-0.023 220 31 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.023 220 31(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
-0.023 220 31(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:

|-0.023 220 31| = 0.023 220 31


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: 0.023 220 31.

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) 0.023 220 31 × 2 = 0 + 0.046 440 62;
  • 2) 0.046 440 62 × 2 = 0 + 0.092 881 24;
  • 3) 0.092 881 24 × 2 = 0 + 0.185 762 48;
  • 4) 0.185 762 48 × 2 = 0 + 0.371 524 96;
  • 5) 0.371 524 96 × 2 = 0 + 0.743 049 92;
  • 6) 0.743 049 92 × 2 = 1 + 0.486 099 84;
  • 7) 0.486 099 84 × 2 = 0 + 0.972 199 68;
  • 8) 0.972 199 68 × 2 = 1 + 0.944 399 36;
  • 9) 0.944 399 36 × 2 = 1 + 0.888 798 72;
  • 10) 0.888 798 72 × 2 = 1 + 0.777 597 44;
  • 11) 0.777 597 44 × 2 = 1 + 0.555 194 88;
  • 12) 0.555 194 88 × 2 = 1 + 0.110 389 76;
  • 13) 0.110 389 76 × 2 = 0 + 0.220 779 52;
  • 14) 0.220 779 52 × 2 = 0 + 0.441 559 04;
  • 15) 0.441 559 04 × 2 = 0 + 0.883 118 08;
  • 16) 0.883 118 08 × 2 = 1 + 0.766 236 16;
  • 17) 0.766 236 16 × 2 = 1 + 0.532 472 32;
  • 18) 0.532 472 32 × 2 = 1 + 0.064 944 64;
  • 19) 0.064 944 64 × 2 = 0 + 0.129 889 28;
  • 20) 0.129 889 28 × 2 = 0 + 0.259 778 56;
  • 21) 0.259 778 56 × 2 = 0 + 0.519 557 12;
  • 22) 0.519 557 12 × 2 = 1 + 0.039 114 24;
  • 23) 0.039 114 24 × 2 = 0 + 0.078 228 48;
  • 24) 0.078 228 48 × 2 = 0 + 0.156 456 96;
  • 25) 0.156 456 96 × 2 = 0 + 0.312 913 92;
  • 26) 0.312 913 92 × 2 = 0 + 0.625 827 84;
  • 27) 0.625 827 84 × 2 = 1 + 0.251 655 68;
  • 28) 0.251 655 68 × 2 = 0 + 0.503 311 36;
  • 29) 0.503 311 36 × 2 = 1 + 0.006 622 72;
  • 30) 0.006 622 72 × 2 = 0 + 0.013 245 44;
  • 31) 0.013 245 44 × 2 = 0 + 0.026 490 88;
  • 32) 0.026 490 88 × 2 = 0 + 0.052 981 76;
  • 33) 0.052 981 76 × 2 = 0 + 0.105 963 52;
  • 34) 0.105 963 52 × 2 = 0 + 0.211 927 04;
  • 35) 0.211 927 04 × 2 = 0 + 0.423 854 08;
  • 36) 0.423 854 08 × 2 = 0 + 0.847 708 16;
  • 37) 0.847 708 16 × 2 = 1 + 0.695 416 32;
  • 38) 0.695 416 32 × 2 = 1 + 0.390 832 64;
  • 39) 0.390 832 64 × 2 = 0 + 0.781 665 28;
  • 40) 0.781 665 28 × 2 = 1 + 0.563 330 56;
  • 41) 0.563 330 56 × 2 = 1 + 0.126 661 12;
  • 42) 0.126 661 12 × 2 = 0 + 0.253 322 24;
  • 43) 0.253 322 24 × 2 = 0 + 0.506 644 48;
  • 44) 0.506 644 48 × 2 = 1 + 0.013 288 96;
  • 45) 0.013 288 96 × 2 = 0 + 0.026 577 92;
  • 46) 0.026 577 92 × 2 = 0 + 0.053 155 84;
  • 47) 0.053 155 84 × 2 = 0 + 0.106 311 68;
  • 48) 0.106 311 68 × 2 = 0 + 0.212 623 36;
  • 49) 0.212 623 36 × 2 = 0 + 0.425 246 72;
  • 50) 0.425 246 72 × 2 = 0 + 0.850 493 44;
  • 51) 0.850 493 44 × 2 = 1 + 0.700 986 88;
  • 52) 0.700 986 88 × 2 = 1 + 0.401 973 76;
  • 53) 0.401 973 76 × 2 = 0 + 0.803 947 52;
  • 54) 0.803 947 52 × 2 = 1 + 0.607 895 04;
  • 55) 0.607 895 04 × 2 = 1 + 0.215 790 08;
  • 56) 0.215 790 08 × 2 = 0 + 0.431 580 16;
  • 57) 0.431 580 16 × 2 = 0 + 0.863 160 32;
  • 58) 0.863 160 32 × 2 = 1 + 0.726 320 64;

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:


0.023 220 31(10) =

0.0000 0101 1111 0001 1100 0100 0010 1000 0000 1101 1001 0000 0011 0110 01(2)

6. Positive number before normalization:

0.023 220 31(10) =

0.0000 0101 1111 0001 1100 0100 0010 1000 0000 1101 1001 0000 0011 0110 01(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 6 positions to the right, so that only one non zero digit remains to the left of it:


0.023 220 31(10) =

0.0000 0101 1111 0001 1100 0100 0010 1000 0000 1101 1001 0000 0011 0110 01(2) =

0.0000 0101 1111 0001 1100 0100 0010 1000 0000 1101 1001 0000 0011 0110 01(2) × 20 =

1.0111 1100 0111 0001 0000 1010 0000 0011 0110 0100 0000 1101 1001(2) × 2-6


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): -6

Mantissa (not normalized):
1.0111 1100 0111 0001 0000 1010 0000 0011 0110 0100 0000 1101 1001


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-6 + 2(11-1) - 1 =

(-6 + 1 023)(10) =

1 017(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;
  • 1 017 ÷ 2 = 508 + 1;
  • 508 ÷ 2 = 254 + 0;
  • 254 ÷ 2 = 127 + 0;
  • 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) =

1017(10) =

011 1111 1001(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 1100 0111 0001 0000 1010 0000 0011 0110 0100 0000 1101 1001 =

0111 1100 0111 0001 0000 1010 0000 0011 0110 0100 0000 1101 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 1001

Mantissa (52 bits) =
0111 1100 0111 0001 0000 1010 0000 0011 0110 0100 0000 1101 1001


Decimal number -0.023 220 31 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1001 - 0111 1100 0111 0001 0000 1010 0000 0011 0110 0100 0000 1101 1001

» Convert decimal -0.023 220 39 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