-0.000 282 436 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 436(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
-0.000 282 436(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:

|-0.000 282 436| = 0.000 282 436


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: 0.000 282 436.

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) 0.000 282 436 × 2 = 0 + 0.000 564 872;
  • 2) 0.000 564 872 × 2 = 0 + 0.001 129 744;
  • 3) 0.001 129 744 × 2 = 0 + 0.002 259 488;
  • 4) 0.002 259 488 × 2 = 0 + 0.004 518 976;
  • 5) 0.004 518 976 × 2 = 0 + 0.009 037 952;
  • 6) 0.009 037 952 × 2 = 0 + 0.018 075 904;
  • 7) 0.018 075 904 × 2 = 0 + 0.036 151 808;
  • 8) 0.036 151 808 × 2 = 0 + 0.072 303 616;
  • 9) 0.072 303 616 × 2 = 0 + 0.144 607 232;
  • 10) 0.144 607 232 × 2 = 0 + 0.289 214 464;
  • 11) 0.289 214 464 × 2 = 0 + 0.578 428 928;
  • 12) 0.578 428 928 × 2 = 1 + 0.156 857 856;
  • 13) 0.156 857 856 × 2 = 0 + 0.313 715 712;
  • 14) 0.313 715 712 × 2 = 0 + 0.627 431 424;
  • 15) 0.627 431 424 × 2 = 1 + 0.254 862 848;
  • 16) 0.254 862 848 × 2 = 0 + 0.509 725 696;
  • 17) 0.509 725 696 × 2 = 1 + 0.019 451 392;
  • 18) 0.019 451 392 × 2 = 0 + 0.038 902 784;
  • 19) 0.038 902 784 × 2 = 0 + 0.077 805 568;
  • 20) 0.077 805 568 × 2 = 0 + 0.155 611 136;
  • 21) 0.155 611 136 × 2 = 0 + 0.311 222 272;
  • 22) 0.311 222 272 × 2 = 0 + 0.622 444 544;
  • 23) 0.622 444 544 × 2 = 1 + 0.244 889 088;
  • 24) 0.244 889 088 × 2 = 0 + 0.489 778 176;
  • 25) 0.489 778 176 × 2 = 0 + 0.979 556 352;
  • 26) 0.979 556 352 × 2 = 1 + 0.959 112 704;
  • 27) 0.959 112 704 × 2 = 1 + 0.918 225 408;
  • 28) 0.918 225 408 × 2 = 1 + 0.836 450 816;
  • 29) 0.836 450 816 × 2 = 1 + 0.672 901 632;
  • 30) 0.672 901 632 × 2 = 1 + 0.345 803 264;
  • 31) 0.345 803 264 × 2 = 0 + 0.691 606 528;
  • 32) 0.691 606 528 × 2 = 1 + 0.383 213 056;
  • 33) 0.383 213 056 × 2 = 0 + 0.766 426 112;
  • 34) 0.766 426 112 × 2 = 1 + 0.532 852 224;
  • 35) 0.532 852 224 × 2 = 1 + 0.065 704 448;
  • 36) 0.065 704 448 × 2 = 0 + 0.131 408 896;
  • 37) 0.131 408 896 × 2 = 0 + 0.262 817 792;
  • 38) 0.262 817 792 × 2 = 0 + 0.525 635 584;
  • 39) 0.525 635 584 × 2 = 1 + 0.051 271 168;
  • 40) 0.051 271 168 × 2 = 0 + 0.102 542 336;
  • 41) 0.102 542 336 × 2 = 0 + 0.205 084 672;
  • 42) 0.205 084 672 × 2 = 0 + 0.410 169 344;
  • 43) 0.410 169 344 × 2 = 0 + 0.820 338 688;
  • 44) 0.820 338 688 × 2 = 1 + 0.640 677 376;
  • 45) 0.640 677 376 × 2 = 1 + 0.281 354 752;
  • 46) 0.281 354 752 × 2 = 0 + 0.562 709 504;
  • 47) 0.562 709 504 × 2 = 1 + 0.125 419 008;
  • 48) 0.125 419 008 × 2 = 0 + 0.250 838 016;
  • 49) 0.250 838 016 × 2 = 0 + 0.501 676 032;
  • 50) 0.501 676 032 × 2 = 1 + 0.003 352 064;
  • 51) 0.003 352 064 × 2 = 0 + 0.006 704 128;
  • 52) 0.006 704 128 × 2 = 0 + 0.013 408 256;
  • 53) 0.013 408 256 × 2 = 0 + 0.026 816 512;
  • 54) 0.026 816 512 × 2 = 0 + 0.053 633 024;
  • 55) 0.053 633 024 × 2 = 0 + 0.107 266 048;
  • 56) 0.107 266 048 × 2 = 0 + 0.214 532 096;
  • 57) 0.214 532 096 × 2 = 0 + 0.429 064 192;
  • 58) 0.429 064 192 × 2 = 0 + 0.858 128 384;
  • 59) 0.858 128 384 × 2 = 1 + 0.716 256 768;
  • 60) 0.716 256 768 × 2 = 1 + 0.432 513 536;
  • 61) 0.432 513 536 × 2 = 0 + 0.865 027 072;
  • 62) 0.865 027 072 × 2 = 1 + 0.730 054 144;
  • 63) 0.730 054 144 × 2 = 1 + 0.460 108 288;
  • 64) 0.460 108 288 × 2 = 0 + 0.920 216 576;

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:


0.000 282 436(10) =

0.0000 0000 0001 0010 1000 0010 0111 1101 0110 0010 0001 1010 0100 0000 0011 0110(2)

6. Positive number before normalization:

0.000 282 436(10) =

0.0000 0000 0001 0010 1000 0010 0111 1101 0110 0010 0001 1010 0100 0000 0011 0110(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:


0.000 282 436(10) =

0.0000 0000 0001 0010 1000 0010 0111 1101 0110 0010 0001 1010 0100 0000 0011 0110(2) =

0.0000 0000 0001 0010 1000 0010 0111 1101 0110 0010 0001 1010 0100 0000 0011 0110(2) × 20 =

1.0010 1000 0010 0111 1101 0110 0010 0001 1010 0100 0000 0011 0110(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 1000 0010 0111 1101 0110 0010 0001 1010 0100 0000 0011 0110


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 1000 0010 0111 1101 0110 0010 0001 1010 0100 0000 0011 0110 =

0010 1000 0010 0111 1101 0110 0010 0001 1010 0100 0000 0011 0110


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 1000 0010 0111 1101 0110 0010 0001 1010 0100 0000 0011 0110


Decimal number -0.000 282 436 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 1000 0010 0111 1101 0110 0010 0001 1010 0100 0000 0011 0110

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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