-0.000 281 999 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 281 999(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
-0.000 281 999(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:

|-0.000 281 999| = 0.000 281 999


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: 0.000 281 999.

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) 0.000 281 999 × 2 = 0 + 0.000 563 998;
  • 2) 0.000 563 998 × 2 = 0 + 0.001 127 996;
  • 3) 0.001 127 996 × 2 = 0 + 0.002 255 992;
  • 4) 0.002 255 992 × 2 = 0 + 0.004 511 984;
  • 5) 0.004 511 984 × 2 = 0 + 0.009 023 968;
  • 6) 0.009 023 968 × 2 = 0 + 0.018 047 936;
  • 7) 0.018 047 936 × 2 = 0 + 0.036 095 872;
  • 8) 0.036 095 872 × 2 = 0 + 0.072 191 744;
  • 9) 0.072 191 744 × 2 = 0 + 0.144 383 488;
  • 10) 0.144 383 488 × 2 = 0 + 0.288 766 976;
  • 11) 0.288 766 976 × 2 = 0 + 0.577 533 952;
  • 12) 0.577 533 952 × 2 = 1 + 0.155 067 904;
  • 13) 0.155 067 904 × 2 = 0 + 0.310 135 808;
  • 14) 0.310 135 808 × 2 = 0 + 0.620 271 616;
  • 15) 0.620 271 616 × 2 = 1 + 0.240 543 232;
  • 16) 0.240 543 232 × 2 = 0 + 0.481 086 464;
  • 17) 0.481 086 464 × 2 = 0 + 0.962 172 928;
  • 18) 0.962 172 928 × 2 = 1 + 0.924 345 856;
  • 19) 0.924 345 856 × 2 = 1 + 0.848 691 712;
  • 20) 0.848 691 712 × 2 = 1 + 0.697 383 424;
  • 21) 0.697 383 424 × 2 = 1 + 0.394 766 848;
  • 22) 0.394 766 848 × 2 = 0 + 0.789 533 696;
  • 23) 0.789 533 696 × 2 = 1 + 0.579 067 392;
  • 24) 0.579 067 392 × 2 = 1 + 0.158 134 784;
  • 25) 0.158 134 784 × 2 = 0 + 0.316 269 568;
  • 26) 0.316 269 568 × 2 = 0 + 0.632 539 136;
  • 27) 0.632 539 136 × 2 = 1 + 0.265 078 272;
  • 28) 0.265 078 272 × 2 = 0 + 0.530 156 544;
  • 29) 0.530 156 544 × 2 = 1 + 0.060 313 088;
  • 30) 0.060 313 088 × 2 = 0 + 0.120 626 176;
  • 31) 0.120 626 176 × 2 = 0 + 0.241 252 352;
  • 32) 0.241 252 352 × 2 = 0 + 0.482 504 704;
  • 33) 0.482 504 704 × 2 = 0 + 0.965 009 408;
  • 34) 0.965 009 408 × 2 = 1 + 0.930 018 816;
  • 35) 0.930 018 816 × 2 = 1 + 0.860 037 632;
  • 36) 0.860 037 632 × 2 = 1 + 0.720 075 264;
  • 37) 0.720 075 264 × 2 = 1 + 0.440 150 528;
  • 38) 0.440 150 528 × 2 = 0 + 0.880 301 056;
  • 39) 0.880 301 056 × 2 = 1 + 0.760 602 112;
  • 40) 0.760 602 112 × 2 = 1 + 0.521 204 224;
  • 41) 0.521 204 224 × 2 = 1 + 0.042 408 448;
  • 42) 0.042 408 448 × 2 = 0 + 0.084 816 896;
  • 43) 0.084 816 896 × 2 = 0 + 0.169 633 792;
  • 44) 0.169 633 792 × 2 = 0 + 0.339 267 584;
  • 45) 0.339 267 584 × 2 = 0 + 0.678 535 168;
  • 46) 0.678 535 168 × 2 = 1 + 0.357 070 336;
  • 47) 0.357 070 336 × 2 = 0 + 0.714 140 672;
  • 48) 0.714 140 672 × 2 = 1 + 0.428 281 344;
  • 49) 0.428 281 344 × 2 = 0 + 0.856 562 688;
  • 50) 0.856 562 688 × 2 = 1 + 0.713 125 376;
  • 51) 0.713 125 376 × 2 = 1 + 0.426 250 752;
  • 52) 0.426 250 752 × 2 = 0 + 0.852 501 504;
  • 53) 0.852 501 504 × 2 = 1 + 0.705 003 008;
  • 54) 0.705 003 008 × 2 = 1 + 0.410 006 016;
  • 55) 0.410 006 016 × 2 = 0 + 0.820 012 032;
  • 56) 0.820 012 032 × 2 = 1 + 0.640 024 064;
  • 57) 0.640 024 064 × 2 = 1 + 0.280 048 128;
  • 58) 0.280 048 128 × 2 = 0 + 0.560 096 256;
  • 59) 0.560 096 256 × 2 = 1 + 0.120 192 512;
  • 60) 0.120 192 512 × 2 = 0 + 0.240 385 024;
  • 61) 0.240 385 024 × 2 = 0 + 0.480 770 048;
  • 62) 0.480 770 048 × 2 = 0 + 0.961 540 096;
  • 63) 0.961 540 096 × 2 = 1 + 0.923 080 192;
  • 64) 0.923 080 192 × 2 = 1 + 0.846 160 384;

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:


0.000 281 999(10) =

0.0000 0000 0001 0010 0111 1011 0010 1000 0111 1011 1000 0101 0110 1101 1010 0011(2)

6. Positive number before normalization:

0.000 281 999(10) =

0.0000 0000 0001 0010 0111 1011 0010 1000 0111 1011 1000 0101 0110 1101 1010 0011(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:


0.000 281 999(10) =

0.0000 0000 0001 0010 0111 1011 0010 1000 0111 1011 1000 0101 0110 1101 1010 0011(2) =

0.0000 0000 0001 0010 0111 1011 0010 1000 0111 1011 1000 0101 0110 1101 1010 0011(2) × 20 =

1.0010 0111 1011 0010 1000 0111 1011 1000 0101 0110 1101 1010 0011(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 0010 1000 0111 1011 1000 0101 0110 1101 1010 0011


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 0010 1000 0111 1011 1000 0101 0110 1101 1010 0011 =

0010 0111 1011 0010 1000 0111 1011 1000 0101 0110 1101 1010 0011


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 0010 1000 0111 1011 1000 0101 0110 1101 1010 0011


Decimal number -0.000 281 999 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0010 1000 0111 1011 1000 0101 0110 1101 1010 0011

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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