-0.000 282 005 915 62 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 915 62(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
-0.000 282 005 915 62(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:

|-0.000 282 005 915 62| = 0.000 282 005 915 62


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: 0.000 282 005 915 62.

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) 0.000 282 005 915 62 × 2 = 0 + 0.000 564 011 831 24;
  • 2) 0.000 564 011 831 24 × 2 = 0 + 0.001 128 023 662 48;
  • 3) 0.001 128 023 662 48 × 2 = 0 + 0.002 256 047 324 96;
  • 4) 0.002 256 047 324 96 × 2 = 0 + 0.004 512 094 649 92;
  • 5) 0.004 512 094 649 92 × 2 = 0 + 0.009 024 189 299 84;
  • 6) 0.009 024 189 299 84 × 2 = 0 + 0.018 048 378 599 68;
  • 7) 0.018 048 378 599 68 × 2 = 0 + 0.036 096 757 199 36;
  • 8) 0.036 096 757 199 36 × 2 = 0 + 0.072 193 514 398 72;
  • 9) 0.072 193 514 398 72 × 2 = 0 + 0.144 387 028 797 44;
  • 10) 0.144 387 028 797 44 × 2 = 0 + 0.288 774 057 594 88;
  • 11) 0.288 774 057 594 88 × 2 = 0 + 0.577 548 115 189 76;
  • 12) 0.577 548 115 189 76 × 2 = 1 + 0.155 096 230 379 52;
  • 13) 0.155 096 230 379 52 × 2 = 0 + 0.310 192 460 759 04;
  • 14) 0.310 192 460 759 04 × 2 = 0 + 0.620 384 921 518 08;
  • 15) 0.620 384 921 518 08 × 2 = 1 + 0.240 769 843 036 16;
  • 16) 0.240 769 843 036 16 × 2 = 0 + 0.481 539 686 072 32;
  • 17) 0.481 539 686 072 32 × 2 = 0 + 0.963 079 372 144 64;
  • 18) 0.963 079 372 144 64 × 2 = 1 + 0.926 158 744 289 28;
  • 19) 0.926 158 744 289 28 × 2 = 1 + 0.852 317 488 578 56;
  • 20) 0.852 317 488 578 56 × 2 = 1 + 0.704 634 977 157 12;
  • 21) 0.704 634 977 157 12 × 2 = 1 + 0.409 269 954 314 24;
  • 22) 0.409 269 954 314 24 × 2 = 0 + 0.818 539 908 628 48;
  • 23) 0.818 539 908 628 48 × 2 = 1 + 0.637 079 817 256 96;
  • 24) 0.637 079 817 256 96 × 2 = 1 + 0.274 159 634 513 92;
  • 25) 0.274 159 634 513 92 × 2 = 0 + 0.548 319 269 027 84;
  • 26) 0.548 319 269 027 84 × 2 = 1 + 0.096 638 538 055 68;
  • 27) 0.096 638 538 055 68 × 2 = 0 + 0.193 277 076 111 36;
  • 28) 0.193 277 076 111 36 × 2 = 0 + 0.386 554 152 222 72;
  • 29) 0.386 554 152 222 72 × 2 = 0 + 0.773 108 304 445 44;
  • 30) 0.773 108 304 445 44 × 2 = 1 + 0.546 216 608 890 88;
  • 31) 0.546 216 608 890 88 × 2 = 1 + 0.092 433 217 781 76;
  • 32) 0.092 433 217 781 76 × 2 = 0 + 0.184 866 435 563 52;
  • 33) 0.184 866 435 563 52 × 2 = 0 + 0.369 732 871 127 04;
  • 34) 0.369 732 871 127 04 × 2 = 0 + 0.739 465 742 254 08;
  • 35) 0.739 465 742 254 08 × 2 = 1 + 0.478 931 484 508 16;
  • 36) 0.478 931 484 508 16 × 2 = 0 + 0.957 862 969 016 32;
  • 37) 0.957 862 969 016 32 × 2 = 1 + 0.915 725 938 032 64;
  • 38) 0.915 725 938 032 64 × 2 = 1 + 0.831 451 876 065 28;
  • 39) 0.831 451 876 065 28 × 2 = 1 + 0.662 903 752 130 56;
  • 40) 0.662 903 752 130 56 × 2 = 1 + 0.325 807 504 261 12;
  • 41) 0.325 807 504 261 12 × 2 = 0 + 0.651 615 008 522 24;
  • 42) 0.651 615 008 522 24 × 2 = 1 + 0.303 230 017 044 48;
  • 43) 0.303 230 017 044 48 × 2 = 0 + 0.606 460 034 088 96;
  • 44) 0.606 460 034 088 96 × 2 = 1 + 0.212 920 068 177 92;
  • 45) 0.212 920 068 177 92 × 2 = 0 + 0.425 840 136 355 84;
  • 46) 0.425 840 136 355 84 × 2 = 0 + 0.851 680 272 711 68;
  • 47) 0.851 680 272 711 68 × 2 = 1 + 0.703 360 545 423 36;
  • 48) 0.703 360 545 423 36 × 2 = 1 + 0.406 721 090 846 72;
  • 49) 0.406 721 090 846 72 × 2 = 0 + 0.813 442 181 693 44;
  • 50) 0.813 442 181 693 44 × 2 = 1 + 0.626 884 363 386 88;
  • 51) 0.626 884 363 386 88 × 2 = 1 + 0.253 768 726 773 76;
  • 52) 0.253 768 726 773 76 × 2 = 0 + 0.507 537 453 547 52;
  • 53) 0.507 537 453 547 52 × 2 = 1 + 0.015 074 907 095 04;
  • 54) 0.015 074 907 095 04 × 2 = 0 + 0.030 149 814 190 08;
  • 55) 0.030 149 814 190 08 × 2 = 0 + 0.060 299 628 380 16;
  • 56) 0.060 299 628 380 16 × 2 = 0 + 0.120 599 256 760 32;
  • 57) 0.120 599 256 760 32 × 2 = 0 + 0.241 198 513 520 64;
  • 58) 0.241 198 513 520 64 × 2 = 0 + 0.482 397 027 041 28;
  • 59) 0.482 397 027 041 28 × 2 = 0 + 0.964 794 054 082 56;
  • 60) 0.964 794 054 082 56 × 2 = 1 + 0.929 588 108 165 12;
  • 61) 0.929 588 108 165 12 × 2 = 1 + 0.859 176 216 330 24;
  • 62) 0.859 176 216 330 24 × 2 = 1 + 0.718 352 432 660 48;
  • 63) 0.718 352 432 660 48 × 2 = 1 + 0.436 704 865 320 96;
  • 64) 0.436 704 865 320 96 × 2 = 0 + 0.873 409 730 641 92;

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:


0.000 282 005 915 62(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1111 0101 0011 0110 1000 0001 1110(2)

6. Positive number before normalization:

0.000 282 005 915 62(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1111 0101 0011 0110 1000 0001 1110(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:


0.000 282 005 915 62(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1111 0101 0011 0110 1000 0001 1110(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1111 0101 0011 0110 1000 0001 1110(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1111 0101 0011 0110 1000 0001 1110(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 0010 1111 0101 0011 0110 1000 0001 1110


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 0010 1111 0101 0011 0110 1000 0001 1110 =

0010 0111 1011 0100 0110 0010 1111 0101 0011 0110 1000 0001 1110


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 0010 1111 0101 0011 0110 1000 0001 1110


Decimal number -0.000 282 005 915 62 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1111 0101 0011 0110 1000 0001 1110

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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