-0.000 282 005 876 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 876 3(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
-0.000 282 005 876 3(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:

|-0.000 282 005 876 3| = 0.000 282 005 876 3


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: 0.000 282 005 876 3.

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) 0.000 282 005 876 3 × 2 = 0 + 0.000 564 011 752 6;
  • 2) 0.000 564 011 752 6 × 2 = 0 + 0.001 128 023 505 2;
  • 3) 0.001 128 023 505 2 × 2 = 0 + 0.002 256 047 010 4;
  • 4) 0.002 256 047 010 4 × 2 = 0 + 0.004 512 094 020 8;
  • 5) 0.004 512 094 020 8 × 2 = 0 + 0.009 024 188 041 6;
  • 6) 0.009 024 188 041 6 × 2 = 0 + 0.018 048 376 083 2;
  • 7) 0.018 048 376 083 2 × 2 = 0 + 0.036 096 752 166 4;
  • 8) 0.036 096 752 166 4 × 2 = 0 + 0.072 193 504 332 8;
  • 9) 0.072 193 504 332 8 × 2 = 0 + 0.144 387 008 665 6;
  • 10) 0.144 387 008 665 6 × 2 = 0 + 0.288 774 017 331 2;
  • 11) 0.288 774 017 331 2 × 2 = 0 + 0.577 548 034 662 4;
  • 12) 0.577 548 034 662 4 × 2 = 1 + 0.155 096 069 324 8;
  • 13) 0.155 096 069 324 8 × 2 = 0 + 0.310 192 138 649 6;
  • 14) 0.310 192 138 649 6 × 2 = 0 + 0.620 384 277 299 2;
  • 15) 0.620 384 277 299 2 × 2 = 1 + 0.240 768 554 598 4;
  • 16) 0.240 768 554 598 4 × 2 = 0 + 0.481 537 109 196 8;
  • 17) 0.481 537 109 196 8 × 2 = 0 + 0.963 074 218 393 6;
  • 18) 0.963 074 218 393 6 × 2 = 1 + 0.926 148 436 787 2;
  • 19) 0.926 148 436 787 2 × 2 = 1 + 0.852 296 873 574 4;
  • 20) 0.852 296 873 574 4 × 2 = 1 + 0.704 593 747 148 8;
  • 21) 0.704 593 747 148 8 × 2 = 1 + 0.409 187 494 297 6;
  • 22) 0.409 187 494 297 6 × 2 = 0 + 0.818 374 988 595 2;
  • 23) 0.818 374 988 595 2 × 2 = 1 + 0.636 749 977 190 4;
  • 24) 0.636 749 977 190 4 × 2 = 1 + 0.273 499 954 380 8;
  • 25) 0.273 499 954 380 8 × 2 = 0 + 0.546 999 908 761 6;
  • 26) 0.546 999 908 761 6 × 2 = 1 + 0.093 999 817 523 2;
  • 27) 0.093 999 817 523 2 × 2 = 0 + 0.187 999 635 046 4;
  • 28) 0.187 999 635 046 4 × 2 = 0 + 0.375 999 270 092 8;
  • 29) 0.375 999 270 092 8 × 2 = 0 + 0.751 998 540 185 6;
  • 30) 0.751 998 540 185 6 × 2 = 1 + 0.503 997 080 371 2;
  • 31) 0.503 997 080 371 2 × 2 = 1 + 0.007 994 160 742 4;
  • 32) 0.007 994 160 742 4 × 2 = 0 + 0.015 988 321 484 8;
  • 33) 0.015 988 321 484 8 × 2 = 0 + 0.031 976 642 969 6;
  • 34) 0.031 976 642 969 6 × 2 = 0 + 0.063 953 285 939 2;
  • 35) 0.063 953 285 939 2 × 2 = 0 + 0.127 906 571 878 4;
  • 36) 0.127 906 571 878 4 × 2 = 0 + 0.255 813 143 756 8;
  • 37) 0.255 813 143 756 8 × 2 = 0 + 0.511 626 287 513 6;
  • 38) 0.511 626 287 513 6 × 2 = 1 + 0.023 252 575 027 2;
  • 39) 0.023 252 575 027 2 × 2 = 0 + 0.046 505 150 054 4;
  • 40) 0.046 505 150 054 4 × 2 = 0 + 0.093 010 300 108 8;
  • 41) 0.093 010 300 108 8 × 2 = 0 + 0.186 020 600 217 6;
  • 42) 0.186 020 600 217 6 × 2 = 0 + 0.372 041 200 435 2;
  • 43) 0.372 041 200 435 2 × 2 = 0 + 0.744 082 400 870 4;
  • 44) 0.744 082 400 870 4 × 2 = 1 + 0.488 164 801 740 8;
  • 45) 0.488 164 801 740 8 × 2 = 0 + 0.976 329 603 481 6;
  • 46) 0.976 329 603 481 6 × 2 = 1 + 0.952 659 206 963 2;
  • 47) 0.952 659 206 963 2 × 2 = 1 + 0.905 318 413 926 4;
  • 48) 0.905 318 413 926 4 × 2 = 1 + 0.810 636 827 852 8;
  • 49) 0.810 636 827 852 8 × 2 = 1 + 0.621 273 655 705 6;
  • 50) 0.621 273 655 705 6 × 2 = 1 + 0.242 547 311 411 2;
  • 51) 0.242 547 311 411 2 × 2 = 0 + 0.485 094 622 822 4;
  • 52) 0.485 094 622 822 4 × 2 = 0 + 0.970 189 245 644 8;
  • 53) 0.970 189 245 644 8 × 2 = 1 + 0.940 378 491 289 6;
  • 54) 0.940 378 491 289 6 × 2 = 1 + 0.880 756 982 579 2;
  • 55) 0.880 756 982 579 2 × 2 = 1 + 0.761 513 965 158 4;
  • 56) 0.761 513 965 158 4 × 2 = 1 + 0.523 027 930 316 8;
  • 57) 0.523 027 930 316 8 × 2 = 1 + 0.046 055 860 633 6;
  • 58) 0.046 055 860 633 6 × 2 = 0 + 0.092 111 721 267 2;
  • 59) 0.092 111 721 267 2 × 2 = 0 + 0.184 223 442 534 4;
  • 60) 0.184 223 442 534 4 × 2 = 0 + 0.368 446 885 068 8;
  • 61) 0.368 446 885 068 8 × 2 = 0 + 0.736 893 770 137 6;
  • 62) 0.736 893 770 137 6 × 2 = 1 + 0.473 787 540 275 2;
  • 63) 0.473 787 540 275 2 × 2 = 0 + 0.947 575 080 550 4;
  • 64) 0.947 575 080 550 4 × 2 = 1 + 0.895 150 161 100 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:


0.000 282 005 876 3(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0000 0100 0001 0111 1100 1111 1000 0101(2)

6. Positive number before normalization:

0.000 282 005 876 3(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0000 0100 0001 0111 1100 1111 1000 0101(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:


0.000 282 005 876 3(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0000 0100 0001 0111 1100 1111 1000 0101(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0000 0100 0001 0111 1100 1111 1000 0101(2) × 20 =

1.0010 0111 1011 0100 0110 0000 0100 0001 0111 1100 1111 1000 0101(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 0000 0100 0001 0111 1100 1111 1000 0101


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 0000 0100 0001 0111 1100 1111 1000 0101 =

0010 0111 1011 0100 0110 0000 0100 0001 0111 1100 1111 1000 0101


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 0000 0100 0001 0111 1100 1111 1000 0101


Decimal number -0.000 282 005 876 3 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0000 0100 0001 0111 1100 1111 1000 0101

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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