0.974 013 321 19 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.974 013 321 19(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 19(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 19.

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 19 × 2 = 1 + 0.948 026 642 38;
  • 2) 0.948 026 642 38 × 2 = 1 + 0.896 053 284 76;
  • 3) 0.896 053 284 76 × 2 = 1 + 0.792 106 569 52;
  • 4) 0.792 106 569 52 × 2 = 1 + 0.584 213 139 04;
  • 5) 0.584 213 139 04 × 2 = 1 + 0.168 426 278 08;
  • 6) 0.168 426 278 08 × 2 = 0 + 0.336 852 556 16;
  • 7) 0.336 852 556 16 × 2 = 0 + 0.673 705 112 32;
  • 8) 0.673 705 112 32 × 2 = 1 + 0.347 410 224 64;
  • 9) 0.347 410 224 64 × 2 = 0 + 0.694 820 449 28;
  • 10) 0.694 820 449 28 × 2 = 1 + 0.389 640 898 56;
  • 11) 0.389 640 898 56 × 2 = 0 + 0.779 281 797 12;
  • 12) 0.779 281 797 12 × 2 = 1 + 0.558 563 594 24;
  • 13) 0.558 563 594 24 × 2 = 1 + 0.117 127 188 48;
  • 14) 0.117 127 188 48 × 2 = 0 + 0.234 254 376 96;
  • 15) 0.234 254 376 96 × 2 = 0 + 0.468 508 753 92;
  • 16) 0.468 508 753 92 × 2 = 0 + 0.937 017 507 84;
  • 17) 0.937 017 507 84 × 2 = 1 + 0.874 035 015 68;
  • 18) 0.874 035 015 68 × 2 = 1 + 0.748 070 031 36;
  • 19) 0.748 070 031 36 × 2 = 1 + 0.496 140 062 72;
  • 20) 0.496 140 062 72 × 2 = 0 + 0.992 280 125 44;
  • 21) 0.992 280 125 44 × 2 = 1 + 0.984 560 250 88;
  • 22) 0.984 560 250 88 × 2 = 1 + 0.969 120 501 76;
  • 23) 0.969 120 501 76 × 2 = 1 + 0.938 241 003 52;
  • 24) 0.938 241 003 52 × 2 = 1 + 0.876 482 007 04;
  • 25) 0.876 482 007 04 × 2 = 1 + 0.752 964 014 08;
  • 26) 0.752 964 014 08 × 2 = 1 + 0.505 928 028 16;
  • 27) 0.505 928 028 16 × 2 = 1 + 0.011 856 056 32;
  • 28) 0.011 856 056 32 × 2 = 0 + 0.023 712 112 64;
  • 29) 0.023 712 112 64 × 2 = 0 + 0.047 424 225 28;
  • 30) 0.047 424 225 28 × 2 = 0 + 0.094 848 450 56;
  • 31) 0.094 848 450 56 × 2 = 0 + 0.189 696 901 12;
  • 32) 0.189 696 901 12 × 2 = 0 + 0.379 393 802 24;
  • 33) 0.379 393 802 24 × 2 = 0 + 0.758 787 604 48;
  • 34) 0.758 787 604 48 × 2 = 1 + 0.517 575 208 96;
  • 35) 0.517 575 208 96 × 2 = 1 + 0.035 150 417 92;
  • 36) 0.035 150 417 92 × 2 = 0 + 0.070 300 835 84;
  • 37) 0.070 300 835 84 × 2 = 0 + 0.140 601 671 68;
  • 38) 0.140 601 671 68 × 2 = 0 + 0.281 203 343 36;
  • 39) 0.281 203 343 36 × 2 = 0 + 0.562 406 686 72;
  • 40) 0.562 406 686 72 × 2 = 1 + 0.124 813 373 44;
  • 41) 0.124 813 373 44 × 2 = 0 + 0.249 626 746 88;
  • 42) 0.249 626 746 88 × 2 = 0 + 0.499 253 493 76;
  • 43) 0.499 253 493 76 × 2 = 0 + 0.998 506 987 52;
  • 44) 0.998 506 987 52 × 2 = 1 + 0.997 013 975 04;
  • 45) 0.997 013 975 04 × 2 = 1 + 0.994 027 950 08;
  • 46) 0.994 027 950 08 × 2 = 1 + 0.988 055 900 16;
  • 47) 0.988 055 900 16 × 2 = 1 + 0.976 111 800 32;
  • 48) 0.976 111 800 32 × 2 = 1 + 0.952 223 600 64;
  • 49) 0.952 223 600 64 × 2 = 1 + 0.904 447 201 28;
  • 50) 0.904 447 201 28 × 2 = 1 + 0.808 894 402 56;
  • 51) 0.808 894 402 56 × 2 = 1 + 0.617 788 805 12;
  • 52) 0.617 788 805 12 × 2 = 1 + 0.235 577 610 24;
  • 53) 0.235 577 610 24 × 2 = 0 + 0.471 155 220 48;

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 19(10) =

0.1111 1001 0101 1000 1110 1111 1110 0000 0110 0001 0001 1111 1111 0(2)

5. Positive number before normalization:

0.974 013 321 19(10) =

0.1111 1001 0101 1000 1110 1111 1110 0000 0110 0001 0001 1111 1111 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 19(10) =

0.1111 1001 0101 1000 1110 1111 1110 0000 0110 0001 0001 1111 1111 0(2) =

0.1111 1001 0101 1000 1110 1111 1110 0000 0110 0001 0001 1111 1111 0(2) × 20 =

1.1111 0010 1011 0001 1101 1111 1100 0000 1100 0010 0011 1111 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 0000 1100 0010 0011 1111 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 0000 1100 0010 0011 1111 1110 =

1111 0010 1011 0001 1101 1111 1100 0000 1100 0010 0011 1111 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 0000 1100 0010 0011 1111 1110


Decimal number 0.974 013 321 19 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1110 - 1111 0010 1011 0001 1101 1111 1100 0000 1100 0010 0011 1111 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