-0.000 282 005 906 5 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 906 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
-0.000 282 005 906 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:

|-0.000 282 005 906 5| = 0.000 282 005 906 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: 0.000 282 005 906 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) 0.000 282 005 906 5 × 2 = 0 + 0.000 564 011 813;
  • 2) 0.000 564 011 813 × 2 = 0 + 0.001 128 023 626;
  • 3) 0.001 128 023 626 × 2 = 0 + 0.002 256 047 252;
  • 4) 0.002 256 047 252 × 2 = 0 + 0.004 512 094 504;
  • 5) 0.004 512 094 504 × 2 = 0 + 0.009 024 189 008;
  • 6) 0.009 024 189 008 × 2 = 0 + 0.018 048 378 016;
  • 7) 0.018 048 378 016 × 2 = 0 + 0.036 096 756 032;
  • 8) 0.036 096 756 032 × 2 = 0 + 0.072 193 512 064;
  • 9) 0.072 193 512 064 × 2 = 0 + 0.144 387 024 128;
  • 10) 0.144 387 024 128 × 2 = 0 + 0.288 774 048 256;
  • 11) 0.288 774 048 256 × 2 = 0 + 0.577 548 096 512;
  • 12) 0.577 548 096 512 × 2 = 1 + 0.155 096 193 024;
  • 13) 0.155 096 193 024 × 2 = 0 + 0.310 192 386 048;
  • 14) 0.310 192 386 048 × 2 = 0 + 0.620 384 772 096;
  • 15) 0.620 384 772 096 × 2 = 1 + 0.240 769 544 192;
  • 16) 0.240 769 544 192 × 2 = 0 + 0.481 539 088 384;
  • 17) 0.481 539 088 384 × 2 = 0 + 0.963 078 176 768;
  • 18) 0.963 078 176 768 × 2 = 1 + 0.926 156 353 536;
  • 19) 0.926 156 353 536 × 2 = 1 + 0.852 312 707 072;
  • 20) 0.852 312 707 072 × 2 = 1 + 0.704 625 414 144;
  • 21) 0.704 625 414 144 × 2 = 1 + 0.409 250 828 288;
  • 22) 0.409 250 828 288 × 2 = 0 + 0.818 501 656 576;
  • 23) 0.818 501 656 576 × 2 = 1 + 0.637 003 313 152;
  • 24) 0.637 003 313 152 × 2 = 1 + 0.274 006 626 304;
  • 25) 0.274 006 626 304 × 2 = 0 + 0.548 013 252 608;
  • 26) 0.548 013 252 608 × 2 = 1 + 0.096 026 505 216;
  • 27) 0.096 026 505 216 × 2 = 0 + 0.192 053 010 432;
  • 28) 0.192 053 010 432 × 2 = 0 + 0.384 106 020 864;
  • 29) 0.384 106 020 864 × 2 = 0 + 0.768 212 041 728;
  • 30) 0.768 212 041 728 × 2 = 1 + 0.536 424 083 456;
  • 31) 0.536 424 083 456 × 2 = 1 + 0.072 848 166 912;
  • 32) 0.072 848 166 912 × 2 = 0 + 0.145 696 333 824;
  • 33) 0.145 696 333 824 × 2 = 0 + 0.291 392 667 648;
  • 34) 0.291 392 667 648 × 2 = 0 + 0.582 785 335 296;
  • 35) 0.582 785 335 296 × 2 = 1 + 0.165 570 670 592;
  • 36) 0.165 570 670 592 × 2 = 0 + 0.331 141 341 184;
  • 37) 0.331 141 341 184 × 2 = 0 + 0.662 282 682 368;
  • 38) 0.662 282 682 368 × 2 = 1 + 0.324 565 364 736;
  • 39) 0.324 565 364 736 × 2 = 0 + 0.649 130 729 472;
  • 40) 0.649 130 729 472 × 2 = 1 + 0.298 261 458 944;
  • 41) 0.298 261 458 944 × 2 = 0 + 0.596 522 917 888;
  • 42) 0.596 522 917 888 × 2 = 1 + 0.193 045 835 776;
  • 43) 0.193 045 835 776 × 2 = 0 + 0.386 091 671 552;
  • 44) 0.386 091 671 552 × 2 = 0 + 0.772 183 343 104;
  • 45) 0.772 183 343 104 × 2 = 1 + 0.544 366 686 208;
  • 46) 0.544 366 686 208 × 2 = 1 + 0.088 733 372 416;
  • 47) 0.088 733 372 416 × 2 = 0 + 0.177 466 744 832;
  • 48) 0.177 466 744 832 × 2 = 0 + 0.354 933 489 664;
  • 49) 0.354 933 489 664 × 2 = 0 + 0.709 866 979 328;
  • 50) 0.709 866 979 328 × 2 = 1 + 0.419 733 958 656;
  • 51) 0.419 733 958 656 × 2 = 0 + 0.839 467 917 312;
  • 52) 0.839 467 917 312 × 2 = 1 + 0.678 935 834 624;
  • 53) 0.678 935 834 624 × 2 = 1 + 0.357 871 669 248;
  • 54) 0.357 871 669 248 × 2 = 0 + 0.715 743 338 496;
  • 55) 0.715 743 338 496 × 2 = 1 + 0.431 486 676 992;
  • 56) 0.431 486 676 992 × 2 = 0 + 0.862 973 353 984;
  • 57) 0.862 973 353 984 × 2 = 1 + 0.725 946 707 968;
  • 58) 0.725 946 707 968 × 2 = 1 + 0.451 893 415 936;
  • 59) 0.451 893 415 936 × 2 = 0 + 0.903 786 831 872;
  • 60) 0.903 786 831 872 × 2 = 1 + 0.807 573 663 744;
  • 61) 0.807 573 663 744 × 2 = 1 + 0.615 147 327 488;
  • 62) 0.615 147 327 488 × 2 = 1 + 0.230 294 654 976;
  • 63) 0.230 294 654 976 × 2 = 0 + 0.460 589 309 952;
  • 64) 0.460 589 309 952 × 2 = 0 + 0.921 178 619 904;

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:


0.000 282 005 906 5(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 0101 0100 1100 0101 1010 1101 1100(2)

6. Positive number before normalization:

0.000 282 005 906 5(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 0101 0100 1100 0101 1010 1101 1100(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:


0.000 282 005 906 5(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 0101 0100 1100 0101 1010 1101 1100(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 0101 0100 1100 0101 1010 1101 1100(2) × 20 =

1.0010 0111 1011 0100 0110 0010 0101 0100 1100 0101 1010 1101 1100(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 0010 0101 0100 1100 0101 1010 1101 1100


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 0010 0101 0100 1100 0101 1010 1101 1100 =

0010 0111 1011 0100 0110 0010 0101 0100 1100 0101 1010 1101 1100


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 0010 0101 0100 1100 0101 1010 1101 1100


Decimal number -0.000 282 005 906 5 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 0101 0100 1100 0101 1010 1101 1100

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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