Decimal to 64 Bit IEEE 754 Binary: Convert Number -0.085 969 999 999 992 8 to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From Base Ten Decimal System

Number -0.085 969 999 999 992 8(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:

|-0.085 969 999 999 992 8| = 0.085 969 999 999 992 8


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: 0.085 969 999 999 992 8.

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) 0.085 969 999 999 992 8 × 2 = 0 + 0.171 939 999 999 985 6;
  • 2) 0.171 939 999 999 985 6 × 2 = 0 + 0.343 879 999 999 971 2;
  • 3) 0.343 879 999 999 971 2 × 2 = 0 + 0.687 759 999 999 942 4;
  • 4) 0.687 759 999 999 942 4 × 2 = 1 + 0.375 519 999 999 884 8;
  • 5) 0.375 519 999 999 884 8 × 2 = 0 + 0.751 039 999 999 769 6;
  • 6) 0.751 039 999 999 769 6 × 2 = 1 + 0.502 079 999 999 539 2;
  • 7) 0.502 079 999 999 539 2 × 2 = 1 + 0.004 159 999 999 078 4;
  • 8) 0.004 159 999 999 078 4 × 2 = 0 + 0.008 319 999 998 156 8;
  • 9) 0.008 319 999 998 156 8 × 2 = 0 + 0.016 639 999 996 313 6;
  • 10) 0.016 639 999 996 313 6 × 2 = 0 + 0.033 279 999 992 627 2;
  • 11) 0.033 279 999 992 627 2 × 2 = 0 + 0.066 559 999 985 254 4;
  • 12) 0.066 559 999 985 254 4 × 2 = 0 + 0.133 119 999 970 508 8;
  • 13) 0.133 119 999 970 508 8 × 2 = 0 + 0.266 239 999 941 017 6;
  • 14) 0.266 239 999 941 017 6 × 2 = 0 + 0.532 479 999 882 035 2;
  • 15) 0.532 479 999 882 035 2 × 2 = 1 + 0.064 959 999 764 070 4;
  • 16) 0.064 959 999 764 070 4 × 2 = 0 + 0.129 919 999 528 140 8;
  • 17) 0.129 919 999 528 140 8 × 2 = 0 + 0.259 839 999 056 281 6;
  • 18) 0.259 839 999 056 281 6 × 2 = 0 + 0.519 679 998 112 563 2;
  • 19) 0.519 679 998 112 563 2 × 2 = 1 + 0.039 359 996 225 126 4;
  • 20) 0.039 359 996 225 126 4 × 2 = 0 + 0.078 719 992 450 252 8;
  • 21) 0.078 719 992 450 252 8 × 2 = 0 + 0.157 439 984 900 505 6;
  • 22) 0.157 439 984 900 505 6 × 2 = 0 + 0.314 879 969 801 011 2;
  • 23) 0.314 879 969 801 011 2 × 2 = 0 + 0.629 759 939 602 022 4;
  • 24) 0.629 759 939 602 022 4 × 2 = 1 + 0.259 519 879 204 044 8;
  • 25) 0.259 519 879 204 044 8 × 2 = 0 + 0.519 039 758 408 089 6;
  • 26) 0.519 039 758 408 089 6 × 2 = 1 + 0.038 079 516 816 179 2;
  • 27) 0.038 079 516 816 179 2 × 2 = 0 + 0.076 159 033 632 358 4;
  • 28) 0.076 159 033 632 358 4 × 2 = 0 + 0.152 318 067 264 716 8;
  • 29) 0.152 318 067 264 716 8 × 2 = 0 + 0.304 636 134 529 433 6;
  • 30) 0.304 636 134 529 433 6 × 2 = 0 + 0.609 272 269 058 867 2;
  • 31) 0.609 272 269 058 867 2 × 2 = 1 + 0.218 544 538 117 734 4;
  • 32) 0.218 544 538 117 734 4 × 2 = 0 + 0.437 089 076 235 468 8;
  • 33) 0.437 089 076 235 468 8 × 2 = 0 + 0.874 178 152 470 937 6;
  • 34) 0.874 178 152 470 937 6 × 2 = 1 + 0.748 356 304 941 875 2;
  • 35) 0.748 356 304 941 875 2 × 2 = 1 + 0.496 712 609 883 750 4;
  • 36) 0.496 712 609 883 750 4 × 2 = 0 + 0.993 425 219 767 500 8;
  • 37) 0.993 425 219 767 500 8 × 2 = 1 + 0.986 850 439 535 001 6;
  • 38) 0.986 850 439 535 001 6 × 2 = 1 + 0.973 700 879 070 003 2;
  • 39) 0.973 700 879 070 003 2 × 2 = 1 + 0.947 401 758 140 006 4;
  • 40) 0.947 401 758 140 006 4 × 2 = 1 + 0.894 803 516 280 012 8;
  • 41) 0.894 803 516 280 012 8 × 2 = 1 + 0.789 607 032 560 025 6;
  • 42) 0.789 607 032 560 025 6 × 2 = 1 + 0.579 214 065 120 051 2;
  • 43) 0.579 214 065 120 051 2 × 2 = 1 + 0.158 428 130 240 102 4;
  • 44) 0.158 428 130 240 102 4 × 2 = 0 + 0.316 856 260 480 204 8;
  • 45) 0.316 856 260 480 204 8 × 2 = 0 + 0.633 712 520 960 409 6;
  • 46) 0.633 712 520 960 409 6 × 2 = 1 + 0.267 425 041 920 819 2;
  • 47) 0.267 425 041 920 819 2 × 2 = 0 + 0.534 850 083 841 638 4;
  • 48) 0.534 850 083 841 638 4 × 2 = 1 + 0.069 700 167 683 276 8;
  • 49) 0.069 700 167 683 276 8 × 2 = 0 + 0.139 400 335 366 553 6;
  • 50) 0.139 400 335 366 553 6 × 2 = 0 + 0.278 800 670 733 107 2;
  • 51) 0.278 800 670 733 107 2 × 2 = 0 + 0.557 601 341 466 214 4;
  • 52) 0.557 601 341 466 214 4 × 2 = 1 + 0.115 202 682 932 428 8;
  • 53) 0.115 202 682 932 428 8 × 2 = 0 + 0.230 405 365 864 857 6;
  • 54) 0.230 405 365 864 857 6 × 2 = 0 + 0.460 810 731 729 715 2;
  • 55) 0.460 810 731 729 715 2 × 2 = 0 + 0.921 621 463 459 430 4;
  • 56) 0.921 621 463 459 430 4 × 2 = 1 + 0.843 242 926 918 860 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:


0.085 969 999 999 992 8(10) =

0.0001 0110 0000 0010 0010 0001 0100 0010 0110 1111 1110 0101 0001 0001(2)

6. Positive number before normalization:

0.085 969 999 999 992 8(10) =

0.0001 0110 0000 0010 0010 0001 0100 0010 0110 1111 1110 0101 0001 0001(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 4 positions to the right, so that only one non zero digit remains to the left of it:


0.085 969 999 999 992 8(10) =

0.0001 0110 0000 0010 0010 0001 0100 0010 0110 1111 1110 0101 0001 0001(2) =

0.0001 0110 0000 0010 0010 0001 0100 0010 0110 1111 1110 0101 0001 0001(2) × 20 =

1.0110 0000 0010 0010 0001 0100 0010 0110 1111 1110 0101 0001 0001(2) × 2-4


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):
1.0110 0000 0010 0010 0001 0100 0010 0110 1111 1110 0101 0001 0001


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-4 + 2(11-1) - 1 =

(-4 + 1 023)(10) =

1 019(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;
  • 1 019 ÷ 2 = 509 + 1;
  • 509 ÷ 2 = 254 + 1;
  • 254 ÷ 2 = 127 + 0;
  • 127 ÷ 2 = 63 + 1;
  • 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) =

1019(10) =

011 1111 1011(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. 0110 0000 0010 0010 0001 0100 0010 0110 1111 1110 0101 0001 0001 =

0110 0000 0010 0010 0001 0100 0010 0110 1111 1110 0101 0001 0001


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 1011

Mantissa (52 bits) =
0110 0000 0010 0010 0001 0100 0010 0110 1111 1110 0101 0001 0001


The base ten decimal number -0.085 969 999 999 992 8 converted and written in 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1011 - 0110 0000 0010 0010 0001 0100 0010 0110 1111 1110 0101 0001 0001

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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