-0.000 281 986 5 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 281 986 5(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
-0.000 281 986 5(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:

|-0.000 281 986 5| = 0.000 281 986 5


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: 0.000 281 986 5.

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) 0.000 281 986 5 × 2 = 0 + 0.000 563 973;
  • 2) 0.000 563 973 × 2 = 0 + 0.001 127 946;
  • 3) 0.001 127 946 × 2 = 0 + 0.002 255 892;
  • 4) 0.002 255 892 × 2 = 0 + 0.004 511 784;
  • 5) 0.004 511 784 × 2 = 0 + 0.009 023 568;
  • 6) 0.009 023 568 × 2 = 0 + 0.018 047 136;
  • 7) 0.018 047 136 × 2 = 0 + 0.036 094 272;
  • 8) 0.036 094 272 × 2 = 0 + 0.072 188 544;
  • 9) 0.072 188 544 × 2 = 0 + 0.144 377 088;
  • 10) 0.144 377 088 × 2 = 0 + 0.288 754 176;
  • 11) 0.288 754 176 × 2 = 0 + 0.577 508 352;
  • 12) 0.577 508 352 × 2 = 1 + 0.155 016 704;
  • 13) 0.155 016 704 × 2 = 0 + 0.310 033 408;
  • 14) 0.310 033 408 × 2 = 0 + 0.620 066 816;
  • 15) 0.620 066 816 × 2 = 1 + 0.240 133 632;
  • 16) 0.240 133 632 × 2 = 0 + 0.480 267 264;
  • 17) 0.480 267 264 × 2 = 0 + 0.960 534 528;
  • 18) 0.960 534 528 × 2 = 1 + 0.921 069 056;
  • 19) 0.921 069 056 × 2 = 1 + 0.842 138 112;
  • 20) 0.842 138 112 × 2 = 1 + 0.684 276 224;
  • 21) 0.684 276 224 × 2 = 1 + 0.368 552 448;
  • 22) 0.368 552 448 × 2 = 0 + 0.737 104 896;
  • 23) 0.737 104 896 × 2 = 1 + 0.474 209 792;
  • 24) 0.474 209 792 × 2 = 0 + 0.948 419 584;
  • 25) 0.948 419 584 × 2 = 1 + 0.896 839 168;
  • 26) 0.896 839 168 × 2 = 1 + 0.793 678 336;
  • 27) 0.793 678 336 × 2 = 1 + 0.587 356 672;
  • 28) 0.587 356 672 × 2 = 1 + 0.174 713 344;
  • 29) 0.174 713 344 × 2 = 0 + 0.349 426 688;
  • 30) 0.349 426 688 × 2 = 0 + 0.698 853 376;
  • 31) 0.698 853 376 × 2 = 1 + 0.397 706 752;
  • 32) 0.397 706 752 × 2 = 0 + 0.795 413 504;
  • 33) 0.795 413 504 × 2 = 1 + 0.590 827 008;
  • 34) 0.590 827 008 × 2 = 1 + 0.181 654 016;
  • 35) 0.181 654 016 × 2 = 0 + 0.363 308 032;
  • 36) 0.363 308 032 × 2 = 0 + 0.726 616 064;
  • 37) 0.726 616 064 × 2 = 1 + 0.453 232 128;
  • 38) 0.453 232 128 × 2 = 0 + 0.906 464 256;
  • 39) 0.906 464 256 × 2 = 1 + 0.812 928 512;
  • 40) 0.812 928 512 × 2 = 1 + 0.625 857 024;
  • 41) 0.625 857 024 × 2 = 1 + 0.251 714 048;
  • 42) 0.251 714 048 × 2 = 0 + 0.503 428 096;
  • 43) 0.503 428 096 × 2 = 1 + 0.006 856 192;
  • 44) 0.006 856 192 × 2 = 0 + 0.013 712 384;
  • 45) 0.013 712 384 × 2 = 0 + 0.027 424 768;
  • 46) 0.027 424 768 × 2 = 0 + 0.054 849 536;
  • 47) 0.054 849 536 × 2 = 0 + 0.109 699 072;
  • 48) 0.109 699 072 × 2 = 0 + 0.219 398 144;
  • 49) 0.219 398 144 × 2 = 0 + 0.438 796 288;
  • 50) 0.438 796 288 × 2 = 0 + 0.877 592 576;
  • 51) 0.877 592 576 × 2 = 1 + 0.755 185 152;
  • 52) 0.755 185 152 × 2 = 1 + 0.510 370 304;
  • 53) 0.510 370 304 × 2 = 1 + 0.020 740 608;
  • 54) 0.020 740 608 × 2 = 0 + 0.041 481 216;
  • 55) 0.041 481 216 × 2 = 0 + 0.082 962 432;
  • 56) 0.082 962 432 × 2 = 0 + 0.165 924 864;
  • 57) 0.165 924 864 × 2 = 0 + 0.331 849 728;
  • 58) 0.331 849 728 × 2 = 0 + 0.663 699 456;
  • 59) 0.663 699 456 × 2 = 1 + 0.327 398 912;
  • 60) 0.327 398 912 × 2 = 0 + 0.654 797 824;
  • 61) 0.654 797 824 × 2 = 1 + 0.309 595 648;
  • 62) 0.309 595 648 × 2 = 0 + 0.619 191 296;
  • 63) 0.619 191 296 × 2 = 1 + 0.238 382 592;
  • 64) 0.238 382 592 × 2 = 0 + 0.476 765 184;

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:


0.000 281 986 5(10) =

0.0000 0000 0001 0010 0111 1010 1111 0010 1100 1011 1010 0000 0011 1000 0010 1010(2)

6. Positive number before normalization:

0.000 281 986 5(10) =

0.0000 0000 0001 0010 0111 1010 1111 0010 1100 1011 1010 0000 0011 1000 0010 1010(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:


0.000 281 986 5(10) =

0.0000 0000 0001 0010 0111 1010 1111 0010 1100 1011 1010 0000 0011 1000 0010 1010(2) =

0.0000 0000 0001 0010 0111 1010 1111 0010 1100 1011 1010 0000 0011 1000 0010 1010(2) × 20 =

1.0010 0111 1010 1111 0010 1100 1011 1010 0000 0011 1000 0010 1010(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 1010 1111 0010 1100 1011 1010 0000 0011 1000 0010 1010


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 1010 1111 0010 1100 1011 1010 0000 0011 1000 0010 1010 =

0010 0111 1010 1111 0010 1100 1011 1010 0000 0011 1000 0010 1010


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 1010 1111 0010 1100 1011 1010 0000 0011 1000 0010 1010


Decimal number -0.000 281 986 5 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1010 1111 0010 1100 1011 1010 0000 0011 1000 0010 1010

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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