0.738 413 072 969 749 654 83 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.738 413 072 969 749 654 83(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
0.738 413 072 969 749 654 83(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: 0.738 413 072 969 749 654 83.

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) 0.738 413 072 969 749 654 83 × 2 = 1 + 0.476 826 145 939 499 309 66;
  • 2) 0.476 826 145 939 499 309 66 × 2 = 0 + 0.953 652 291 878 998 619 32;
  • 3) 0.953 652 291 878 998 619 32 × 2 = 1 + 0.907 304 583 757 997 238 64;
  • 4) 0.907 304 583 757 997 238 64 × 2 = 1 + 0.814 609 167 515 994 477 28;
  • 5) 0.814 609 167 515 994 477 28 × 2 = 1 + 0.629 218 335 031 988 954 56;
  • 6) 0.629 218 335 031 988 954 56 × 2 = 1 + 0.258 436 670 063 977 909 12;
  • 7) 0.258 436 670 063 977 909 12 × 2 = 0 + 0.516 873 340 127 955 818 24;
  • 8) 0.516 873 340 127 955 818 24 × 2 = 1 + 0.033 746 680 255 911 636 48;
  • 9) 0.033 746 680 255 911 636 48 × 2 = 0 + 0.067 493 360 511 823 272 96;
  • 10) 0.067 493 360 511 823 272 96 × 2 = 0 + 0.134 986 721 023 646 545 92;
  • 11) 0.134 986 721 023 646 545 92 × 2 = 0 + 0.269 973 442 047 293 091 84;
  • 12) 0.269 973 442 047 293 091 84 × 2 = 0 + 0.539 946 884 094 586 183 68;
  • 13) 0.539 946 884 094 586 183 68 × 2 = 1 + 0.079 893 768 189 172 367 36;
  • 14) 0.079 893 768 189 172 367 36 × 2 = 0 + 0.159 787 536 378 344 734 72;
  • 15) 0.159 787 536 378 344 734 72 × 2 = 0 + 0.319 575 072 756 689 469 44;
  • 16) 0.319 575 072 756 689 469 44 × 2 = 0 + 0.639 150 145 513 378 938 88;
  • 17) 0.639 150 145 513 378 938 88 × 2 = 1 + 0.278 300 291 026 757 877 76;
  • 18) 0.278 300 291 026 757 877 76 × 2 = 0 + 0.556 600 582 053 515 755 52;
  • 19) 0.556 600 582 053 515 755 52 × 2 = 1 + 0.113 201 164 107 031 511 04;
  • 20) 0.113 201 164 107 031 511 04 × 2 = 0 + 0.226 402 328 214 063 022 08;
  • 21) 0.226 402 328 214 063 022 08 × 2 = 0 + 0.452 804 656 428 126 044 16;
  • 22) 0.452 804 656 428 126 044 16 × 2 = 0 + 0.905 609 312 856 252 088 32;
  • 23) 0.905 609 312 856 252 088 32 × 2 = 1 + 0.811 218 625 712 504 176 64;
  • 24) 0.811 218 625 712 504 176 64 × 2 = 1 + 0.622 437 251 425 008 353 28;
  • 25) 0.622 437 251 425 008 353 28 × 2 = 1 + 0.244 874 502 850 016 706 56;
  • 26) 0.244 874 502 850 016 706 56 × 2 = 0 + 0.489 749 005 700 033 413 12;
  • 27) 0.489 749 005 700 033 413 12 × 2 = 0 + 0.979 498 011 400 066 826 24;
  • 28) 0.979 498 011 400 066 826 24 × 2 = 1 + 0.958 996 022 800 133 652 48;
  • 29) 0.958 996 022 800 133 652 48 × 2 = 1 + 0.917 992 045 600 267 304 96;
  • 30) 0.917 992 045 600 267 304 96 × 2 = 1 + 0.835 984 091 200 534 609 92;
  • 31) 0.835 984 091 200 534 609 92 × 2 = 1 + 0.671 968 182 401 069 219 84;
  • 32) 0.671 968 182 401 069 219 84 × 2 = 1 + 0.343 936 364 802 138 439 68;
  • 33) 0.343 936 364 802 138 439 68 × 2 = 0 + 0.687 872 729 604 276 879 36;
  • 34) 0.687 872 729 604 276 879 36 × 2 = 1 + 0.375 745 459 208 553 758 72;
  • 35) 0.375 745 459 208 553 758 72 × 2 = 0 + 0.751 490 918 417 107 517 44;
  • 36) 0.751 490 918 417 107 517 44 × 2 = 1 + 0.502 981 836 834 215 034 88;
  • 37) 0.502 981 836 834 215 034 88 × 2 = 1 + 0.005 963 673 668 430 069 76;
  • 38) 0.005 963 673 668 430 069 76 × 2 = 0 + 0.011 927 347 336 860 139 52;
  • 39) 0.011 927 347 336 860 139 52 × 2 = 0 + 0.023 854 694 673 720 279 04;
  • 40) 0.023 854 694 673 720 279 04 × 2 = 0 + 0.047 709 389 347 440 558 08;
  • 41) 0.047 709 389 347 440 558 08 × 2 = 0 + 0.095 418 778 694 881 116 16;
  • 42) 0.095 418 778 694 881 116 16 × 2 = 0 + 0.190 837 557 389 762 232 32;
  • 43) 0.190 837 557 389 762 232 32 × 2 = 0 + 0.381 675 114 779 524 464 64;
  • 44) 0.381 675 114 779 524 464 64 × 2 = 0 + 0.763 350 229 559 048 929 28;
  • 45) 0.763 350 229 559 048 929 28 × 2 = 1 + 0.526 700 459 118 097 858 56;
  • 46) 0.526 700 459 118 097 858 56 × 2 = 1 + 0.053 400 918 236 195 717 12;
  • 47) 0.053 400 918 236 195 717 12 × 2 = 0 + 0.106 801 836 472 391 434 24;
  • 48) 0.106 801 836 472 391 434 24 × 2 = 0 + 0.213 603 672 944 782 868 48;
  • 49) 0.213 603 672 944 782 868 48 × 2 = 0 + 0.427 207 345 889 565 736 96;
  • 50) 0.427 207 345 889 565 736 96 × 2 = 0 + 0.854 414 691 779 131 473 92;
  • 51) 0.854 414 691 779 131 473 92 × 2 = 1 + 0.708 829 383 558 262 947 84;
  • 52) 0.708 829 383 558 262 947 84 × 2 = 1 + 0.417 658 767 116 525 895 68;
  • 53) 0.417 658 767 116 525 895 68 × 2 = 0 + 0.835 317 534 233 051 791 36;

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:


0.738 413 072 969 749 654 83(10) =

0.1011 1101 0000 1000 1010 0011 1001 1111 0101 1000 0000 1100 0011 0(2)

5. Positive number before normalization:

0.738 413 072 969 749 654 83(10) =

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


0.738 413 072 969 749 654 83(10) =

0.1011 1101 0000 1000 1010 0011 1001 1111 0101 1000 0000 1100 0011 0(2) =

0.1011 1101 0000 1000 1010 0011 1001 1111 0101 1000 0000 1100 0011 0(2) × 20 =

1.0111 1010 0001 0001 0100 0111 0011 1110 1011 0000 0001 1000 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):
1.0111 1010 0001 0001 0100 0111 0011 1110 1011 0000 0001 1000 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. 0111 1010 0001 0001 0100 0111 0011 1110 1011 0000 0001 1000 0110 =

0111 1010 0001 0001 0100 0111 0011 1110 1011 0000 0001 1000 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) =
0111 1010 0001 0001 0100 0111 0011 1110 1011 0000 0001 1000 0110


Decimal number 0.738 413 072 969 749 654 83 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 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