0.974 013 321 42 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.974 013 321 42(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 42(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 42.

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 42 × 2 = 1 + 0.948 026 642 84;
  • 2) 0.948 026 642 84 × 2 = 1 + 0.896 053 285 68;
  • 3) 0.896 053 285 68 × 2 = 1 + 0.792 106 571 36;
  • 4) 0.792 106 571 36 × 2 = 1 + 0.584 213 142 72;
  • 5) 0.584 213 142 72 × 2 = 1 + 0.168 426 285 44;
  • 6) 0.168 426 285 44 × 2 = 0 + 0.336 852 570 88;
  • 7) 0.336 852 570 88 × 2 = 0 + 0.673 705 141 76;
  • 8) 0.673 705 141 76 × 2 = 1 + 0.347 410 283 52;
  • 9) 0.347 410 283 52 × 2 = 0 + 0.694 820 567 04;
  • 10) 0.694 820 567 04 × 2 = 1 + 0.389 641 134 08;
  • 11) 0.389 641 134 08 × 2 = 0 + 0.779 282 268 16;
  • 12) 0.779 282 268 16 × 2 = 1 + 0.558 564 536 32;
  • 13) 0.558 564 536 32 × 2 = 1 + 0.117 129 072 64;
  • 14) 0.117 129 072 64 × 2 = 0 + 0.234 258 145 28;
  • 15) 0.234 258 145 28 × 2 = 0 + 0.468 516 290 56;
  • 16) 0.468 516 290 56 × 2 = 0 + 0.937 032 581 12;
  • 17) 0.937 032 581 12 × 2 = 1 + 0.874 065 162 24;
  • 18) 0.874 065 162 24 × 2 = 1 + 0.748 130 324 48;
  • 19) 0.748 130 324 48 × 2 = 1 + 0.496 260 648 96;
  • 20) 0.496 260 648 96 × 2 = 0 + 0.992 521 297 92;
  • 21) 0.992 521 297 92 × 2 = 1 + 0.985 042 595 84;
  • 22) 0.985 042 595 84 × 2 = 1 + 0.970 085 191 68;
  • 23) 0.970 085 191 68 × 2 = 1 + 0.940 170 383 36;
  • 24) 0.940 170 383 36 × 2 = 1 + 0.880 340 766 72;
  • 25) 0.880 340 766 72 × 2 = 1 + 0.760 681 533 44;
  • 26) 0.760 681 533 44 × 2 = 1 + 0.521 363 066 88;
  • 27) 0.521 363 066 88 × 2 = 1 + 0.042 726 133 76;
  • 28) 0.042 726 133 76 × 2 = 0 + 0.085 452 267 52;
  • 29) 0.085 452 267 52 × 2 = 0 + 0.170 904 535 04;
  • 30) 0.170 904 535 04 × 2 = 0 + 0.341 809 070 08;
  • 31) 0.341 809 070 08 × 2 = 0 + 0.683 618 140 16;
  • 32) 0.683 618 140 16 × 2 = 1 + 0.367 236 280 32;
  • 33) 0.367 236 280 32 × 2 = 0 + 0.734 472 560 64;
  • 34) 0.734 472 560 64 × 2 = 1 + 0.468 945 121 28;
  • 35) 0.468 945 121 28 × 2 = 0 + 0.937 890 242 56;
  • 36) 0.937 890 242 56 × 2 = 1 + 0.875 780 485 12;
  • 37) 0.875 780 485 12 × 2 = 1 + 0.751 560 970 24;
  • 38) 0.751 560 970 24 × 2 = 1 + 0.503 121 940 48;
  • 39) 0.503 121 940 48 × 2 = 1 + 0.006 243 880 96;
  • 40) 0.006 243 880 96 × 2 = 0 + 0.012 487 761 92;
  • 41) 0.012 487 761 92 × 2 = 0 + 0.024 975 523 84;
  • 42) 0.024 975 523 84 × 2 = 0 + 0.049 951 047 68;
  • 43) 0.049 951 047 68 × 2 = 0 + 0.099 902 095 36;
  • 44) 0.099 902 095 36 × 2 = 0 + 0.199 804 190 72;
  • 45) 0.199 804 190 72 × 2 = 0 + 0.399 608 381 44;
  • 46) 0.399 608 381 44 × 2 = 0 + 0.799 216 762 88;
  • 47) 0.799 216 762 88 × 2 = 1 + 0.598 433 525 76;
  • 48) 0.598 433 525 76 × 2 = 1 + 0.196 867 051 52;
  • 49) 0.196 867 051 52 × 2 = 0 + 0.393 734 103 04;
  • 50) 0.393 734 103 04 × 2 = 0 + 0.787 468 206 08;
  • 51) 0.787 468 206 08 × 2 = 1 + 0.574 936 412 16;
  • 52) 0.574 936 412 16 × 2 = 1 + 0.149 872 824 32;
  • 53) 0.149 872 824 32 × 2 = 0 + 0.299 745 648 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).


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 42(10) =

0.1111 1001 0101 1000 1110 1111 1110 0001 0101 1110 0000 0011 0011 0(2)

5. Positive number before normalization:

0.974 013 321 42(10) =

0.1111 1001 0101 1000 1110 1111 1110 0001 0101 1110 0000 0011 0011 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 42(10) =

0.1111 1001 0101 1000 1110 1111 1110 0001 0101 1110 0000 0011 0011 0(2) =

0.1111 1001 0101 1000 1110 1111 1110 0001 0101 1110 0000 0011 0011 0(2) × 20 =

1.1111 0010 1011 0001 1101 1111 1100 0010 1011 1100 0000 0110 0110(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 1011 1100 0000 0110 0110


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 1011 1100 0000 0110 0110 =

1111 0010 1011 0001 1101 1111 1100 0010 1011 1100 0000 0110 0110


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 1011 1100 0000 0110 0110


Decimal number 0.974 013 321 42 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1110 - 1111 0010 1011 0001 1101 1111 1100 0010 1011 1100 0000 0110 0110

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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