0.738 413 072 969 749 655 19 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.738 413 072 969 749 655 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
0.738 413 072 969 749 655 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: 0.738 413 072 969 749 655 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) 0.738 413 072 969 749 655 19 × 2 = 1 + 0.476 826 145 939 499 310 38;
  • 2) 0.476 826 145 939 499 310 38 × 2 = 0 + 0.953 652 291 878 998 620 76;
  • 3) 0.953 652 291 878 998 620 76 × 2 = 1 + 0.907 304 583 757 997 241 52;
  • 4) 0.907 304 583 757 997 241 52 × 2 = 1 + 0.814 609 167 515 994 483 04;
  • 5) 0.814 609 167 515 994 483 04 × 2 = 1 + 0.629 218 335 031 988 966 08;
  • 6) 0.629 218 335 031 988 966 08 × 2 = 1 + 0.258 436 670 063 977 932 16;
  • 7) 0.258 436 670 063 977 932 16 × 2 = 0 + 0.516 873 340 127 955 864 32;
  • 8) 0.516 873 340 127 955 864 32 × 2 = 1 + 0.033 746 680 255 911 728 64;
  • 9) 0.033 746 680 255 911 728 64 × 2 = 0 + 0.067 493 360 511 823 457 28;
  • 10) 0.067 493 360 511 823 457 28 × 2 = 0 + 0.134 986 721 023 646 914 56;
  • 11) 0.134 986 721 023 646 914 56 × 2 = 0 + 0.269 973 442 047 293 829 12;
  • 12) 0.269 973 442 047 293 829 12 × 2 = 0 + 0.539 946 884 094 587 658 24;
  • 13) 0.539 946 884 094 587 658 24 × 2 = 1 + 0.079 893 768 189 175 316 48;
  • 14) 0.079 893 768 189 175 316 48 × 2 = 0 + 0.159 787 536 378 350 632 96;
  • 15) 0.159 787 536 378 350 632 96 × 2 = 0 + 0.319 575 072 756 701 265 92;
  • 16) 0.319 575 072 756 701 265 92 × 2 = 0 + 0.639 150 145 513 402 531 84;
  • 17) 0.639 150 145 513 402 531 84 × 2 = 1 + 0.278 300 291 026 805 063 68;
  • 18) 0.278 300 291 026 805 063 68 × 2 = 0 + 0.556 600 582 053 610 127 36;
  • 19) 0.556 600 582 053 610 127 36 × 2 = 1 + 0.113 201 164 107 220 254 72;
  • 20) 0.113 201 164 107 220 254 72 × 2 = 0 + 0.226 402 328 214 440 509 44;
  • 21) 0.226 402 328 214 440 509 44 × 2 = 0 + 0.452 804 656 428 881 018 88;
  • 22) 0.452 804 656 428 881 018 88 × 2 = 0 + 0.905 609 312 857 762 037 76;
  • 23) 0.905 609 312 857 762 037 76 × 2 = 1 + 0.811 218 625 715 524 075 52;
  • 24) 0.811 218 625 715 524 075 52 × 2 = 1 + 0.622 437 251 431 048 151 04;
  • 25) 0.622 437 251 431 048 151 04 × 2 = 1 + 0.244 874 502 862 096 302 08;
  • 26) 0.244 874 502 862 096 302 08 × 2 = 0 + 0.489 749 005 724 192 604 16;
  • 27) 0.489 749 005 724 192 604 16 × 2 = 0 + 0.979 498 011 448 385 208 32;
  • 28) 0.979 498 011 448 385 208 32 × 2 = 1 + 0.958 996 022 896 770 416 64;
  • 29) 0.958 996 022 896 770 416 64 × 2 = 1 + 0.917 992 045 793 540 833 28;
  • 30) 0.917 992 045 793 540 833 28 × 2 = 1 + 0.835 984 091 587 081 666 56;
  • 31) 0.835 984 091 587 081 666 56 × 2 = 1 + 0.671 968 183 174 163 333 12;
  • 32) 0.671 968 183 174 163 333 12 × 2 = 1 + 0.343 936 366 348 326 666 24;
  • 33) 0.343 936 366 348 326 666 24 × 2 = 0 + 0.687 872 732 696 653 332 48;
  • 34) 0.687 872 732 696 653 332 48 × 2 = 1 + 0.375 745 465 393 306 664 96;
  • 35) 0.375 745 465 393 306 664 96 × 2 = 0 + 0.751 490 930 786 613 329 92;
  • 36) 0.751 490 930 786 613 329 92 × 2 = 1 + 0.502 981 861 573 226 659 84;
  • 37) 0.502 981 861 573 226 659 84 × 2 = 1 + 0.005 963 723 146 453 319 68;
  • 38) 0.005 963 723 146 453 319 68 × 2 = 0 + 0.011 927 446 292 906 639 36;
  • 39) 0.011 927 446 292 906 639 36 × 2 = 0 + 0.023 854 892 585 813 278 72;
  • 40) 0.023 854 892 585 813 278 72 × 2 = 0 + 0.047 709 785 171 626 557 44;
  • 41) 0.047 709 785 171 626 557 44 × 2 = 0 + 0.095 419 570 343 253 114 88;
  • 42) 0.095 419 570 343 253 114 88 × 2 = 0 + 0.190 839 140 686 506 229 76;
  • 43) 0.190 839 140 686 506 229 76 × 2 = 0 + 0.381 678 281 373 012 459 52;
  • 44) 0.381 678 281 373 012 459 52 × 2 = 0 + 0.763 356 562 746 024 919 04;
  • 45) 0.763 356 562 746 024 919 04 × 2 = 1 + 0.526 713 125 492 049 838 08;
  • 46) 0.526 713 125 492 049 838 08 × 2 = 1 + 0.053 426 250 984 099 676 16;
  • 47) 0.053 426 250 984 099 676 16 × 2 = 0 + 0.106 852 501 968 199 352 32;
  • 48) 0.106 852 501 968 199 352 32 × 2 = 0 + 0.213 705 003 936 398 704 64;
  • 49) 0.213 705 003 936 398 704 64 × 2 = 0 + 0.427 410 007 872 797 409 28;
  • 50) 0.427 410 007 872 797 409 28 × 2 = 0 + 0.854 820 015 745 594 818 56;
  • 51) 0.854 820 015 745 594 818 56 × 2 = 1 + 0.709 640 031 491 189 637 12;
  • 52) 0.709 640 031 491 189 637 12 × 2 = 1 + 0.419 280 062 982 379 274 24;
  • 53) 0.419 280 062 982 379 274 24 × 2 = 0 + 0.838 560 125 964 758 548 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:


0.738 413 072 969 749 655 19(10) =

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

5. Positive number before normalization:

0.738 413 072 969 749 655 19(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:


0.738 413 072 969 749 655 19(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 0.738 413 072 969 749 655 19 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