0.974 013 318 43 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.974 013 318 43(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
0.974 013 318 43(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: 0.974 013 318 43.

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) 0.974 013 318 43 × 2 = 1 + 0.948 026 636 86;
  • 2) 0.948 026 636 86 × 2 = 1 + 0.896 053 273 72;
  • 3) 0.896 053 273 72 × 2 = 1 + 0.792 106 547 44;
  • 4) 0.792 106 547 44 × 2 = 1 + 0.584 213 094 88;
  • 5) 0.584 213 094 88 × 2 = 1 + 0.168 426 189 76;
  • 6) 0.168 426 189 76 × 2 = 0 + 0.336 852 379 52;
  • 7) 0.336 852 379 52 × 2 = 0 + 0.673 704 759 04;
  • 8) 0.673 704 759 04 × 2 = 1 + 0.347 409 518 08;
  • 9) 0.347 409 518 08 × 2 = 0 + 0.694 819 036 16;
  • 10) 0.694 819 036 16 × 2 = 1 + 0.389 638 072 32;
  • 11) 0.389 638 072 32 × 2 = 0 + 0.779 276 144 64;
  • 12) 0.779 276 144 64 × 2 = 1 + 0.558 552 289 28;
  • 13) 0.558 552 289 28 × 2 = 1 + 0.117 104 578 56;
  • 14) 0.117 104 578 56 × 2 = 0 + 0.234 209 157 12;
  • 15) 0.234 209 157 12 × 2 = 0 + 0.468 418 314 24;
  • 16) 0.468 418 314 24 × 2 = 0 + 0.936 836 628 48;
  • 17) 0.936 836 628 48 × 2 = 1 + 0.873 673 256 96;
  • 18) 0.873 673 256 96 × 2 = 1 + 0.747 346 513 92;
  • 19) 0.747 346 513 92 × 2 = 1 + 0.494 693 027 84;
  • 20) 0.494 693 027 84 × 2 = 0 + 0.989 386 055 68;
  • 21) 0.989 386 055 68 × 2 = 1 + 0.978 772 111 36;
  • 22) 0.978 772 111 36 × 2 = 1 + 0.957 544 222 72;
  • 23) 0.957 544 222 72 × 2 = 1 + 0.915 088 445 44;
  • 24) 0.915 088 445 44 × 2 = 1 + 0.830 176 890 88;
  • 25) 0.830 176 890 88 × 2 = 1 + 0.660 353 781 76;
  • 26) 0.660 353 781 76 × 2 = 1 + 0.320 707 563 52;
  • 27) 0.320 707 563 52 × 2 = 0 + 0.641 415 127 04;
  • 28) 0.641 415 127 04 × 2 = 1 + 0.282 830 254 08;
  • 29) 0.282 830 254 08 × 2 = 0 + 0.565 660 508 16;
  • 30) 0.565 660 508 16 × 2 = 1 + 0.131 321 016 32;
  • 31) 0.131 321 016 32 × 2 = 0 + 0.262 642 032 64;
  • 32) 0.262 642 032 64 × 2 = 0 + 0.525 284 065 28;
  • 33) 0.525 284 065 28 × 2 = 1 + 0.050 568 130 56;
  • 34) 0.050 568 130 56 × 2 = 0 + 0.101 136 261 12;
  • 35) 0.101 136 261 12 × 2 = 0 + 0.202 272 522 24;
  • 36) 0.202 272 522 24 × 2 = 0 + 0.404 545 044 48;
  • 37) 0.404 545 044 48 × 2 = 0 + 0.809 090 088 96;
  • 38) 0.809 090 088 96 × 2 = 1 + 0.618 180 177 92;
  • 39) 0.618 180 177 92 × 2 = 1 + 0.236 360 355 84;
  • 40) 0.236 360 355 84 × 2 = 0 + 0.472 720 711 68;
  • 41) 0.472 720 711 68 × 2 = 0 + 0.945 441 423 36;
  • 42) 0.945 441 423 36 × 2 = 1 + 0.890 882 846 72;
  • 43) 0.890 882 846 72 × 2 = 1 + 0.781 765 693 44;
  • 44) 0.781 765 693 44 × 2 = 1 + 0.563 531 386 88;
  • 45) 0.563 531 386 88 × 2 = 1 + 0.127 062 773 76;
  • 46) 0.127 062 773 76 × 2 = 0 + 0.254 125 547 52;
  • 47) 0.254 125 547 52 × 2 = 0 + 0.508 251 095 04;
  • 48) 0.508 251 095 04 × 2 = 1 + 0.016 502 190 08;
  • 49) 0.016 502 190 08 × 2 = 0 + 0.033 004 380 16;
  • 50) 0.033 004 380 16 × 2 = 0 + 0.066 008 760 32;
  • 51) 0.066 008 760 32 × 2 = 0 + 0.132 017 520 64;
  • 52) 0.132 017 520 64 × 2 = 0 + 0.264 035 041 28;
  • 53) 0.264 035 041 28 × 2 = 0 + 0.528 070 082 56;

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:


0.974 013 318 43(10) =

0.1111 1001 0101 1000 1110 1111 1101 0100 1000 0110 0111 1001 0000 0(2)

5. Positive number before normalization:

0.974 013 318 43(10) =

0.1111 1001 0101 1000 1110 1111 1101 0100 1000 0110 0111 1001 0000 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:


0.974 013 318 43(10) =

0.1111 1001 0101 1000 1110 1111 1101 0100 1000 0110 0111 1001 0000 0(2) =

0.1111 1001 0101 1000 1110 1111 1101 0100 1000 0110 0111 1001 0000 0(2) × 20 =

1.1111 0010 1011 0001 1101 1111 1010 1001 0000 1100 1111 0010 0000(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 1010 1001 0000 1100 1111 0010 0000


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 1010 1001 0000 1100 1111 0010 0000 =

1111 0010 1011 0001 1101 1111 1010 1001 0000 1100 1111 0010 0000


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 1010 1001 0000 1100 1111 0010 0000


Decimal number 0.974 013 318 43 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1110 - 1111 0010 1011 0001 1101 1111 1010 1001 0000 1100 1111 0010 0000

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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