 8;
  • 63) 0.757 493 452 8 × 2 = 1 + 0.514 986 905 6;
  • 64) 0.514 986 905 6 × 2 = 1 + 0.029 973 811 2;
  • 65) 0.029 973 811 2 × 2 = 0 + 0.059 947 622 4;
  • 66) 0.059 947 622 4 × 2 = 0 + 0.119 895 244 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).


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:


0.000 105 918 2(10) =

0.0000 0000 0000 0110 1111 0001 0000 0011 0011 0100 0111 1110 0011 1100 0000 1111 00(2)

6. Positive number before normalization:

0.000 105 918 2(10) =

0.0000 0000 0000 0110 1111 0001 0000 0011 0011 0100 0111 1110 0011 1100 0000 1111 00(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:


0.000 105 918 2(10) =

0.0000 0000 0000 0110 1111 0001 0000 0011 0011 0100 0111 1110 0011 1100 0000 1111 00(2) =

0.0000 0000 0000 0110 1111 0001 0000 0011 0011 0100 0111 1110 0011 1100 0000 1111 00(2) × 20 =

1.1011 1100 0100 0000 1100 1101 0001 1111 1000 1111 0000 0011 1100(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 0000 1100 1101 0001 1111 1000 1111 0000 0011 1100


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 0000 1100 1101 0001 1111 1000 1111 0000 0011 1100 =

1011 1100 0100 0000 1100 1101 0001 1111 1000 1111 0000 0011 1100


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 0000 1100 1101 0001 1111 1000 1111 0000 0011 1100


Decimal number -0.000 105 918 2 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0001 - 1011 1100 0100 0000 1100 1101 0001 1111 1000 1111 0000 0011 1100

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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