2.356 194 490 192 344 928 855 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 2.356 194 490 192 344 928 855(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
2.356 194 490 192 344 928 855(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: 2.
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;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

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.

2(10) =

10(2)


3. Convert to binary (base 2) the fractional part: 0.356 194 490 192 344 928 855.

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) 0.356 194 490 192 344 928 855 × 2 = 0 + 0.712 388 980 384 689 857 71;
  • 2) 0.712 388 980 384 689 857 71 × 2 = 1 + 0.424 777 960 769 379 715 42;
  • 3) 0.424 777 960 769 379 715 42 × 2 = 0 + 0.849 555 921 538 759 430 84;
  • 4) 0.849 555 921 538 759 430 84 × 2 = 1 + 0.699 111 843 077 518 861 68;
  • 5) 0.699 111 843 077 518 861 68 × 2 = 1 + 0.398 223 686 155 037 723 36;
  • 6) 0.398 223 686 155 037 723 36 × 2 = 0 + 0.796 447 372 310 075 446 72;
  • 7) 0.796 447 372 310 075 446 72 × 2 = 1 + 0.592 894 744 620 150 893 44;
  • 8) 0.592 894 744 620 150 893 44 × 2 = 1 + 0.185 789 489 240 301 786 88;
  • 9) 0.185 789 489 240 301 786 88 × 2 = 0 + 0.371 578 978 480 603 573 76;
  • 10) 0.371 578 978 480 603 573 76 × 2 = 0 + 0.743 157 956 961 207 147 52;
  • 11) 0.743 157 956 961 207 147 52 × 2 = 1 + 0.486 315 913 922 414 295 04;
  • 12) 0.486 315 913 922 414 295 04 × 2 = 0 + 0.972 631 827 844 828 590 08;
  • 13) 0.972 631 827 844 828 590 08 × 2 = 1 + 0.945 263 655 689 657 180 16;
  • 14) 0.945 263 655 689 657 180 16 × 2 = 1 + 0.890 527 311 379 314 360 32;
  • 15) 0.890 527 311 379 314 360 32 × 2 = 1 + 0.781 054 622 758 628 720 64;
  • 16) 0.781 054 622 758 628 720 64 × 2 = 1 + 0.562 109 245 517 257 441 28;
  • 17) 0.562 109 245 517 257 441 28 × 2 = 1 + 0.124 218 491 034 514 882 56;
  • 18) 0.124 218 491 034 514 882 56 × 2 = 0 + 0.248 436 982 069 029 765 12;
  • 19) 0.248 436 982 069 029 765 12 × 2 = 0 + 0.496 873 964 138 059 530 24;
  • 20) 0.496 873 964 138 059 530 24 × 2 = 0 + 0.993 747 928 276 119 060 48;
  • 21) 0.993 747 928 276 119 060 48 × 2 = 1 + 0.987 495 856 552 238 120 96;
  • 22) 0.987 495 856 552 238 120 96 × 2 = 1 + 0.974 991 713 104 476 241 92;
  • 23) 0.974 991 713 104 476 241 92 × 2 = 1 + 0.949 983 426 208 952 483 84;
  • 24) 0.949 983 426 208 952 483 84 × 2 = 1 + 0.899 966 852 417 904 967 68;
  • 25) 0.899 966 852 417 904 967 68 × 2 = 1 + 0.799 933 704 835 809 935 36;
  • 26) 0.799 933 704 835 809 935 36 × 2 = 1 + 0.599 867 409 671 619 870 72;
  • 27) 0.599 867 409 671 619 870 72 × 2 = 1 + 0.199 734 819 343 239 741 44;
  • 28) 0.199 734 819 343 239 741 44 × 2 = 0 + 0.399 469 638 686 479 482 88;
  • 29) 0.399 469 638 686 479 482 88 × 2 = 0 + 0.798 939 277 372 958 965 76;
  • 30) 0.798 939 277 372 958 965 76 × 2 = 1 + 0.597 878 554 745 917 931 52;
  • 31) 0.597 878 554 745 917 931 52 × 2 = 1 + 0.195 757 109 491 835 863 04;
  • 32) 0.195 757 109 491 835 863 04 × 2 = 0 + 0.391 514 218 983 671 726 08;
  • 33) 0.391 514 218 983 671 726 08 × 2 = 0 + 0.783 028 437 967 343 452 16;
  • 34) 0.783 028 437 967 343 452 16 × 2 = 1 + 0.566 056 875 934 686 904 32;
  • 35) 0.566 056 875 934 686 904 32 × 2 = 1 + 0.132 113 751 869 373 808 64;
  • 36) 0.132 113 751 869 373 808 64 × 2 = 0 + 0.264 227 503 738 747 617 28;
  • 37) 0.264 227 503 738 747 617 28 × 2 = 0 + 0.528 455 007 477 495 234 56;
  • 38) 0.528 455 007 477 495 234 56 × 2 = 1 + 0.056 910 014 954 990 469 12;
  • 39) 0.056 910 014 954 990 469 12 × 2 = 0 + 0.113 820 029 909 980 938 24;
  • 40) 0.113 820 029 909 980 938 24 × 2 = 0 + 0.227 640 059 819 961 876 48;
  • 41) 0.227 640 059 819 961 876 48 × 2 = 0 + 0.455 280 119 639 923 752 96;
  • 42) 0.455 280 119 639 923 752 96 × 2 = 0 + 0.910 560 239 279 847 505 92;
  • 43) 0.910 560 239 279 847 505 92 × 2 = 1 + 0.821 120 478 559 695 011 84;
  • 44) 0.821 120 478 559 695 011 84 × 2 = 1 + 0.642 240 957 119 390 023 68;
  • 45) 0.642 240 957 119 390 023 68 × 2 = 1 + 0.284 481 914 238 780 047 36;
  • 46) 0.284 481 914 238 780 047 36 × 2 = 0 + 0.568 963 828 477 560 094 72;
  • 47) 0.568 963 828 477 560 094 72 × 2 = 1 + 0.137 927 656 955 120 189 44;
  • 48) 0.137 927 656 955 120 189 44 × 2 = 0 + 0.275 855 313 910 240 378 88;
  • 49) 0.275 855 313 910 240 378 88 × 2 = 0 + 0.551 710 627 820 480 757 76;
  • 50) 0.551 710 627 820 480 757 76 × 2 = 1 + 0.103 421 255 640 961 515 52;
  • 51) 0.103 421 255 640 961 515 52 × 2 = 0 + 0.206 842 511 281 923 031 04;
  • 52) 0.206 842 511 281 923 031 04 × 2 = 0 + 0.413 685 022 563 846 062 08;
  • 53) 0.413 685 022 563 846 062 08 × 2 = 0 + 0.827 370 045 127 692 124 16;

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:


0.356 194 490 192 344 928 855(10) =

0.0101 1011 0010 1111 1000 1111 1110 0110 0110 0100 0011 1010 0100 0(2)

5. Positive number before normalization:

2.356 194 490 192 344 928 855(10) =

10.0101 1011 0010 1111 1000 1111 1110 0110 0110 0100 0011 1010 0100 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 1 positions to the left, so that only one non zero digit remains to the left of it:


2.356 194 490 192 344 928 855(10) =

10.0101 1011 0010 1111 1000 1111 1110 0110 0110 0100 0011 1010 0100 0(2) =

10.0101 1011 0010 1111 1000 1111 1110 0110 0110 0100 0011 1010 0100 0(2) × 20 =

1.0010 1101 1001 0111 1100 0111 1111 0011 0011 0010 0001 1101 0010 00(2) × 21


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):
1.0010 1101 1001 0111 1100 0111 1111 0011 0011 0010 0001 1101 0010 00


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) =

1 024(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;
  • 1 024 ÷ 2 = 512 + 0;
  • 512 ÷ 2 = 256 + 0;
  • 256 ÷ 2 = 128 + 0;
  • 128 ÷ 2 = 64 + 0;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 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) =

1024(10) =

100 0000 0000(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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =

1. 0010 1101 1001 0111 1100 0111 1111 0011 0011 0010 0001 1101 0010 00 =

0010 1101 1001 0111 1100 0111 1111 0011 0011 0010 0001 1101 0010


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) =
100 0000 0000

Mantissa (52 bits) =
0010 1101 1001 0111 1100 0111 1111 0011 0011 0010 0001 1101 0010


Decimal number 2.356 194 490 192 344 928 855 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0000 - 0010 1101 1001 0111 1100 0111 1111 0011 0011 0010 0001 1101 0010

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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