-0.000 282 005 889 6 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 889 6(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
-0.000 282 005 889 6(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:

|-0.000 282 005 889 6| = 0.000 282 005 889 6


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: 0.000 282 005 889 6.

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) 0.000 282 005 889 6 × 2 = 0 + 0.000 564 011 779 2;
  • 2) 0.000 564 011 779 2 × 2 = 0 + 0.001 128 023 558 4;
  • 3) 0.001 128 023 558 4 × 2 = 0 + 0.002 256 047 116 8;
  • 4) 0.002 256 047 116 8 × 2 = 0 + 0.004 512 094 233 6;
  • 5) 0.004 512 094 233 6 × 2 = 0 + 0.009 024 188 467 2;
  • 6) 0.009 024 188 467 2 × 2 = 0 + 0.018 048 376 934 4;
  • 7) 0.018 048 376 934 4 × 2 = 0 + 0.036 096 753 868 8;
  • 8) 0.036 096 753 868 8 × 2 = 0 + 0.072 193 507 737 6;
  • 9) 0.072 193 507 737 6 × 2 = 0 + 0.144 387 015 475 2;
  • 10) 0.144 387 015 475 2 × 2 = 0 + 0.288 774 030 950 4;
  • 11) 0.288 774 030 950 4 × 2 = 0 + 0.577 548 061 900 8;
  • 12) 0.577 548 061 900 8 × 2 = 1 + 0.155 096 123 801 6;
  • 13) 0.155 096 123 801 6 × 2 = 0 + 0.310 192 247 603 2;
  • 14) 0.310 192 247 603 2 × 2 = 0 + 0.620 384 495 206 4;
  • 15) 0.620 384 495 206 4 × 2 = 1 + 0.240 768 990 412 8;
  • 16) 0.240 768 990 412 8 × 2 = 0 + 0.481 537 980 825 6;
  • 17) 0.481 537 980 825 6 × 2 = 0 + 0.963 075 961 651 2;
  • 18) 0.963 075 961 651 2 × 2 = 1 + 0.926 151 923 302 4;
  • 19) 0.926 151 923 302 4 × 2 = 1 + 0.852 303 846 604 8;
  • 20) 0.852 303 846 604 8 × 2 = 1 + 0.704 607 693 209 6;
  • 21) 0.704 607 693 209 6 × 2 = 1 + 0.409 215 386 419 2;
  • 22) 0.409 215 386 419 2 × 2 = 0 + 0.818 430 772 838 4;
  • 23) 0.818 430 772 838 4 × 2 = 1 + 0.636 861 545 676 8;
  • 24) 0.636 861 545 676 8 × 2 = 1 + 0.273 723 091 353 6;
  • 25) 0.273 723 091 353 6 × 2 = 0 + 0.547 446 182 707 2;
  • 26) 0.547 446 182 707 2 × 2 = 1 + 0.094 892 365 414 4;
  • 27) 0.094 892 365 414 4 × 2 = 0 + 0.189 784 730 828 8;
  • 28) 0.189 784 730 828 8 × 2 = 0 + 0.379 569 461 657 6;
  • 29) 0.379 569 461 657 6 × 2 = 0 + 0.759 138 923 315 2;
  • 30) 0.759 138 923 315 2 × 2 = 1 + 0.518 277 846 630 4;
  • 31) 0.518 277 846 630 4 × 2 = 1 + 0.036 555 693 260 8;
  • 32) 0.036 555 693 260 8 × 2 = 0 + 0.073 111 386 521 6;
  • 33) 0.073 111 386 521 6 × 2 = 0 + 0.146 222 773 043 2;
  • 34) 0.146 222 773 043 2 × 2 = 0 + 0.292 445 546 086 4;
  • 35) 0.292 445 546 086 4 × 2 = 0 + 0.584 891 092 172 8;
  • 36) 0.584 891 092 172 8 × 2 = 1 + 0.169 782 184 345 6;
  • 37) 0.169 782 184 345 6 × 2 = 0 + 0.339 564 368 691 2;
  • 38) 0.339 564 368 691 2 × 2 = 0 + 0.679 128 737 382 4;
  • 39) 0.679 128 737 382 4 × 2 = 1 + 0.358 257 474 764 8;
  • 40) 0.358 257 474 764 8 × 2 = 0 + 0.716 514 949 529 6;
  • 41) 0.716 514 949 529 6 × 2 = 1 + 0.433 029 899 059 2;
  • 42) 0.433 029 899 059 2 × 2 = 0 + 0.866 059 798 118 4;
  • 43) 0.866 059 798 118 4 × 2 = 1 + 0.732 119 596 236 8;
  • 44) 0.732 119 596 236 8 × 2 = 1 + 0.464 239 192 473 6;
  • 45) 0.464 239 192 473 6 × 2 = 0 + 0.928 478 384 947 2;
  • 46) 0.928 478 384 947 2 × 2 = 1 + 0.856 956 769 894 4;
  • 47) 0.856 956 769 894 4 × 2 = 1 + 0.713 913 539 788 8;
  • 48) 0.713 913 539 788 8 × 2 = 1 + 0.427 827 079 577 6;
  • 49) 0.427 827 079 577 6 × 2 = 0 + 0.855 654 159 155 2;
  • 50) 0.855 654 159 155 2 × 2 = 1 + 0.711 308 318 310 4;
  • 51) 0.711 308 318 310 4 × 2 = 1 + 0.422 616 636 620 8;
  • 52) 0.422 616 636 620 8 × 2 = 0 + 0.845 233 273 241 6;
  • 53) 0.845 233 273 241 6 × 2 = 1 + 0.690 466 546 483 2;
  • 54) 0.690 466 546 483 2 × 2 = 1 + 0.380 933 092 966 4;
  • 55) 0.380 933 092 966 4 × 2 = 0 + 0.761 866 185 932 8;
  • 56) 0.761 866 185 932 8 × 2 = 1 + 0.523 732 371 865 6;
  • 57) 0.523 732 371 865 6 × 2 = 1 + 0.047 464 743 731 2;
  • 58) 0.047 464 743 731 2 × 2 = 0 + 0.094 929 487 462 4;
  • 59) 0.094 929 487 462 4 × 2 = 0 + 0.189 858 974 924 8;
  • 60) 0.189 858 974 924 8 × 2 = 0 + 0.379 717 949 849 6;
  • 61) 0.379 717 949 849 6 × 2 = 0 + 0.759 435 899 699 2;
  • 62) 0.759 435 899 699 2 × 2 = 1 + 0.518 871 799 398 4;
  • 63) 0.518 871 799 398 4 × 2 = 1 + 0.037 743 598 796 8;
  • 64) 0.037 743 598 796 8 × 2 = 0 + 0.075 487 197 593 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:


0.000 282 005 889 6(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0001 0010 1011 0111 0110 1101 1000 0110(2)

6. Positive number before normalization:

0.000 282 005 889 6(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0001 0010 1011 0111 0110 1101 1000 0110(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 12 positions to the right, so that only one non zero digit remains to the left of it:


0.000 282 005 889 6(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0001 0010 1011 0111 0110 1101 1000 0110(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0001 0010 1011 0111 0110 1101 1000 0110(2) × 20 =

1.0010 0111 1011 0100 0110 0001 0010 1011 0111 0110 1101 1000 0110(2) × 2-12


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): -12

Mantissa (not normalized):
1.0010 0111 1011 0100 0110 0001 0010 1011 0111 0110 1101 1000 0110


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-12 + 2(11-1) - 1 =

(-12 + 1 023)(10) =

1 011(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;
  • 1 011 ÷ 2 = 505 + 1;
  • 505 ÷ 2 = 252 + 1;
  • 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) =

1011(10) =

011 1111 0011(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. 0010 0111 1011 0100 0110 0001 0010 1011 0111 0110 1101 1000 0110 =

0010 0111 1011 0100 0110 0001 0010 1011 0111 0110 1101 1000 0110


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 0011

Mantissa (52 bits) =
0010 0111 1011 0100 0110 0001 0010 1011 0111 0110 1101 1000 0110


Decimal number -0.000 282 005 889 6 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0001 0010 1011 0111 0110 1101 1000 0110

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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