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

Convert decimal -0.000 282 005 906(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(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| = 0.000 282 005 906


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.

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 × 2 = 0 + 0.000 564 011 812;
  • 2) 0.000 564 011 812 × 2 = 0 + 0.001 128 023 624;
  • 3) 0.001 128 023 624 × 2 = 0 + 0.002 256 047 248;
  • 4) 0.002 256 047 248 × 2 = 0 + 0.004 512 094 496;
  • 5) 0.004 512 094 496 × 2 = 0 + 0.009 024 188 992;
  • 6) 0.009 024 188 992 × 2 = 0 + 0.018 048 377 984;
  • 7) 0.018 048 377 984 × 2 = 0 + 0.036 096 755 968;
  • 8) 0.036 096 755 968 × 2 = 0 + 0.072 193 511 936;
  • 9) 0.072 193 511 936 × 2 = 0 + 0.144 387 023 872;
  • 10) 0.144 387 023 872 × 2 = 0 + 0.288 774 047 744;
  • 11) 0.288 774 047 744 × 2 = 0 + 0.577 548 095 488;
  • 12) 0.577 548 095 488 × 2 = 1 + 0.155 096 190 976;
  • 13) 0.155 096 190 976 × 2 = 0 + 0.310 192 381 952;
  • 14) 0.310 192 381 952 × 2 = 0 + 0.620 384 763 904;
  • 15) 0.620 384 763 904 × 2 = 1 + 0.240 769 527 808;
  • 16) 0.240 769 527 808 × 2 = 0 + 0.481 539 055 616;
  • 17) 0.481 539 055 616 × 2 = 0 + 0.963 078 111 232;
  • 18) 0.963 078 111 232 × 2 = 1 + 0.926 156 222 464;
  • 19) 0.926 156 222 464 × 2 = 1 + 0.852 312 444 928;
  • 20) 0.852 312 444 928 × 2 = 1 + 0.704 624 889 856;
  • 21) 0.704 624 889 856 × 2 = 1 + 0.409 249 779 712;
  • 22) 0.409 249 779 712 × 2 = 0 + 0.818 499 559 424;
  • 23) 0.818 499 559 424 × 2 = 1 + 0.636 999 118 848;
  • 24) 0.636 999 118 848 × 2 = 1 + 0.273 998 237 696;
  • 25) 0.273 998 237 696 × 2 = 0 + 0.547 996 475 392;
  • 26) 0.547 996 475 392 × 2 = 1 + 0.095 992 950 784;
  • 27) 0.095 992 950 784 × 2 = 0 + 0.191 985 901 568;
  • 28) 0.191 985 901 568 × 2 = 0 + 0.383 971 803 136;
  • 29) 0.383 971 803 136 × 2 = 0 + 0.767 943 606 272;
  • 30) 0.767 943 606 272 × 2 = 1 + 0.535 887 212 544;
  • 31) 0.535 887 212 544 × 2 = 1 + 0.071 774 425 088;
  • 32) 0.071 774 425 088 × 2 = 0 + 0.143 548 850 176;
  • 33) 0.143 548 850 176 × 2 = 0 + 0.287 097 700 352;
  • 34) 0.287 097 700 352 × 2 = 0 + 0.574 195 400 704;
  • 35) 0.574 195 400 704 × 2 = 1 + 0.148 390 801 408;
  • 36) 0.148 390 801 408 × 2 = 0 + 0.296 781 602 816;
  • 37) 0.296 781 602 816 × 2 = 0 + 0.593 563 205 632;
  • 38) 0.593 563 205 632 × 2 = 1 + 0.187 126 411 264;
  • 39) 0.187 126 411 264 × 2 = 0 + 0.374 252 822 528;
  • 40) 0.374 252 822 528 × 2 = 0 + 0.748 505 645 056;
  • 41) 0.748 505 645 056 × 2 = 1 + 0.497 011 290 112;
  • 42) 0.497 011 290 112 × 2 = 0 + 0.994 022 580 224;
  • 43) 0.994 022 580 224 × 2 = 1 + 0.988 045 160 448;
  • 44) 0.988 045 160 448 × 2 = 1 + 0.976 090 320 896;
  • 45) 0.976 090 320 896 × 2 = 1 + 0.952 180 641 792;
  • 46) 0.952 180 641 792 × 2 = 1 + 0.904 361 283 584;
  • 47) 0.904 361 283 584 × 2 = 1 + 0.808 722 567 168;
  • 48) 0.808 722 567 168 × 2 = 1 + 0.617 445 134 336;
  • 49) 0.617 445 134 336 × 2 = 1 + 0.234 890 268 672;
  • 50) 0.234 890 268 672 × 2 = 0 + 0.469 780 537 344;
  • 51) 0.469 780 537 344 × 2 = 0 + 0.939 561 074 688;
  • 52) 0.939 561 074 688 × 2 = 1 + 0.879 122 149 376;
  • 53) 0.879 122 149 376 × 2 = 1 + 0.758 244 298 752;
  • 54) 0.758 244 298 752 × 2 = 1 + 0.516 488 597 504;
  • 55) 0.516 488 597 504 × 2 = 1 + 0.032 977 195 008;
  • 56) 0.032 977 195 008 × 2 = 0 + 0.065 954 390 016;
  • 57) 0.065 954 390 016 × 2 = 0 + 0.131 908 780 032;
  • 58) 0.131 908 780 032 × 2 = 0 + 0.263 817 560 064;
  • 59) 0.263 817 560 064 × 2 = 0 + 0.527 635 120 128;
  • 60) 0.527 635 120 128 × 2 = 1 + 0.055 270 240 256;
  • 61) 0.055 270 240 256 × 2 = 0 + 0.110 540 480 512;
  • 62) 0.110 540 480 512 × 2 = 0 + 0.221 080 961 024;
  • 63) 0.221 080 961 024 × 2 = 0 + 0.442 161 922 048;
  • 64) 0.442 161 922 048 × 2 = 0 + 0.884 323 844 096;

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(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 0100 1011 1111 1001 1110 0001 0000(2)

6. Positive number before normalization:

0.000 282 005 906(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 0100 1011 1111 1001 1110 0001 0000(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(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 0100 1011 1111 1001 1110 0001 0000(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 0100 1011 1111 1001 1110 0001 0000(2) × 20 =

1.0010 0111 1011 0100 0110 0010 0100 1011 1111 1001 1110 0001 0000(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 0100 1011 1111 1001 1110 0001 0000


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 0100 1011 1111 1001 1110 0001 0000 =

0010 0111 1011 0100 0110 0010 0100 1011 1111 1001 1110 0001 0000


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 0100 1011 1111 1001 1110 0001 0000


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

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 0100 1011 1111 1001 1110 0001 0000

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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