-0.000 282 621 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 621(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
-0.000 282 621(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:

|-0.000 282 621| = 0.000 282 621


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: 0.000 282 621.

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) 0.000 282 621 × 2 = 0 + 0.000 565 242;
  • 2) 0.000 565 242 × 2 = 0 + 0.001 130 484;
  • 3) 0.001 130 484 × 2 = 0 + 0.002 260 968;
  • 4) 0.002 260 968 × 2 = 0 + 0.004 521 936;
  • 5) 0.004 521 936 × 2 = 0 + 0.009 043 872;
  • 6) 0.009 043 872 × 2 = 0 + 0.018 087 744;
  • 7) 0.018 087 744 × 2 = 0 + 0.036 175 488;
  • 8) 0.036 175 488 × 2 = 0 + 0.072 350 976;
  • 9) 0.072 350 976 × 2 = 0 + 0.144 701 952;
  • 10) 0.144 701 952 × 2 = 0 + 0.289 403 904;
  • 11) 0.289 403 904 × 2 = 0 + 0.578 807 808;
  • 12) 0.578 807 808 × 2 = 1 + 0.157 615 616;
  • 13) 0.157 615 616 × 2 = 0 + 0.315 231 232;
  • 14) 0.315 231 232 × 2 = 0 + 0.630 462 464;
  • 15) 0.630 462 464 × 2 = 1 + 0.260 924 928;
  • 16) 0.260 924 928 × 2 = 0 + 0.521 849 856;
  • 17) 0.521 849 856 × 2 = 1 + 0.043 699 712;
  • 18) 0.043 699 712 × 2 = 0 + 0.087 399 424;
  • 19) 0.087 399 424 × 2 = 0 + 0.174 798 848;
  • 20) 0.174 798 848 × 2 = 0 + 0.349 597 696;
  • 21) 0.349 597 696 × 2 = 0 + 0.699 195 392;
  • 22) 0.699 195 392 × 2 = 1 + 0.398 390 784;
  • 23) 0.398 390 784 × 2 = 0 + 0.796 781 568;
  • 24) 0.796 781 568 × 2 = 1 + 0.593 563 136;
  • 25) 0.593 563 136 × 2 = 1 + 0.187 126 272;
  • 26) 0.187 126 272 × 2 = 0 + 0.374 252 544;
  • 27) 0.374 252 544 × 2 = 0 + 0.748 505 088;
  • 28) 0.748 505 088 × 2 = 1 + 0.497 010 176;
  • 29) 0.497 010 176 × 2 = 0 + 0.994 020 352;
  • 30) 0.994 020 352 × 2 = 1 + 0.988 040 704;
  • 31) 0.988 040 704 × 2 = 1 + 0.976 081 408;
  • 32) 0.976 081 408 × 2 = 1 + 0.952 162 816;
  • 33) 0.952 162 816 × 2 = 1 + 0.904 325 632;
  • 34) 0.904 325 632 × 2 = 1 + 0.808 651 264;
  • 35) 0.808 651 264 × 2 = 1 + 0.617 302 528;
  • 36) 0.617 302 528 × 2 = 1 + 0.234 605 056;
  • 37) 0.234 605 056 × 2 = 0 + 0.469 210 112;
  • 38) 0.469 210 112 × 2 = 0 + 0.938 420 224;
  • 39) 0.938 420 224 × 2 = 1 + 0.876 840 448;
  • 40) 0.876 840 448 × 2 = 1 + 0.753 680 896;
  • 41) 0.753 680 896 × 2 = 1 + 0.507 361 792;
  • 42) 0.507 361 792 × 2 = 1 + 0.014 723 584;
  • 43) 0.014 723 584 × 2 = 0 + 0.029 447 168;
  • 44) 0.029 447 168 × 2 = 0 + 0.058 894 336;
  • 45) 0.058 894 336 × 2 = 0 + 0.117 788 672;
  • 46) 0.117 788 672 × 2 = 0 + 0.235 577 344;
  • 47) 0.235 577 344 × 2 = 0 + 0.471 154 688;
  • 48) 0.471 154 688 × 2 = 0 + 0.942 309 376;
  • 49) 0.942 309 376 × 2 = 1 + 0.884 618 752;
  • 50) 0.884 618 752 × 2 = 1 + 0.769 237 504;
  • 51) 0.769 237 504 × 2 = 1 + 0.538 475 008;
  • 52) 0.538 475 008 × 2 = 1 + 0.076 950 016;
  • 53) 0.076 950 016 × 2 = 0 + 0.153 900 032;
  • 54) 0.153 900 032 × 2 = 0 + 0.307 800 064;
  • 55) 0.307 800 064 × 2 = 0 + 0.615 600 128;
  • 56) 0.615 600 128 × 2 = 1 + 0.231 200 256;
  • 57) 0.231 200 256 × 2 = 0 + 0.462 400 512;
  • 58) 0.462 400 512 × 2 = 0 + 0.924 801 024;
  • 59) 0.924 801 024 × 2 = 1 + 0.849 602 048;
  • 60) 0.849 602 048 × 2 = 1 + 0.699 204 096;
  • 61) 0.699 204 096 × 2 = 1 + 0.398 408 192;
  • 62) 0.398 408 192 × 2 = 0 + 0.796 816 384;
  • 63) 0.796 816 384 × 2 = 1 + 0.593 632 768;
  • 64) 0.593 632 768 × 2 = 1 + 0.187 265 536;

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:


0.000 282 621(10) =

0.0000 0000 0001 0010 1000 0101 1001 0111 1111 0011 1100 0000 1111 0001 0011 1011(2)

6. Positive number before normalization:

0.000 282 621(10) =

0.0000 0000 0001 0010 1000 0101 1001 0111 1111 0011 1100 0000 1111 0001 0011 1011(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:


0.000 282 621(10) =

0.0000 0000 0001 0010 1000 0101 1001 0111 1111 0011 1100 0000 1111 0001 0011 1011(2) =

0.0000 0000 0001 0010 1000 0101 1001 0111 1111 0011 1100 0000 1111 0001 0011 1011(2) × 20 =

1.0010 1000 0101 1001 0111 1111 0011 1100 0000 1111 0001 0011 1011(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 1000 0101 1001 0111 1111 0011 1100 0000 1111 0001 0011 1011


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 1000 0101 1001 0111 1111 0011 1100 0000 1111 0001 0011 1011 =

0010 1000 0101 1001 0111 1111 0011 1100 0000 1111 0001 0011 1011


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 1000 0101 1001 0111 1111 0011 1100 0000 1111 0001 0011 1011


Decimal number -0.000 282 621 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 1000 0101 1001 0111 1111 0011 1100 0000 1111 0001 0011 1011

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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