0.738 413 072 969 749 685 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.738 413 072 969 749 685 3(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
0.738 413 072 969 749 685 3(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: 0.738 413 072 969 749 685 3.

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) 0.738 413 072 969 749 685 3 × 2 = 1 + 0.476 826 145 939 499 370 6;
  • 2) 0.476 826 145 939 499 370 6 × 2 = 0 + 0.953 652 291 878 998 741 2;
  • 3) 0.953 652 291 878 998 741 2 × 2 = 1 + 0.907 304 583 757 997 482 4;
  • 4) 0.907 304 583 757 997 482 4 × 2 = 1 + 0.814 609 167 515 994 964 8;
  • 5) 0.814 609 167 515 994 964 8 × 2 = 1 + 0.629 218 335 031 989 929 6;
  • 6) 0.629 218 335 031 989 929 6 × 2 = 1 + 0.258 436 670 063 979 859 2;
  • 7) 0.258 436 670 063 979 859 2 × 2 = 0 + 0.516 873 340 127 959 718 4;
  • 8) 0.516 873 340 127 959 718 4 × 2 = 1 + 0.033 746 680 255 919 436 8;
  • 9) 0.033 746 680 255 919 436 8 × 2 = 0 + 0.067 493 360 511 838 873 6;
  • 10) 0.067 493 360 511 838 873 6 × 2 = 0 + 0.134 986 721 023 677 747 2;
  • 11) 0.134 986 721 023 677 747 2 × 2 = 0 + 0.269 973 442 047 355 494 4;
  • 12) 0.269 973 442 047 355 494 4 × 2 = 0 + 0.539 946 884 094 710 988 8;
  • 13) 0.539 946 884 094 710 988 8 × 2 = 1 + 0.079 893 768 189 421 977 6;
  • 14) 0.079 893 768 189 421 977 6 × 2 = 0 + 0.159 787 536 378 843 955 2;
  • 15) 0.159 787 536 378 843 955 2 × 2 = 0 + 0.319 575 072 757 687 910 4;
  • 16) 0.319 575 072 757 687 910 4 × 2 = 0 + 0.639 150 145 515 375 820 8;
  • 17) 0.639 150 145 515 375 820 8 × 2 = 1 + 0.278 300 291 030 751 641 6;
  • 18) 0.278 300 291 030 751 641 6 × 2 = 0 + 0.556 600 582 061 503 283 2;
  • 19) 0.556 600 582 061 503 283 2 × 2 = 1 + 0.113 201 164 123 006 566 4;
  • 20) 0.113 201 164 123 006 566 4 × 2 = 0 + 0.226 402 328 246 013 132 8;
  • 21) 0.226 402 328 246 013 132 8 × 2 = 0 + 0.452 804 656 492 026 265 6;
  • 22) 0.452 804 656 492 026 265 6 × 2 = 0 + 0.905 609 312 984 052 531 2;
  • 23) 0.905 609 312 984 052 531 2 × 2 = 1 + 0.811 218 625 968 105 062 4;
  • 24) 0.811 218 625 968 105 062 4 × 2 = 1 + 0.622 437 251 936 210 124 8;
  • 25) 0.622 437 251 936 210 124 8 × 2 = 1 + 0.244 874 503 872 420 249 6;
  • 26) 0.244 874 503 872 420 249 6 × 2 = 0 + 0.489 749 007 744 840 499 2;
  • 27) 0.489 749 007 744 840 499 2 × 2 = 0 + 0.979 498 015 489 680 998 4;
  • 28) 0.979 498 015 489 680 998 4 × 2 = 1 + 0.958 996 030 979 361 996 8;
  • 29) 0.958 996 030 979 361 996 8 × 2 = 1 + 0.917 992 061 958 723 993 6;
  • 30) 0.917 992 061 958 723 993 6 × 2 = 1 + 0.835 984 123 917 447 987 2;
  • 31) 0.835 984 123 917 447 987 2 × 2 = 1 + 0.671 968 247 834 895 974 4;
  • 32) 0.671 968 247 834 895 974 4 × 2 = 1 + 0.343 936 495 669 791 948 8;
  • 33) 0.343 936 495 669 791 948 8 × 2 = 0 + 0.687 872 991 339 583 897 6;
  • 34) 0.687 872 991 339 583 897 6 × 2 = 1 + 0.375 745 982 679 167 795 2;
  • 35) 0.375 745 982 679 167 795 2 × 2 = 0 + 0.751 491 965 358 335 590 4;
  • 36) 0.751 491 965 358 335 590 4 × 2 = 1 + 0.502 983 930 716 671 180 8;
  • 37) 0.502 983 930 716 671 180 8 × 2 = 1 + 0.005 967 861 433 342 361 6;
  • 38) 0.005 967 861 433 342 361 6 × 2 = 0 + 0.011 935 722 866 684 723 2;
  • 39) 0.011 935 722 866 684 723 2 × 2 = 0 + 0.023 871 445 733 369 446 4;
  • 40) 0.023 871 445 733 369 446 4 × 2 = 0 + 0.047 742 891 466 738 892 8;
  • 41) 0.047 742 891 466 738 892 8 × 2 = 0 + 0.095 485 782 933 477 785 6;
  • 42) 0.095 485 782 933 477 785 6 × 2 = 0 + 0.190 971 565 866 955 571 2;
  • 43) 0.190 971 565 866 955 571 2 × 2 = 0 + 0.381 943 131 733 911 142 4;
  • 44) 0.381 943 131 733 911 142 4 × 2 = 0 + 0.763 886 263 467 822 284 8;
  • 45) 0.763 886 263 467 822 284 8 × 2 = 1 + 0.527 772 526 935 644 569 6;
  • 46) 0.527 772 526 935 644 569 6 × 2 = 1 + 0.055 545 053 871 289 139 2;
  • 47) 0.055 545 053 871 289 139 2 × 2 = 0 + 0.111 090 107 742 578 278 4;
  • 48) 0.111 090 107 742 578 278 4 × 2 = 0 + 0.222 180 215 485 156 556 8;
  • 49) 0.222 180 215 485 156 556 8 × 2 = 0 + 0.444 360 430 970 313 113 6;
  • 50) 0.444 360 430 970 313 113 6 × 2 = 0 + 0.888 720 861 940 626 227 2;
  • 51) 0.888 720 861 940 626 227 2 × 2 = 1 + 0.777 441 723 881 252 454 4;
  • 52) 0.777 441 723 881 252 454 4 × 2 = 1 + 0.554 883 447 762 504 908 8;
  • 53) 0.554 883 447 762 504 908 8 × 2 = 1 + 0.109 766 895 525 009 817 6;

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:


0.738 413 072 969 749 685 3(10) =

0.1011 1101 0000 1000 1010 0011 1001 1111 0101 1000 0000 1100 0011 1(2)

5. Positive number before normalization:

0.738 413 072 969 749 685 3(10) =

0.1011 1101 0000 1000 1010 0011 1001 1111 0101 1000 0000 1100 0011 1(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:


0.738 413 072 969 749 685 3(10) =

0.1011 1101 0000 1000 1010 0011 1001 1111 0101 1000 0000 1100 0011 1(2) =

0.1011 1101 0000 1000 1010 0011 1001 1111 0101 1000 0000 1100 0011 1(2) × 20 =

1.0111 1010 0001 0001 0100 0111 0011 1110 1011 0000 0001 1000 0111(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):
1.0111 1010 0001 0001 0100 0111 0011 1110 1011 0000 0001 1000 0111


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. 0111 1010 0001 0001 0100 0111 0011 1110 1011 0000 0001 1000 0111 =

0111 1010 0001 0001 0100 0111 0011 1110 1011 0000 0001 1000 0111


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) =
0111 1010 0001 0001 0100 0111 0011 1110 1011 0000 0001 1000 0111


Decimal number 0.738 413 072 969 749 685 3 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1110 - 0111 1010 0001 0001 0100 0111 0011 1110 1011 0000 0001 1000 0111

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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