0.974 013 321 37 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.974 013 321 37(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
0.974 013 321 37(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. 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;

2. 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)


3. Convert to binary (base 2) the fractional part: 0.974 013 321 37.

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) 0.974 013 321 37 × 2 = 1 + 0.948 026 642 74;
  • 2) 0.948 026 642 74 × 2 = 1 + 0.896 053 285 48;
  • 3) 0.896 053 285 48 × 2 = 1 + 0.792 106 570 96;
  • 4) 0.792 106 570 96 × 2 = 1 + 0.584 213 141 92;
  • 5) 0.584 213 141 92 × 2 = 1 + 0.168 426 283 84;
  • 6) 0.168 426 283 84 × 2 = 0 + 0.336 852 567 68;
  • 7) 0.336 852 567 68 × 2 = 0 + 0.673 705 135 36;
  • 8) 0.673 705 135 36 × 2 = 1 + 0.347 410 270 72;
  • 9) 0.347 410 270 72 × 2 = 0 + 0.694 820 541 44;
  • 10) 0.694 820 541 44 × 2 = 1 + 0.389 641 082 88;
  • 11) 0.389 641 082 88 × 2 = 0 + 0.779 282 165 76;
  • 12) 0.779 282 165 76 × 2 = 1 + 0.558 564 331 52;
  • 13) 0.558 564 331 52 × 2 = 1 + 0.117 128 663 04;
  • 14) 0.117 128 663 04 × 2 = 0 + 0.234 257 326 08;
  • 15) 0.234 257 326 08 × 2 = 0 + 0.468 514 652 16;
  • 16) 0.468 514 652 16 × 2 = 0 + 0.937 029 304 32;
  • 17) 0.937 029 304 32 × 2 = 1 + 0.874 058 608 64;
  • 18) 0.874 058 608 64 × 2 = 1 + 0.748 117 217 28;
  • 19) 0.748 117 217 28 × 2 = 1 + 0.496 234 434 56;
  • 20) 0.496 234 434 56 × 2 = 0 + 0.992 468 869 12;
  • 21) 0.992 468 869 12 × 2 = 1 + 0.984 937 738 24;
  • 22) 0.984 937 738 24 × 2 = 1 + 0.969 875 476 48;
  • 23) 0.969 875 476 48 × 2 = 1 + 0.939 750 952 96;
  • 24) 0.939 750 952 96 × 2 = 1 + 0.879 501 905 92;
  • 25) 0.879 501 905 92 × 2 = 1 + 0.759 003 811 84;
  • 26) 0.759 003 811 84 × 2 = 1 + 0.518 007 623 68;
  • 27) 0.518 007 623 68 × 2 = 1 + 0.036 015 247 36;
  • 28) 0.036 015 247 36 × 2 = 0 + 0.072 030 494 72;
  • 29) 0.072 030 494 72 × 2 = 0 + 0.144 060 989 44;
  • 30) 0.144 060 989 44 × 2 = 0 + 0.288 121 978 88;
  • 31) 0.288 121 978 88 × 2 = 0 + 0.576 243 957 76;
  • 32) 0.576 243 957 76 × 2 = 1 + 0.152 487 915 52;
  • 33) 0.152 487 915 52 × 2 = 0 + 0.304 975 831 04;
  • 34) 0.304 975 831 04 × 2 = 0 + 0.609 951 662 08;
  • 35) 0.609 951 662 08 × 2 = 1 + 0.219 903 324 16;
  • 36) 0.219 903 324 16 × 2 = 0 + 0.439 806 648 32;
  • 37) 0.439 806 648 32 × 2 = 0 + 0.879 613 296 64;
  • 38) 0.879 613 296 64 × 2 = 1 + 0.759 226 593 28;
  • 39) 0.759 226 593 28 × 2 = 1 + 0.518 453 186 56;
  • 40) 0.518 453 186 56 × 2 = 1 + 0.036 906 373 12;
  • 41) 0.036 906 373 12 × 2 = 0 + 0.073 812 746 24;
  • 42) 0.073 812 746 24 × 2 = 0 + 0.147 625 492 48;
  • 43) 0.147 625 492 48 × 2 = 0 + 0.295 250 984 96;
  • 44) 0.295 250 984 96 × 2 = 0 + 0.590 501 969 92;
  • 45) 0.590 501 969 92 × 2 = 1 + 0.181 003 939 84;
  • 46) 0.181 003 939 84 × 2 = 0 + 0.362 007 879 68;
  • 47) 0.362 007 879 68 × 2 = 0 + 0.724 015 759 36;
  • 48) 0.724 015 759 36 × 2 = 1 + 0.448 031 518 72;
  • 49) 0.448 031 518 72 × 2 = 0 + 0.896 063 037 44;
  • 50) 0.896 063 037 44 × 2 = 1 + 0.792 126 074 88;
  • 51) 0.792 126 074 88 × 2 = 1 + 0.584 252 149 76;
  • 52) 0.584 252 149 76 × 2 = 1 + 0.168 504 299 52;
  • 53) 0.168 504 299 52 × 2 = 0 + 0.337 008 599 04;

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).


4. 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:


0.974 013 321 37(10) =

0.1111 1001 0101 1000 1110 1111 1110 0001 0010 0111 0000 1001 0111 0(2)

5. Positive number before normalization:

0.974 013 321 37(10) =

0.1111 1001 0101 1000 1110 1111 1110 0001 0010 0111 0000 1001 0111 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 1 positions to the right, so that only one non zero digit remains to the left of it:


0.974 013 321 37(10) =

0.1111 1001 0101 1000 1110 1111 1110 0001 0010 0111 0000 1001 0111 0(2) =

0.1111 1001 0101 1000 1110 1111 1110 0001 0010 0111 0000 1001 0111 0(2) × 20 =

1.1111 0010 1011 0001 1101 1111 1100 0010 0100 1110 0001 0010 1110(2) × 2-1


7. 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 0 (a positive number)

Exponent (unadjusted): -1

Mantissa (not normalized):
1.1111 0010 1011 0001 1101 1111 1100 0010 0100 1110 0001 0010 1110


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

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

(-1 + 1 023)(10) =

1 022(10)


9. 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;
  • 1 022 ÷ 2 = 511 + 0;
  • 511 ÷ 2 = 255 + 1;
  • 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;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

1022(10) =

011 1111 1110(2)


11. 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. 1111 0010 1011 0001 1101 1111 1100 0010 0100 1110 0001 0010 1110 =

1111 0010 1011 0001 1101 1111 1100 0010 0100 1110 0001 0010 1110


12. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (11 bits) =
011 1111 1110

Mantissa (52 bits) =
1111 0010 1011 0001 1101 1111 1100 0010 0100 1110 0001 0010 1110


Decimal number 0.974 013 321 37 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1110 - 1111 0010 1011 0001 1101 1111 1100 0010 0100 1110 0001 0010 1110

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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