5.919 999 999 999 999 928 954 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 5.919 999 999 999 999 928 954(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
5.919 999 999 999 999 928 954(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: 5.
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;
  • 5 ÷ 2 = 2 + 1;
  • 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.

5(10) =

101(2)


3. Convert to binary (base 2) the fractional part: 0.919 999 999 999 999 928 954.

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) 0.919 999 999 999 999 928 954 × 2 = 1 + 0.839 999 999 999 999 857 908;
  • 2) 0.839 999 999 999 999 857 908 × 2 = 1 + 0.679 999 999 999 999 715 816;
  • 3) 0.679 999 999 999 999 715 816 × 2 = 1 + 0.359 999 999 999 999 431 632;
  • 4) 0.359 999 999 999 999 431 632 × 2 = 0 + 0.719 999 999 999 998 863 264;
  • 5) 0.719 999 999 999 998 863 264 × 2 = 1 + 0.439 999 999 999 997 726 528;
  • 6) 0.439 999 999 999 997 726 528 × 2 = 0 + 0.879 999 999 999 995 453 056;
  • 7) 0.879 999 999 999 995 453 056 × 2 = 1 + 0.759 999 999 999 990 906 112;
  • 8) 0.759 999 999 999 990 906 112 × 2 = 1 + 0.519 999 999 999 981 812 224;
  • 9) 0.519 999 999 999 981 812 224 × 2 = 1 + 0.039 999 999 999 963 624 448;
  • 10) 0.039 999 999 999 963 624 448 × 2 = 0 + 0.079 999 999 999 927 248 896;
  • 11) 0.079 999 999 999 927 248 896 × 2 = 0 + 0.159 999 999 999 854 497 792;
  • 12) 0.159 999 999 999 854 497 792 × 2 = 0 + 0.319 999 999 999 708 995 584;
  • 13) 0.319 999 999 999 708 995 584 × 2 = 0 + 0.639 999 999 999 417 991 168;
  • 14) 0.639 999 999 999 417 991 168 × 2 = 1 + 0.279 999 999 998 835 982 336;
  • 15) 0.279 999 999 998 835 982 336 × 2 = 0 + 0.559 999 999 997 671 964 672;
  • 16) 0.559 999 999 997 671 964 672 × 2 = 1 + 0.119 999 999 995 343 929 344;
  • 17) 0.119 999 999 995 343 929 344 × 2 = 0 + 0.239 999 999 990 687 858 688;
  • 18) 0.239 999 999 990 687 858 688 × 2 = 0 + 0.479 999 999 981 375 717 376;
  • 19) 0.479 999 999 981 375 717 376 × 2 = 0 + 0.959 999 999 962 751 434 752;
  • 20) 0.959 999 999 962 751 434 752 × 2 = 1 + 0.919 999 999 925 502 869 504;
  • 21) 0.919 999 999 925 502 869 504 × 2 = 1 + 0.839 999 999 851 005 739 008;
  • 22) 0.839 999 999 851 005 739 008 × 2 = 1 + 0.679 999 999 702 011 478 016;
  • 23) 0.679 999 999 702 011 478 016 × 2 = 1 + 0.359 999 999 404 022 956 032;
  • 24) 0.359 999 999 404 022 956 032 × 2 = 0 + 0.719 999 998 808 045 912 064;
  • 25) 0.719 999 998 808 045 912 064 × 2 = 1 + 0.439 999 997 616 091 824 128;
  • 26) 0.439 999 997 616 091 824 128 × 2 = 0 + 0.879 999 995 232 183 648 256;
  • 27) 0.879 999 995 232 183 648 256 × 2 = 1 + 0.759 999 990 464 367 296 512;
  • 28) 0.759 999 990 464 367 296 512 × 2 = 1 + 0.519 999 980 928 734 593 024;
  • 29) 0.519 999 980 928 734 593 024 × 2 = 1 + 0.039 999 961 857 469 186 048;
  • 30) 0.039 999 961 857 469 186 048 × 2 = 0 + 0.079 999 923 714 938 372 096;
  • 31) 0.079 999 923 714 938 372 096 × 2 = 0 + 0.159 999 847 429 876 744 192;
  • 32) 0.159 999 847 429 876 744 192 × 2 = 0 + 0.319 999 694 859 753 488 384;
  • 33) 0.319 999 694 859 753 488 384 × 2 = 0 + 0.639 999 389 719 506 976 768;
  • 34) 0.639 999 389 719 506 976 768 × 2 = 1 + 0.279 998 779 439 013 953 536;
  • 35) 0.279 998 779 439 013 953 536 × 2 = 0 + 0.559 997 558 878 027 907 072;
  • 36) 0.559 997 558 878 027 907 072 × 2 = 1 + 0.119 995 117 756 055 814 144;
  • 37) 0.119 995 117 756 055 814 144 × 2 = 0 + 0.239 990 235 512 111 628 288;
  • 38) 0.239 990 235 512 111 628 288 × 2 = 0 + 0.479 980 471 024 223 256 576;
  • 39) 0.479 980 471 024 223 256 576 × 2 = 0 + 0.959 960 942 048 446 513 152;
  • 40) 0.959 960 942 048 446 513 152 × 2 = 1 + 0.919 921 884 096 893 026 304;
  • 41) 0.919 921 884 096 893 026 304 × 2 = 1 + 0.839 843 768 193 786 052 608;
  • 42) 0.839 843 768 193 786 052 608 × 2 = 1 + 0.679 687 536 387 572 105 216;
  • 43) 0.679 687 536 387 572 105 216 × 2 = 1 + 0.359 375 072 775 144 210 432;
  • 44) 0.359 375 072 775 144 210 432 × 2 = 0 + 0.718 750 145 550 288 420 864;
  • 45) 0.718 750 145 550 288 420 864 × 2 = 1 + 0.437 500 291 100 576 841 728;
  • 46) 0.437 500 291 100 576 841 728 × 2 = 0 + 0.875 000 582 201 153 683 456;
  • 47) 0.875 000 582 201 153 683 456 × 2 = 1 + 0.750 001 164 402 307 366 912;
  • 48) 0.750 001 164 402 307 366 912 × 2 = 1 + 0.500 002 328 804 614 733 824;
  • 49) 0.500 002 328 804 614 733 824 × 2 = 1 + 0.000 004 657 609 229 467 648;
  • 50) 0.000 004 657 609 229 467 648 × 2 = 0 + 0.000 009 315 218 458 935 296;
  • 51) 0.000 009 315 218 458 935 296 × 2 = 0 + 0.000 018 630 436 917 870 592;
  • 52) 0.000 018 630 436 917 870 592 × 2 = 0 + 0.000 037 260 873 835 741 184;
  • 53) 0.000 037 260 873 835 741 184 × 2 = 0 + 0.000 074 521 747 671 482 368;

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:


0.919 999 999 999 999 928 954(10) =

0.1110 1011 1000 0101 0001 1110 1011 1000 0101 0001 1110 1011 1000 0(2)

5. Positive number before normalization:

5.919 999 999 999 999 928 954(10) =

101.1110 1011 1000 0101 0001 1110 1011 1000 0101 0001 1110 1011 1000 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 2 positions to the left, so that only one non zero digit remains to the left of it:


5.919 999 999 999 999 928 954(10) =

101.1110 1011 1000 0101 0001 1110 1011 1000 0101 0001 1110 1011 1000 0(2) =

101.1110 1011 1000 0101 0001 1110 1011 1000 0101 0001 1110 1011 1000 0(2) × 20 =

1.0111 1010 1110 0001 0100 0111 1010 1110 0001 0100 0111 1010 1110 000(2) × 22


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): 2

Mantissa (not normalized):
1.0111 1010 1110 0001 0100 0111 1010 1110 0001 0100 0111 1010 1110 000


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

2 + 2(11-1) - 1 =

(2 + 1 023)(10) =

1 025(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;
  • 1 025 ÷ 2 = 512 + 1;
  • 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) =

1025(10) =

100 0000 0001(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. 0111 1010 1110 0001 0100 0111 1010 1110 0001 0100 0111 1010 1110 000 =

0111 1010 1110 0001 0100 0111 1010 1110 0001 0100 0111 1010 1110


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 0001

Mantissa (52 bits) =
0111 1010 1110 0001 0100 0111 1010 1110 0001 0100 0111 1010 1110


Decimal number 5.919 999 999 999 999 928 954 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0001 - 0111 1010 1110 0001 0100 0111 1010 1110 0001 0100 0111 1010 1110

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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