-0.000 282 005 687 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 687(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
-0.000 282 005 687(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:

|-0.000 282 005 687| = 0.000 282 005 687


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: 0.000 282 005 687.

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) 0.000 282 005 687 × 2 = 0 + 0.000 564 011 374;
  • 2) 0.000 564 011 374 × 2 = 0 + 0.001 128 022 748;
  • 3) 0.001 128 022 748 × 2 = 0 + 0.002 256 045 496;
  • 4) 0.002 256 045 496 × 2 = 0 + 0.004 512 090 992;
  • 5) 0.004 512 090 992 × 2 = 0 + 0.009 024 181 984;
  • 6) 0.009 024 181 984 × 2 = 0 + 0.018 048 363 968;
  • 7) 0.018 048 363 968 × 2 = 0 + 0.036 096 727 936;
  • 8) 0.036 096 727 936 × 2 = 0 + 0.072 193 455 872;
  • 9) 0.072 193 455 872 × 2 = 0 + 0.144 386 911 744;
  • 10) 0.144 386 911 744 × 2 = 0 + 0.288 773 823 488;
  • 11) 0.288 773 823 488 × 2 = 0 + 0.577 547 646 976;
  • 12) 0.577 547 646 976 × 2 = 1 + 0.155 095 293 952;
  • 13) 0.155 095 293 952 × 2 = 0 + 0.310 190 587 904;
  • 14) 0.310 190 587 904 × 2 = 0 + 0.620 381 175 808;
  • 15) 0.620 381 175 808 × 2 = 1 + 0.240 762 351 616;
  • 16) 0.240 762 351 616 × 2 = 0 + 0.481 524 703 232;
  • 17) 0.481 524 703 232 × 2 = 0 + 0.963 049 406 464;
  • 18) 0.963 049 406 464 × 2 = 1 + 0.926 098 812 928;
  • 19) 0.926 098 812 928 × 2 = 1 + 0.852 197 625 856;
  • 20) 0.852 197 625 856 × 2 = 1 + 0.704 395 251 712;
  • 21) 0.704 395 251 712 × 2 = 1 + 0.408 790 503 424;
  • 22) 0.408 790 503 424 × 2 = 0 + 0.817 581 006 848;
  • 23) 0.817 581 006 848 × 2 = 1 + 0.635 162 013 696;
  • 24) 0.635 162 013 696 × 2 = 1 + 0.270 324 027 392;
  • 25) 0.270 324 027 392 × 2 = 0 + 0.540 648 054 784;
  • 26) 0.540 648 054 784 × 2 = 1 + 0.081 296 109 568;
  • 27) 0.081 296 109 568 × 2 = 0 + 0.162 592 219 136;
  • 28) 0.162 592 219 136 × 2 = 0 + 0.325 184 438 272;
  • 29) 0.325 184 438 272 × 2 = 0 + 0.650 368 876 544;
  • 30) 0.650 368 876 544 × 2 = 1 + 0.300 737 753 088;
  • 31) 0.300 737 753 088 × 2 = 0 + 0.601 475 506 176;
  • 32) 0.601 475 506 176 × 2 = 1 + 0.202 951 012 352;
  • 33) 0.202 951 012 352 × 2 = 0 + 0.405 902 024 704;
  • 34) 0.405 902 024 704 × 2 = 0 + 0.811 804 049 408;
  • 35) 0.811 804 049 408 × 2 = 1 + 0.623 608 098 816;
  • 36) 0.623 608 098 816 × 2 = 1 + 0.247 216 197 632;
  • 37) 0.247 216 197 632 × 2 = 0 + 0.494 432 395 264;
  • 38) 0.494 432 395 264 × 2 = 0 + 0.988 864 790 528;
  • 39) 0.988 864 790 528 × 2 = 1 + 0.977 729 581 056;
  • 40) 0.977 729 581 056 × 2 = 1 + 0.955 459 162 112;
  • 41) 0.955 459 162 112 × 2 = 1 + 0.910 918 324 224;
  • 42) 0.910 918 324 224 × 2 = 1 + 0.821 836 648 448;
  • 43) 0.821 836 648 448 × 2 = 1 + 0.643 673 296 896;
  • 44) 0.643 673 296 896 × 2 = 1 + 0.287 346 593 792;
  • 45) 0.287 346 593 792 × 2 = 0 + 0.574 693 187 584;
  • 46) 0.574 693 187 584 × 2 = 1 + 0.149 386 375 168;
  • 47) 0.149 386 375 168 × 2 = 0 + 0.298 772 750 336;
  • 48) 0.298 772 750 336 × 2 = 0 + 0.597 545 500 672;
  • 49) 0.597 545 500 672 × 2 = 1 + 0.195 091 001 344;
  • 50) 0.195 091 001 344 × 2 = 0 + 0.390 182 002 688;
  • 51) 0.390 182 002 688 × 2 = 0 + 0.780 364 005 376;
  • 52) 0.780 364 005 376 × 2 = 1 + 0.560 728 010 752;
  • 53) 0.560 728 010 752 × 2 = 1 + 0.121 456 021 504;
  • 54) 0.121 456 021 504 × 2 = 0 + 0.242 912 043 008;
  • 55) 0.242 912 043 008 × 2 = 0 + 0.485 824 086 016;
  • 56) 0.485 824 086 016 × 2 = 0 + 0.971 648 172 032;
  • 57) 0.971 648 172 032 × 2 = 1 + 0.943 296 344 064;
  • 58) 0.943 296 344 064 × 2 = 1 + 0.886 592 688 128;
  • 59) 0.886 592 688 128 × 2 = 1 + 0.773 185 376 256;
  • 60) 0.773 185 376 256 × 2 = 1 + 0.546 370 752 512;
  • 61) 0.546 370 752 512 × 2 = 1 + 0.092 741 505 024;
  • 62) 0.092 741 505 024 × 2 = 0 + 0.185 483 010 048;
  • 63) 0.185 483 010 048 × 2 = 0 + 0.370 966 020 096;
  • 64) 0.370 966 020 096 × 2 = 0 + 0.741 932 040 192;

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:


0.000 282 005 687(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 0011 0011 1111 0100 1001 1000 1111 1000(2)

6. Positive number before normalization:

0.000 282 005 687(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 0011 0011 1111 0100 1001 1000 1111 1000(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:


0.000 282 005 687(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 0011 0011 1111 0100 1001 1000 1111 1000(2) =

0.0000 0000 0001 0010 0111 1011 0100 0101 0011 0011 1111 0100 1001 1000 1111 1000(2) × 20 =

1.0010 0111 1011 0100 0101 0011 0011 1111 0100 1001 1000 1111 1000(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 0101 0011 0011 1111 0100 1001 1000 1111 1000


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 0101 0011 0011 1111 0100 1001 1000 1111 1000 =

0010 0111 1011 0100 0101 0011 0011 1111 0100 1001 1000 1111 1000


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 0101 0011 0011 1111 0100 1001 1000 1111 1000


Decimal number -0.000 282 005 687 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0101 0011 0011 1111 0100 1001 1000 1111 1000

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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