-0.000 282 005 873 4 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 873 4(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
-0.000 282 005 873 4(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:

|-0.000 282 005 873 4| = 0.000 282 005 873 4


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: 0.000 282 005 873 4.

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) 0.000 282 005 873 4 × 2 = 0 + 0.000 564 011 746 8;
  • 2) 0.000 564 011 746 8 × 2 = 0 + 0.001 128 023 493 6;
  • 3) 0.001 128 023 493 6 × 2 = 0 + 0.002 256 046 987 2;
  • 4) 0.002 256 046 987 2 × 2 = 0 + 0.004 512 093 974 4;
  • 5) 0.004 512 093 974 4 × 2 = 0 + 0.009 024 187 948 8;
  • 6) 0.009 024 187 948 8 × 2 = 0 + 0.018 048 375 897 6;
  • 7) 0.018 048 375 897 6 × 2 = 0 + 0.036 096 751 795 2;
  • 8) 0.036 096 751 795 2 × 2 = 0 + 0.072 193 503 590 4;
  • 9) 0.072 193 503 590 4 × 2 = 0 + 0.144 387 007 180 8;
  • 10) 0.144 387 007 180 8 × 2 = 0 + 0.288 774 014 361 6;
  • 11) 0.288 774 014 361 6 × 2 = 0 + 0.577 548 028 723 2;
  • 12) 0.577 548 028 723 2 × 2 = 1 + 0.155 096 057 446 4;
  • 13) 0.155 096 057 446 4 × 2 = 0 + 0.310 192 114 892 8;
  • 14) 0.310 192 114 892 8 × 2 = 0 + 0.620 384 229 785 6;
  • 15) 0.620 384 229 785 6 × 2 = 1 + 0.240 768 459 571 2;
  • 16) 0.240 768 459 571 2 × 2 = 0 + 0.481 536 919 142 4;
  • 17) 0.481 536 919 142 4 × 2 = 0 + 0.963 073 838 284 8;
  • 18) 0.963 073 838 284 8 × 2 = 1 + 0.926 147 676 569 6;
  • 19) 0.926 147 676 569 6 × 2 = 1 + 0.852 295 353 139 2;
  • 20) 0.852 295 353 139 2 × 2 = 1 + 0.704 590 706 278 4;
  • 21) 0.704 590 706 278 4 × 2 = 1 + 0.409 181 412 556 8;
  • 22) 0.409 181 412 556 8 × 2 = 0 + 0.818 362 825 113 6;
  • 23) 0.818 362 825 113 6 × 2 = 1 + 0.636 725 650 227 2;
  • 24) 0.636 725 650 227 2 × 2 = 1 + 0.273 451 300 454 4;
  • 25) 0.273 451 300 454 4 × 2 = 0 + 0.546 902 600 908 8;
  • 26) 0.546 902 600 908 8 × 2 = 1 + 0.093 805 201 817 6;
  • 27) 0.093 805 201 817 6 × 2 = 0 + 0.187 610 403 635 2;
  • 28) 0.187 610 403 635 2 × 2 = 0 + 0.375 220 807 270 4;
  • 29) 0.375 220 807 270 4 × 2 = 0 + 0.750 441 614 540 8;
  • 30) 0.750 441 614 540 8 × 2 = 1 + 0.500 883 229 081 6;
  • 31) 0.500 883 229 081 6 × 2 = 1 + 0.001 766 458 163 2;
  • 32) 0.001 766 458 163 2 × 2 = 0 + 0.003 532 916 326 4;
  • 33) 0.003 532 916 326 4 × 2 = 0 + 0.007 065 832 652 8;
  • 34) 0.007 065 832 652 8 × 2 = 0 + 0.014 131 665 305 6;
  • 35) 0.014 131 665 305 6 × 2 = 0 + 0.028 263 330 611 2;
  • 36) 0.028 263 330 611 2 × 2 = 0 + 0.056 526 661 222 4;
  • 37) 0.056 526 661 222 4 × 2 = 0 + 0.113 053 322 444 8;
  • 38) 0.113 053 322 444 8 × 2 = 0 + 0.226 106 644 889 6;
  • 39) 0.226 106 644 889 6 × 2 = 0 + 0.452 213 289 779 2;
  • 40) 0.452 213 289 779 2 × 2 = 0 + 0.904 426 579 558 4;
  • 41) 0.904 426 579 558 4 × 2 = 1 + 0.808 853 159 116 8;
  • 42) 0.808 853 159 116 8 × 2 = 1 + 0.617 706 318 233 6;
  • 43) 0.617 706 318 233 6 × 2 = 1 + 0.235 412 636 467 2;
  • 44) 0.235 412 636 467 2 × 2 = 0 + 0.470 825 272 934 4;
  • 45) 0.470 825 272 934 4 × 2 = 0 + 0.941 650 545 868 8;
  • 46) 0.941 650 545 868 8 × 2 = 1 + 0.883 301 091 737 6;
  • 47) 0.883 301 091 737 6 × 2 = 1 + 0.766 602 183 475 2;
  • 48) 0.766 602 183 475 2 × 2 = 1 + 0.533 204 366 950 4;
  • 49) 0.533 204 366 950 4 × 2 = 1 + 0.066 408 733 900 8;
  • 50) 0.066 408 733 900 8 × 2 = 0 + 0.132 817 467 801 6;
  • 51) 0.132 817 467 801 6 × 2 = 0 + 0.265 634 935 603 2;
  • 52) 0.265 634 935 603 2 × 2 = 0 + 0.531 269 871 206 4;
  • 53) 0.531 269 871 206 4 × 2 = 1 + 0.062 539 742 412 8;
  • 54) 0.062 539 742 412 8 × 2 = 0 + 0.125 079 484 825 6;
  • 55) 0.125 079 484 825 6 × 2 = 0 + 0.250 158 969 651 2;
  • 56) 0.250 158 969 651 2 × 2 = 0 + 0.500 317 939 302 4;
  • 57) 0.500 317 939 302 4 × 2 = 1 + 0.000 635 878 604 8;
  • 58) 0.000 635 878 604 8 × 2 = 0 + 0.001 271 757 209 6;
  • 59) 0.001 271 757 209 6 × 2 = 0 + 0.002 543 514 419 2;
  • 60) 0.002 543 514 419 2 × 2 = 0 + 0.005 087 028 838 4;
  • 61) 0.005 087 028 838 4 × 2 = 0 + 0.010 174 057 676 8;
  • 62) 0.010 174 057 676 8 × 2 = 0 + 0.020 348 115 353 6;
  • 63) 0.020 348 115 353 6 × 2 = 0 + 0.040 696 230 707 2;
  • 64) 0.040 696 230 707 2 × 2 = 0 + 0.081 392 461 414 4;

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:


0.000 282 005 873 4(10) =

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

6. Positive number before normalization:

0.000 282 005 873 4(10) =

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


0.000 282 005 873 4(10) =

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

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

1.0010 0111 1011 0100 0110 0000 0000 1110 0111 1000 1000 1000 0000(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 0000 1110 0111 1000 1000 1000 0000


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 0000 1110 0111 1000 1000 1000 0000 =

0010 0111 1011 0100 0110 0000 0000 1110 0111 1000 1000 1000 0000


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 0000 1110 0111 1000 1000 1000 0000


Decimal number -0.000 282 005 873 4 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0000 0000 1110 0111 1000 1000 1000 0000

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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