-0.000 233 921 99 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 233 921 99(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
-0.000 233 921 99(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:

|-0.000 233 921 99| = 0.000 233 921 99


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: 0.000 233 921 99.

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) 0.000 233 921 99 × 2 = 0 + 0.000 467 843 98;
  • 2) 0.000 467 843 98 × 2 = 0 + 0.000 935 687 96;
  • 3) 0.000 935 687 96 × 2 = 0 + 0.001 871 375 92;
  • 4) 0.001 871 375 92 × 2 = 0 + 0.003 742 751 84;
  • 5) 0.003 742 751 84 × 2 = 0 + 0.007 485 503 68;
  • 6) 0.007 485 503 68 × 2 = 0 + 0.014 971 007 36;
  • 7) 0.014 971 007 36 × 2 = 0 + 0.029 942 014 72;
  • 8) 0.029 942 014 72 × 2 = 0 + 0.059 884 029 44;
  • 9) 0.059 884 029 44 × 2 = 0 + 0.119 768 058 88;
  • 10) 0.119 768 058 88 × 2 = 0 + 0.239 536 117 76;
  • 11) 0.239 536 117 76 × 2 = 0 + 0.479 072 235 52;
  • 12) 0.479 072 235 52 × 2 = 0 + 0.958 144 471 04;
  • 13) 0.958 144 471 04 × 2 = 1 + 0.916 288 942 08;
  • 14) 0.916 288 942 08 × 2 = 1 + 0.832 577 884 16;
  • 15) 0.832 577 884 16 × 2 = 1 + 0.665 155 768 32;
  • 16) 0.665 155 768 32 × 2 = 1 + 0.330 311 536 64;
  • 17) 0.330 311 536 64 × 2 = 0 + 0.660 623 073 28;
  • 18) 0.660 623 073 28 × 2 = 1 + 0.321 246 146 56;
  • 19) 0.321 246 146 56 × 2 = 0 + 0.642 492 293 12;
  • 20) 0.642 492 293 12 × 2 = 1 + 0.284 984 586 24;
  • 21) 0.284 984 586 24 × 2 = 0 + 0.569 969 172 48;
  • 22) 0.569 969 172 48 × 2 = 1 + 0.139 938 344 96;
  • 23) 0.139 938 344 96 × 2 = 0 + 0.279 876 689 92;
  • 24) 0.279 876 689 92 × 2 = 0 + 0.559 753 379 84;
  • 25) 0.559 753 379 84 × 2 = 1 + 0.119 506 759 68;
  • 26) 0.119 506 759 68 × 2 = 0 + 0.239 013 519 36;
  • 27) 0.239 013 519 36 × 2 = 0 + 0.478 027 038 72;
  • 28) 0.478 027 038 72 × 2 = 0 + 0.956 054 077 44;
  • 29) 0.956 054 077 44 × 2 = 1 + 0.912 108 154 88;
  • 30) 0.912 108 154 88 × 2 = 1 + 0.824 216 309 76;
  • 31) 0.824 216 309 76 × 2 = 1 + 0.648 432 619 52;
  • 32) 0.648 432 619 52 × 2 = 1 + 0.296 865 239 04;
  • 33) 0.296 865 239 04 × 2 = 0 + 0.593 730 478 08;
  • 34) 0.593 730 478 08 × 2 = 1 + 0.187 460 956 16;
  • 35) 0.187 460 956 16 × 2 = 0 + 0.374 921 912 32;
  • 36) 0.374 921 912 32 × 2 = 0 + 0.749 843 824 64;
  • 37) 0.749 843 824 64 × 2 = 1 + 0.499 687 649 28;
  • 38) 0.499 687 649 28 × 2 = 0 + 0.999 375 298 56;
  • 39) 0.999 375 298 56 × 2 = 1 + 0.998 750 597 12;
  • 40) 0.998 750 597 12 × 2 = 1 + 0.997 501 194 24;
  • 41) 0.997 501 194 24 × 2 = 1 + 0.995 002 388 48;
  • 42) 0.995 002 388 48 × 2 = 1 + 0.990 004 776 96;
  • 43) 0.990 004 776 96 × 2 = 1 + 0.980 009 553 92;
  • 44) 0.980 009 553 92 × 2 = 1 + 0.960 019 107 84;
  • 45) 0.960 019 107 84 × 2 = 1 + 0.920 038 215 68;
  • 46) 0.920 038 215 68 × 2 = 1 + 0.840 076 431 36;
  • 47) 0.840 076 431 36 × 2 = 1 + 0.680 152 862 72;
  • 48) 0.680 152 862 72 × 2 = 1 + 0.360 305 725 44;
  • 49) 0.360 305 725 44 × 2 = 0 + 0.720 611 450 88;
  • 50) 0.720 611 450 88 × 2 = 1 + 0.441 222 901 76;
  • 51) 0.441 222 901 76 × 2 = 0 + 0.882 445 803 52;
  • 52) 0.882 445 803 52 × 2 = 1 + 0.764 891 607 04;
  • 53) 0.764 891 607 04 × 2 = 1 + 0.529 783 214 08;
  • 54) 0.529 783 214 08 × 2 = 1 + 0.059 566 428 16;
  • 55) 0.059 566 428 16 × 2 = 0 + 0.119 132 856 32;
  • 56) 0.119 132 856 32 × 2 = 0 + 0.238 265 712 64;
  • 57) 0.238 265 712 64 × 2 = 0 + 0.476 531 425 28;
  • 58) 0.476 531 425 28 × 2 = 0 + 0.953 062 850 56;
  • 59) 0.953 062 850 56 × 2 = 1 + 0.906 125 701 12;
  • 60) 0.906 125 701 12 × 2 = 1 + 0.812 251 402 24;
  • 61) 0.812 251 402 24 × 2 = 1 + 0.624 502 804 48;
  • 62) 0.624 502 804 48 × 2 = 1 + 0.249 005 608 96;
  • 63) 0.249 005 608 96 × 2 = 0 + 0.498 011 217 92;
  • 64) 0.498 011 217 92 × 2 = 0 + 0.996 022 435 84;
  • 65) 0.996 022 435 84 × 2 = 1 + 0.992 044 871 68;

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:


0.000 233 921 99(10) =

0.0000 0000 0000 1111 0101 0100 1000 1111 0100 1011 1111 1111 0101 1100 0011 1100 1(2)

6. Positive number before normalization:

0.000 233 921 99(10) =

0.0000 0000 0000 1111 0101 0100 1000 1111 0100 1011 1111 1111 0101 1100 0011 1100 1(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 13 positions to the right, so that only one non zero digit remains to the left of it:


0.000 233 921 99(10) =

0.0000 0000 0000 1111 0101 0100 1000 1111 0100 1011 1111 1111 0101 1100 0011 1100 1(2) =

0.0000 0000 0000 1111 0101 0100 1000 1111 0100 1011 1111 1111 0101 1100 0011 1100 1(2) × 20 =

1.1110 1010 1001 0001 1110 1001 0111 1111 1110 1011 1000 0111 1001(2) × 2-13


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): -13

Mantissa (not normalized):
1.1110 1010 1001 0001 1110 1001 0111 1111 1110 1011 1000 0111 1001


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-13 + 2(11-1) - 1 =

(-13 + 1 023)(10) =

1 010(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;
  • 1 010 ÷ 2 = 505 + 0;
  • 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) =

1010(10) =

011 1111 0010(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. 1110 1010 1001 0001 1110 1001 0111 1111 1110 1011 1000 0111 1001 =

1110 1010 1001 0001 1110 1001 0111 1111 1110 1011 1000 0111 1001


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 0010

Mantissa (52 bits) =
1110 1010 1001 0001 1110 1001 0111 1111 1110 1011 1000 0111 1001


Decimal number -0.000 233 921 99 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0010 - 1110 1010 1001 0001 1110 1001 0111 1111 1110 1011 1000 0111 1001

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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