-0.000 282 011 59 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 011 59(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
-0.000 282 011 59(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:

|-0.000 282 011 59| = 0.000 282 011 59


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: 0.000 282 011 59.

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) 0.000 282 011 59 × 2 = 0 + 0.000 564 023 18;
  • 2) 0.000 564 023 18 × 2 = 0 + 0.001 128 046 36;
  • 3) 0.001 128 046 36 × 2 = 0 + 0.002 256 092 72;
  • 4) 0.002 256 092 72 × 2 = 0 + 0.004 512 185 44;
  • 5) 0.004 512 185 44 × 2 = 0 + 0.009 024 370 88;
  • 6) 0.009 024 370 88 × 2 = 0 + 0.018 048 741 76;
  • 7) 0.018 048 741 76 × 2 = 0 + 0.036 097 483 52;
  • 8) 0.036 097 483 52 × 2 = 0 + 0.072 194 967 04;
  • 9) 0.072 194 967 04 × 2 = 0 + 0.144 389 934 08;
  • 10) 0.144 389 934 08 × 2 = 0 + 0.288 779 868 16;
  • 11) 0.288 779 868 16 × 2 = 0 + 0.577 559 736 32;
  • 12) 0.577 559 736 32 × 2 = 1 + 0.155 119 472 64;
  • 13) 0.155 119 472 64 × 2 = 0 + 0.310 238 945 28;
  • 14) 0.310 238 945 28 × 2 = 0 + 0.620 477 890 56;
  • 15) 0.620 477 890 56 × 2 = 1 + 0.240 955 781 12;
  • 16) 0.240 955 781 12 × 2 = 0 + 0.481 911 562 24;
  • 17) 0.481 911 562 24 × 2 = 0 + 0.963 823 124 48;
  • 18) 0.963 823 124 48 × 2 = 1 + 0.927 646 248 96;
  • 19) 0.927 646 248 96 × 2 = 1 + 0.855 292 497 92;
  • 20) 0.855 292 497 92 × 2 = 1 + 0.710 584 995 84;
  • 21) 0.710 584 995 84 × 2 = 1 + 0.421 169 991 68;
  • 22) 0.421 169 991 68 × 2 = 0 + 0.842 339 983 36;
  • 23) 0.842 339 983 36 × 2 = 1 + 0.684 679 966 72;
  • 24) 0.684 679 966 72 × 2 = 1 + 0.369 359 933 44;
  • 25) 0.369 359 933 44 × 2 = 0 + 0.738 719 866 88;
  • 26) 0.738 719 866 88 × 2 = 1 + 0.477 439 733 76;
  • 27) 0.477 439 733 76 × 2 = 0 + 0.954 879 467 52;
  • 28) 0.954 879 467 52 × 2 = 1 + 0.909 758 935 04;
  • 29) 0.909 758 935 04 × 2 = 1 + 0.819 517 870 08;
  • 30) 0.819 517 870 08 × 2 = 1 + 0.639 035 740 16;
  • 31) 0.639 035 740 16 × 2 = 1 + 0.278 071 480 32;
  • 32) 0.278 071 480 32 × 2 = 0 + 0.556 142 960 64;
  • 33) 0.556 142 960 64 × 2 = 1 + 0.112 285 921 28;
  • 34) 0.112 285 921 28 × 2 = 0 + 0.224 571 842 56;
  • 35) 0.224 571 842 56 × 2 = 0 + 0.449 143 685 12;
  • 36) 0.449 143 685 12 × 2 = 0 + 0.898 287 370 24;
  • 37) 0.898 287 370 24 × 2 = 1 + 0.796 574 740 48;
  • 38) 0.796 574 740 48 × 2 = 1 + 0.593 149 480 96;
  • 39) 0.593 149 480 96 × 2 = 1 + 0.186 298 961 92;
  • 40) 0.186 298 961 92 × 2 = 0 + 0.372 597 923 84;
  • 41) 0.372 597 923 84 × 2 = 0 + 0.745 195 847 68;
  • 42) 0.745 195 847 68 × 2 = 1 + 0.490 391 695 36;
  • 43) 0.490 391 695 36 × 2 = 0 + 0.980 783 390 72;
  • 44) 0.980 783 390 72 × 2 = 1 + 0.961 566 781 44;
  • 45) 0.961 566 781 44 × 2 = 1 + 0.923 133 562 88;
  • 46) 0.923 133 562 88 × 2 = 1 + 0.846 267 125 76;
  • 47) 0.846 267 125 76 × 2 = 1 + 0.692 534 251 52;
  • 48) 0.692 534 251 52 × 2 = 1 + 0.385 068 503 04;
  • 49) 0.385 068 503 04 × 2 = 0 + 0.770 137 006 08;
  • 50) 0.770 137 006 08 × 2 = 1 + 0.540 274 012 16;
  • 51) 0.540 274 012 16 × 2 = 1 + 0.080 548 024 32;
  • 52) 0.080 548 024 32 × 2 = 0 + 0.161 096 048 64;
  • 53) 0.161 096 048 64 × 2 = 0 + 0.322 192 097 28;
  • 54) 0.322 192 097 28 × 2 = 0 + 0.644 384 194 56;
  • 55) 0.644 384 194 56 × 2 = 1 + 0.288 768 389 12;
  • 56) 0.288 768 389 12 × 2 = 0 + 0.577 536 778 24;
  • 57) 0.577 536 778 24 × 2 = 1 + 0.155 073 556 48;
  • 58) 0.155 073 556 48 × 2 = 0 + 0.310 147 112 96;
  • 59) 0.310 147 112 96 × 2 = 0 + 0.620 294 225 92;
  • 60) 0.620 294 225 92 × 2 = 1 + 0.240 588 451 84;
  • 61) 0.240 588 451 84 × 2 = 0 + 0.481 176 903 68;
  • 62) 0.481 176 903 68 × 2 = 0 + 0.962 353 807 36;
  • 63) 0.962 353 807 36 × 2 = 1 + 0.924 707 614 72;
  • 64) 0.924 707 614 72 × 2 = 1 + 0.849 415 229 44;

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:


0.000 282 011 59(10) =

0.0000 0000 0001 0010 0111 1011 0101 1110 1000 1110 0101 1111 0110 0010 1001 0011(2)

6. Positive number before normalization:

0.000 282 011 59(10) =

0.0000 0000 0001 0010 0111 1011 0101 1110 1000 1110 0101 1111 0110 0010 1001 0011(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:


0.000 282 011 59(10) =

0.0000 0000 0001 0010 0111 1011 0101 1110 1000 1110 0101 1111 0110 0010 1001 0011(2) =

0.0000 0000 0001 0010 0111 1011 0101 1110 1000 1110 0101 1111 0110 0010 1001 0011(2) × 20 =

1.0010 0111 1011 0101 1110 1000 1110 0101 1111 0110 0010 1001 0011(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 0101 1110 1000 1110 0101 1111 0110 0010 1001 0011


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 0101 1110 1000 1110 0101 1111 0110 0010 1001 0011 =

0010 0111 1011 0101 1110 1000 1110 0101 1111 0110 0010 1001 0011


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 0101 1110 1000 1110 0101 1111 0110 0010 1001 0011


Decimal number -0.000 282 011 59 converted to 64 bit double precision IEEE 754 binary floating point representation:

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

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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