-0.000 282 005 892 4 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 892 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
-0.000 282 005 892 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:

|-0.000 282 005 892 4| = 0.000 282 005 892 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: 0.000 282 005 892 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) 0.000 282 005 892 4 × 2 = 0 + 0.000 564 011 784 8;
  • 2) 0.000 564 011 784 8 × 2 = 0 + 0.001 128 023 569 6;
  • 3) 0.001 128 023 569 6 × 2 = 0 + 0.002 256 047 139 2;
  • 4) 0.002 256 047 139 2 × 2 = 0 + 0.004 512 094 278 4;
  • 5) 0.004 512 094 278 4 × 2 = 0 + 0.009 024 188 556 8;
  • 6) 0.009 024 188 556 8 × 2 = 0 + 0.018 048 377 113 6;
  • 7) 0.018 048 377 113 6 × 2 = 0 + 0.036 096 754 227 2;
  • 8) 0.036 096 754 227 2 × 2 = 0 + 0.072 193 508 454 4;
  • 9) 0.072 193 508 454 4 × 2 = 0 + 0.144 387 016 908 8;
  • 10) 0.144 387 016 908 8 × 2 = 0 + 0.288 774 033 817 6;
  • 11) 0.288 774 033 817 6 × 2 = 0 + 0.577 548 067 635 2;
  • 12) 0.577 548 067 635 2 × 2 = 1 + 0.155 096 135 270 4;
  • 13) 0.155 096 135 270 4 × 2 = 0 + 0.310 192 270 540 8;
  • 14) 0.310 192 270 540 8 × 2 = 0 + 0.620 384 541 081 6;
  • 15) 0.620 384 541 081 6 × 2 = 1 + 0.240 769 082 163 2;
  • 16) 0.240 769 082 163 2 × 2 = 0 + 0.481 538 164 326 4;
  • 17) 0.481 538 164 326 4 × 2 = 0 + 0.963 076 328 652 8;
  • 18) 0.963 076 328 652 8 × 2 = 1 + 0.926 152 657 305 6;
  • 19) 0.926 152 657 305 6 × 2 = 1 + 0.852 305 314 611 2;
  • 20) 0.852 305 314 611 2 × 2 = 1 + 0.704 610 629 222 4;
  • 21) 0.704 610 629 222 4 × 2 = 1 + 0.409 221 258 444 8;
  • 22) 0.409 221 258 444 8 × 2 = 0 + 0.818 442 516 889 6;
  • 23) 0.818 442 516 889 6 × 2 = 1 + 0.636 885 033 779 2;
  • 24) 0.636 885 033 779 2 × 2 = 1 + 0.273 770 067 558 4;
  • 25) 0.273 770 067 558 4 × 2 = 0 + 0.547 540 135 116 8;
  • 26) 0.547 540 135 116 8 × 2 = 1 + 0.095 080 270 233 6;
  • 27) 0.095 080 270 233 6 × 2 = 0 + 0.190 160 540 467 2;
  • 28) 0.190 160 540 467 2 × 2 = 0 + 0.380 321 080 934 4;
  • 29) 0.380 321 080 934 4 × 2 = 0 + 0.760 642 161 868 8;
  • 30) 0.760 642 161 868 8 × 2 = 1 + 0.521 284 323 737 6;
  • 31) 0.521 284 323 737 6 × 2 = 1 + 0.042 568 647 475 2;
  • 32) 0.042 568 647 475 2 × 2 = 0 + 0.085 137 294 950 4;
  • 33) 0.085 137 294 950 4 × 2 = 0 + 0.170 274 589 900 8;
  • 34) 0.170 274 589 900 8 × 2 = 0 + 0.340 549 179 801 6;
  • 35) 0.340 549 179 801 6 × 2 = 0 + 0.681 098 359 603 2;
  • 36) 0.681 098 359 603 2 × 2 = 1 + 0.362 196 719 206 4;
  • 37) 0.362 196 719 206 4 × 2 = 0 + 0.724 393 438 412 8;
  • 38) 0.724 393 438 412 8 × 2 = 1 + 0.448 786 876 825 6;
  • 39) 0.448 786 876 825 6 × 2 = 0 + 0.897 573 753 651 2;
  • 40) 0.897 573 753 651 2 × 2 = 1 + 0.795 147 507 302 4;
  • 41) 0.795 147 507 302 4 × 2 = 1 + 0.590 295 014 604 8;
  • 42) 0.590 295 014 604 8 × 2 = 1 + 0.180 590 029 209 6;
  • 43) 0.180 590 029 209 6 × 2 = 0 + 0.361 180 058 419 2;
  • 44) 0.361 180 058 419 2 × 2 = 0 + 0.722 360 116 838 4;
  • 45) 0.722 360 116 838 4 × 2 = 1 + 0.444 720 233 676 8;
  • 46) 0.444 720 233 676 8 × 2 = 0 + 0.889 440 467 353 6;
  • 47) 0.889 440 467 353 6 × 2 = 1 + 0.778 880 934 707 2;
  • 48) 0.778 880 934 707 2 × 2 = 1 + 0.557 761 869 414 4;
  • 49) 0.557 761 869 414 4 × 2 = 1 + 0.115 523 738 828 8;
  • 50) 0.115 523 738 828 8 × 2 = 0 + 0.231 047 477 657 6;
  • 51) 0.231 047 477 657 6 × 2 = 0 + 0.462 094 955 315 2;
  • 52) 0.462 094 955 315 2 × 2 = 0 + 0.924 189 910 630 4;
  • 53) 0.924 189 910 630 4 × 2 = 1 + 0.848 379 821 260 8;
  • 54) 0.848 379 821 260 8 × 2 = 1 + 0.696 759 642 521 6;
  • 55) 0.696 759 642 521 6 × 2 = 1 + 0.393 519 285 043 2;
  • 56) 0.393 519 285 043 2 × 2 = 0 + 0.787 038 570 086 4;
  • 57) 0.787 038 570 086 4 × 2 = 1 + 0.574 077 140 172 8;
  • 58) 0.574 077 140 172 8 × 2 = 1 + 0.148 154 280 345 6;
  • 59) 0.148 154 280 345 6 × 2 = 0 + 0.296 308 560 691 2;
  • 60) 0.296 308 560 691 2 × 2 = 0 + 0.592 617 121 382 4;
  • 61) 0.592 617 121 382 4 × 2 = 1 + 0.185 234 242 764 8;
  • 62) 0.185 234 242 764 8 × 2 = 0 + 0.370 468 485 529 6;
  • 63) 0.370 468 485 529 6 × 2 = 0 + 0.740 936 971 059 2;
  • 64) 0.740 936 971 059 2 × 2 = 1 + 0.481 873 942 118 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:


0.000 282 005 892 4(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0001 0101 1100 1011 1000 1110 1100 1001(2)

6. Positive number before normalization:

0.000 282 005 892 4(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0001 0101 1100 1011 1000 1110 1100 1001(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:


0.000 282 005 892 4(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0001 0101 1100 1011 1000 1110 1100 1001(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0001 0101 1100 1011 1000 1110 1100 1001(2) × 20 =

1.0010 0111 1011 0100 0110 0001 0101 1100 1011 1000 1110 1100 1001(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 0101 1100 1011 1000 1110 1100 1001


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 0101 1100 1011 1000 1110 1100 1001 =

0010 0111 1011 0100 0110 0001 0101 1100 1011 1000 1110 1100 1001


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 0101 1100 1011 1000 1110 1100 1001


Decimal number -0.000 282 005 892 4 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0001 0101 1100 1011 1000 1110 1100 1001

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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