-0.365 481 236 32 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.365 481 236 32(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
-0.365 481 236 32(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:

|-0.365 481 236 32| = 0.365 481 236 32


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: 0.365 481 236 32.

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) 0.365 481 236 32 × 2 = 0 + 0.730 962 472 64;
  • 2) 0.730 962 472 64 × 2 = 1 + 0.461 924 945 28;
  • 3) 0.461 924 945 28 × 2 = 0 + 0.923 849 890 56;
  • 4) 0.923 849 890 56 × 2 = 1 + 0.847 699 781 12;
  • 5) 0.847 699 781 12 × 2 = 1 + 0.695 399 562 24;
  • 6) 0.695 399 562 24 × 2 = 1 + 0.390 799 124 48;
  • 7) 0.390 799 124 48 × 2 = 0 + 0.781 598 248 96;
  • 8) 0.781 598 248 96 × 2 = 1 + 0.563 196 497 92;
  • 9) 0.563 196 497 92 × 2 = 1 + 0.126 392 995 84;
  • 10) 0.126 392 995 84 × 2 = 0 + 0.252 785 991 68;
  • 11) 0.252 785 991 68 × 2 = 0 + 0.505 571 983 36;
  • 12) 0.505 571 983 36 × 2 = 1 + 0.011 143 966 72;
  • 13) 0.011 143 966 72 × 2 = 0 + 0.022 287 933 44;
  • 14) 0.022 287 933 44 × 2 = 0 + 0.044 575 866 88;
  • 15) 0.044 575 866 88 × 2 = 0 + 0.089 151 733 76;
  • 16) 0.089 151 733 76 × 2 = 0 + 0.178 303 467 52;
  • 17) 0.178 303 467 52 × 2 = 0 + 0.356 606 935 04;
  • 18) 0.356 606 935 04 × 2 = 0 + 0.713 213 870 08;
  • 19) 0.713 213 870 08 × 2 = 1 + 0.426 427 740 16;
  • 20) 0.426 427 740 16 × 2 = 0 + 0.852 855 480 32;
  • 21) 0.852 855 480 32 × 2 = 1 + 0.705 710 960 64;
  • 22) 0.705 710 960 64 × 2 = 1 + 0.411 421 921 28;
  • 23) 0.411 421 921 28 × 2 = 0 + 0.822 843 842 56;
  • 24) 0.822 843 842 56 × 2 = 1 + 0.645 687 685 12;
  • 25) 0.645 687 685 12 × 2 = 1 + 0.291 375 370 24;
  • 26) 0.291 375 370 24 × 2 = 0 + 0.582 750 740 48;
  • 27) 0.582 750 740 48 × 2 = 1 + 0.165 501 480 96;
  • 28) 0.165 501 480 96 × 2 = 0 + 0.331 002 961 92;
  • 29) 0.331 002 961 92 × 2 = 0 + 0.662 005 923 84;
  • 30) 0.662 005 923 84 × 2 = 1 + 0.324 011 847 68;
  • 31) 0.324 011 847 68 × 2 = 0 + 0.648 023 695 36;
  • 32) 0.648 023 695 36 × 2 = 1 + 0.296 047 390 72;
  • 33) 0.296 047 390 72 × 2 = 0 + 0.592 094 781 44;
  • 34) 0.592 094 781 44 × 2 = 1 + 0.184 189 562 88;
  • 35) 0.184 189 562 88 × 2 = 0 + 0.368 379 125 76;
  • 36) 0.368 379 125 76 × 2 = 0 + 0.736 758 251 52;
  • 37) 0.736 758 251 52 × 2 = 1 + 0.473 516 503 04;
  • 38) 0.473 516 503 04 × 2 = 0 + 0.947 033 006 08;
  • 39) 0.947 033 006 08 × 2 = 1 + 0.894 066 012 16;
  • 40) 0.894 066 012 16 × 2 = 1 + 0.788 132 024 32;
  • 41) 0.788 132 024 32 × 2 = 1 + 0.576 264 048 64;
  • 42) 0.576 264 048 64 × 2 = 1 + 0.152 528 097 28;
  • 43) 0.152 528 097 28 × 2 = 0 + 0.305 056 194 56;
  • 44) 0.305 056 194 56 × 2 = 0 + 0.610 112 389 12;
  • 45) 0.610 112 389 12 × 2 = 1 + 0.220 224 778 24;
  • 46) 0.220 224 778 24 × 2 = 0 + 0.440 449 556 48;
  • 47) 0.440 449 556 48 × 2 = 0 + 0.880 899 112 96;
  • 48) 0.880 899 112 96 × 2 = 1 + 0.761 798 225 92;
  • 49) 0.761 798 225 92 × 2 = 1 + 0.523 596 451 84;
  • 50) 0.523 596 451 84 × 2 = 1 + 0.047 192 903 68;
  • 51) 0.047 192 903 68 × 2 = 0 + 0.094 385 807 36;
  • 52) 0.094 385 807 36 × 2 = 0 + 0.188 771 614 72;
  • 53) 0.188 771 614 72 × 2 = 0 + 0.377 543 229 44;
  • 54) 0.377 543 229 44 × 2 = 0 + 0.755 086 458 88;

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:


0.365 481 236 32(10) =

0.0101 1101 1001 0000 0010 1101 1010 0101 0100 1011 1100 1001 1100 00(2)

6. Positive number before normalization:

0.365 481 236 32(10) =

0.0101 1101 1001 0000 0010 1101 1010 0101 0100 1011 1100 1001 1100 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:


0.365 481 236 32(10) =

0.0101 1101 1001 0000 0010 1101 1010 0101 0100 1011 1100 1001 1100 00(2) =

0.0101 1101 1001 0000 0010 1101 1010 0101 0100 1011 1100 1001 1100 00(2) × 20 =

1.0111 0110 0100 0000 1011 0110 1001 0101 0010 1111 0010 0111 0000(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):
1.0111 0110 0100 0000 1011 0110 1001 0101 0010 1111 0010 0111 0000


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. 0111 0110 0100 0000 1011 0110 1001 0101 0010 1111 0010 0111 0000 =

0111 0110 0100 0000 1011 0110 1001 0101 0010 1111 0010 0111 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 1101

Mantissa (52 bits) =
0111 0110 0100 0000 1011 0110 1001 0101 0010 1111 0010 0111 0000


Decimal number -0.365 481 236 32 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1101 - 0111 0110 0100 0000 1011 0110 1001 0101 0010 1111 0010 0111 0000

» Convert decimal -0.365 481 235 6 to 64 bit double precision IEEE 754 binary floating point representation standard

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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