-0.000 281 833 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 281 833(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
-0.000 281 833(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:

|-0.000 281 833| = 0.000 281 833


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: 0.000 281 833.

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) 0.000 281 833 × 2 = 0 + 0.000 563 666;
  • 2) 0.000 563 666 × 2 = 0 + 0.001 127 332;
  • 3) 0.001 127 332 × 2 = 0 + 0.002 254 664;
  • 4) 0.002 254 664 × 2 = 0 + 0.004 509 328;
  • 5) 0.004 509 328 × 2 = 0 + 0.009 018 656;
  • 6) 0.009 018 656 × 2 = 0 + 0.018 037 312;
  • 7) 0.018 037 312 × 2 = 0 + 0.036 074 624;
  • 8) 0.036 074 624 × 2 = 0 + 0.072 149 248;
  • 9) 0.072 149 248 × 2 = 0 + 0.144 298 496;
  • 10) 0.144 298 496 × 2 = 0 + 0.288 596 992;
  • 11) 0.288 596 992 × 2 = 0 + 0.577 193 984;
  • 12) 0.577 193 984 × 2 = 1 + 0.154 387 968;
  • 13) 0.154 387 968 × 2 = 0 + 0.308 775 936;
  • 14) 0.308 775 936 × 2 = 0 + 0.617 551 872;
  • 15) 0.617 551 872 × 2 = 1 + 0.235 103 744;
  • 16) 0.235 103 744 × 2 = 0 + 0.470 207 488;
  • 17) 0.470 207 488 × 2 = 0 + 0.940 414 976;
  • 18) 0.940 414 976 × 2 = 1 + 0.880 829 952;
  • 19) 0.880 829 952 × 2 = 1 + 0.761 659 904;
  • 20) 0.761 659 904 × 2 = 1 + 0.523 319 808;
  • 21) 0.523 319 808 × 2 = 1 + 0.046 639 616;
  • 22) 0.046 639 616 × 2 = 0 + 0.093 279 232;
  • 23) 0.093 279 232 × 2 = 0 + 0.186 558 464;
  • 24) 0.186 558 464 × 2 = 0 + 0.373 116 928;
  • 25) 0.373 116 928 × 2 = 0 + 0.746 233 856;
  • 26) 0.746 233 856 × 2 = 1 + 0.492 467 712;
  • 27) 0.492 467 712 × 2 = 0 + 0.984 935 424;
  • 28) 0.984 935 424 × 2 = 1 + 0.969 870 848;
  • 29) 0.969 870 848 × 2 = 1 + 0.939 741 696;
  • 30) 0.939 741 696 × 2 = 1 + 0.879 483 392;
  • 31) 0.879 483 392 × 2 = 1 + 0.758 966 784;
  • 32) 0.758 966 784 × 2 = 1 + 0.517 933 568;
  • 33) 0.517 933 568 × 2 = 1 + 0.035 867 136;
  • 34) 0.035 867 136 × 2 = 0 + 0.071 734 272;
  • 35) 0.071 734 272 × 2 = 0 + 0.143 468 544;
  • 36) 0.143 468 544 × 2 = 0 + 0.286 937 088;
  • 37) 0.286 937 088 × 2 = 0 + 0.573 874 176;
  • 38) 0.573 874 176 × 2 = 1 + 0.147 748 352;
  • 39) 0.147 748 352 × 2 = 0 + 0.295 496 704;
  • 40) 0.295 496 704 × 2 = 0 + 0.590 993 408;
  • 41) 0.590 993 408 × 2 = 1 + 0.181 986 816;
  • 42) 0.181 986 816 × 2 = 0 + 0.363 973 632;
  • 43) 0.363 973 632 × 2 = 0 + 0.727 947 264;
  • 44) 0.727 947 264 × 2 = 1 + 0.455 894 528;
  • 45) 0.455 894 528 × 2 = 0 + 0.911 789 056;
  • 46) 0.911 789 056 × 2 = 1 + 0.823 578 112;
  • 47) 0.823 578 112 × 2 = 1 + 0.647 156 224;
  • 48) 0.647 156 224 × 2 = 1 + 0.294 312 448;
  • 49) 0.294 312 448 × 2 = 0 + 0.588 624 896;
  • 50) 0.588 624 896 × 2 = 1 + 0.177 249 792;
  • 51) 0.177 249 792 × 2 = 0 + 0.354 499 584;
  • 52) 0.354 499 584 × 2 = 0 + 0.708 999 168;
  • 53) 0.708 999 168 × 2 = 1 + 0.417 998 336;
  • 54) 0.417 998 336 × 2 = 0 + 0.835 996 672;
  • 55) 0.835 996 672 × 2 = 1 + 0.671 993 344;
  • 56) 0.671 993 344 × 2 = 1 + 0.343 986 688;
  • 57) 0.343 986 688 × 2 = 0 + 0.687 973 376;
  • 58) 0.687 973 376 × 2 = 1 + 0.375 946 752;
  • 59) 0.375 946 752 × 2 = 0 + 0.751 893 504;
  • 60) 0.751 893 504 × 2 = 1 + 0.503 787 008;
  • 61) 0.503 787 008 × 2 = 1 + 0.007 574 016;
  • 62) 0.007 574 016 × 2 = 0 + 0.015 148 032;
  • 63) 0.015 148 032 × 2 = 0 + 0.030 296 064;
  • 64) 0.030 296 064 × 2 = 0 + 0.060 592 128;

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:


0.000 281 833(10) =

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

6. Positive number before normalization:

0.000 281 833(10) =

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


0.000 281 833(10) =

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

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

1.0010 0111 1000 0101 1111 1000 0100 1001 0111 0100 1011 0101 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 1000 0101 1111 1000 0100 1001 0111 0100 1011 0101 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 1000 0101 1111 1000 0100 1001 0111 0100 1011 0101 1000 =

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


Decimal number -0.000 281 833 converted to 64 bit double precision IEEE 754 binary floating point representation:

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