-0.304 864 729 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.304 864 729(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
-0.304 864 729(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:

|-0.304 864 729| = 0.304 864 729


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: 0.304 864 729.

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) 0.304 864 729 × 2 = 0 + 0.609 729 458;
  • 2) 0.609 729 458 × 2 = 1 + 0.219 458 916;
  • 3) 0.219 458 916 × 2 = 0 + 0.438 917 832;
  • 4) 0.438 917 832 × 2 = 0 + 0.877 835 664;
  • 5) 0.877 835 664 × 2 = 1 + 0.755 671 328;
  • 6) 0.755 671 328 × 2 = 1 + 0.511 342 656;
  • 7) 0.511 342 656 × 2 = 1 + 0.022 685 312;
  • 8) 0.022 685 312 × 2 = 0 + 0.045 370 624;
  • 9) 0.045 370 624 × 2 = 0 + 0.090 741 248;
  • 10) 0.090 741 248 × 2 = 0 + 0.181 482 496;
  • 11) 0.181 482 496 × 2 = 0 + 0.362 964 992;
  • 12) 0.362 964 992 × 2 = 0 + 0.725 929 984;
  • 13) 0.725 929 984 × 2 = 1 + 0.451 859 968;
  • 14) 0.451 859 968 × 2 = 0 + 0.903 719 936;
  • 15) 0.903 719 936 × 2 = 1 + 0.807 439 872;
  • 16) 0.807 439 872 × 2 = 1 + 0.614 879 744;
  • 17) 0.614 879 744 × 2 = 1 + 0.229 759 488;
  • 18) 0.229 759 488 × 2 = 0 + 0.459 518 976;
  • 19) 0.459 518 976 × 2 = 0 + 0.919 037 952;
  • 20) 0.919 037 952 × 2 = 1 + 0.838 075 904;
  • 21) 0.838 075 904 × 2 = 1 + 0.676 151 808;
  • 22) 0.676 151 808 × 2 = 1 + 0.352 303 616;
  • 23) 0.352 303 616 × 2 = 0 + 0.704 607 232;
  • 24) 0.704 607 232 × 2 = 1 + 0.409 214 464;
  • 25) 0.409 214 464 × 2 = 0 + 0.818 428 928;
  • 26) 0.818 428 928 × 2 = 1 + 0.636 857 856;
  • 27) 0.636 857 856 × 2 = 1 + 0.273 715 712;
  • 28) 0.273 715 712 × 2 = 0 + 0.547 431 424;
  • 29) 0.547 431 424 × 2 = 1 + 0.094 862 848;
  • 30) 0.094 862 848 × 2 = 0 + 0.189 725 696;
  • 31) 0.189 725 696 × 2 = 0 + 0.379 451 392;
  • 32) 0.379 451 392 × 2 = 0 + 0.758 902 784;
  • 33) 0.758 902 784 × 2 = 1 + 0.517 805 568;
  • 34) 0.517 805 568 × 2 = 1 + 0.035 611 136;
  • 35) 0.035 611 136 × 2 = 0 + 0.071 222 272;
  • 36) 0.071 222 272 × 2 = 0 + 0.142 444 544;
  • 37) 0.142 444 544 × 2 = 0 + 0.284 889 088;
  • 38) 0.284 889 088 × 2 = 0 + 0.569 778 176;
  • 39) 0.569 778 176 × 2 = 1 + 0.139 556 352;
  • 40) 0.139 556 352 × 2 = 0 + 0.279 112 704;
  • 41) 0.279 112 704 × 2 = 0 + 0.558 225 408;
  • 42) 0.558 225 408 × 2 = 1 + 0.116 450 816;
  • 43) 0.116 450 816 × 2 = 0 + 0.232 901 632;
  • 44) 0.232 901 632 × 2 = 0 + 0.465 803 264;
  • 45) 0.465 803 264 × 2 = 0 + 0.931 606 528;
  • 46) 0.931 606 528 × 2 = 1 + 0.863 213 056;
  • 47) 0.863 213 056 × 2 = 1 + 0.726 426 112;
  • 48) 0.726 426 112 × 2 = 1 + 0.452 852 224;
  • 49) 0.452 852 224 × 2 = 0 + 0.905 704 448;
  • 50) 0.905 704 448 × 2 = 1 + 0.811 408 896;
  • 51) 0.811 408 896 × 2 = 1 + 0.622 817 792;
  • 52) 0.622 817 792 × 2 = 1 + 0.245 635 584;
  • 53) 0.245 635 584 × 2 = 0 + 0.491 271 168;
  • 54) 0.491 271 168 × 2 = 0 + 0.982 542 336;

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:


0.304 864 729(10) =

0.0100 1110 0000 1011 1001 1101 0110 1000 1100 0010 0100 0111 0111 00(2)

6. Positive number before normalization:

0.304 864 729(10) =

0.0100 1110 0000 1011 1001 1101 0110 1000 1100 0010 0100 0111 0111 00(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 2 positions to the right, so that only one non zero digit remains to the left of it:


0.304 864 729(10) =

0.0100 1110 0000 1011 1001 1101 0110 1000 1100 0010 0100 0111 0111 00(2) =

0.0100 1110 0000 1011 1001 1101 0110 1000 1100 0010 0100 0111 0111 00(2) × 20 =

1.0011 1000 0010 1110 0111 0101 1010 0011 0000 1001 0001 1101 1100(2) × 2-2


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): -2

Mantissa (not normalized):
1.0011 1000 0010 1110 0111 0101 1010 0011 0000 1001 0001 1101 1100


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-2 + 2(11-1) - 1 =

(-2 + 1 023)(10) =

1 021(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;
  • 1 021 ÷ 2 = 510 + 1;
  • 510 ÷ 2 = 255 + 0;
  • 255 ÷ 2 = 127 + 1;
  • 127 ÷ 2 = 63 + 1;
  • 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) =

1021(10) =

011 1111 1101(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. 0011 1000 0010 1110 0111 0101 1010 0011 0000 1001 0001 1101 1100 =

0011 1000 0010 1110 0111 0101 1010 0011 0000 1001 0001 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 1101

Mantissa (52 bits) =
0011 1000 0010 1110 0111 0101 1010 0011 0000 1001 0001 1101 1100


Decimal number -0.304 864 729 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1101 - 0011 1000 0010 1110 0111 0101 1010 0011 0000 1001 0001 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