5.919 999 999 999 999 928 971 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 5.919 999 999 999 999 928 971(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
5.919 999 999 999 999 928 971(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: 0.919 999 999 999 999 928 971.

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) 0.919 999 999 999 999 928 971 × 2 = 1 + 0.839 999 999 999 999 857 942;
  • 2) 0.839 999 999 999 999 857 942 × 2 = 1 + 0.679 999 999 999 999 715 884;
  • 3) 0.679 999 999 999 999 715 884 × 2 = 1 + 0.359 999 999 999 999 431 768;
  • 4) 0.359 999 999 999 999 431 768 × 2 = 0 + 0.719 999 999 999 998 863 536;
  • 5) 0.719 999 999 999 998 863 536 × 2 = 1 + 0.439 999 999 999 997 727 072;
  • 6) 0.439 999 999 999 997 727 072 × 2 = 0 + 0.879 999 999 999 995 454 144;
  • 7) 0.879 999 999 999 995 454 144 × 2 = 1 + 0.759 999 999 999 990 908 288;
  • 8) 0.759 999 999 999 990 908 288 × 2 = 1 + 0.519 999 999 999 981 816 576;
  • 9) 0.519 999 999 999 981 816 576 × 2 = 1 + 0.039 999 999 999 963 633 152;
  • 10) 0.039 999 999 999 963 633 152 × 2 = 0 + 0.079 999 999 999 927 266 304;
  • 11) 0.079 999 999 999 927 266 304 × 2 = 0 + 0.159 999 999 999 854 532 608;
  • 12) 0.159 999 999 999 854 532 608 × 2 = 0 + 0.319 999 999 999 709 065 216;
  • 13) 0.319 999 999 999 709 065 216 × 2 = 0 + 0.639 999 999 999 418 130 432;
  • 14) 0.639 999 999 999 418 130 432 × 2 = 1 + 0.279 999 999 998 836 260 864;
  • 15) 0.279 999 999 998 836 260 864 × 2 = 0 + 0.559 999 999 997 672 521 728;
  • 16) 0.559 999 999 997 672 521 728 × 2 = 1 + 0.119 999 999 995 345 043 456;
  • 17) 0.119 999 999 995 345 043 456 × 2 = 0 + 0.239 999 999 990 690 086 912;
  • 18) 0.239 999 999 990 690 086 912 × 2 = 0 + 0.479 999 999 981 380 173 824;
  • 19) 0.479 999 999 981 380 173 824 × 2 = 0 + 0.959 999 999 962 760 347 648;
  • 20) 0.959 999 999 962 760 347 648 × 2 = 1 + 0.919 999 999 925 520 695 296;
  • 21) 0.919 999 999 925 520 695 296 × 2 = 1 + 0.839 999 999 851 041 390 592;
  • 22) 0.839 999 999 851 041 390 592 × 2 = 1 + 0.679 999 999 702 082 781 184;
  • 23) 0.679 999 999 702 082 781 184 × 2 = 1 + 0.359 999 999 404 165 562 368;
  • 24) 0.359 999 999 404 165 562 368 × 2 = 0 + 0.719 999 998 808 331 124 736;
  • 25) 0.719 999 998 808 331 124 736 × 2 = 1 + 0.439 999 997 616 662 249 472;
  • 26) 0.439 999 997 616 662 249 472 × 2 = 0 + 0.879 999 995 233 324 498 944;
  • 27) 0.879 999 995 233 324 498 944 × 2 = 1 + 0.759 999 990 466 648 997 888;
  • 28) 0.759 999 990 466 648 997 888 × 2 = 1 + 0.519 999 980 933 297 995 776;
  • 29) 0.519 999 980 933 297 995 776 × 2 = 1 + 0.039 999 961 866 595 991 552;
  • 30) 0.039 999 961 866 595 991 552 × 2 = 0 + 0.079 999 923 733 191 983 104;
  • 31) 0.079 999 923 733 191 983 104 × 2 = 0 + 0.159 999 847 466 383 966 208;
  • 32) 0.159 999 847 466 383 966 208 × 2 = 0 + 0.319 999 694 932 767 932 416;
  • 33) 0.319 999 694 932 767 932 416 × 2 = 0 + 0.639 999 389 865 535 864 832;
  • 34) 0.639 999 389 865 535 864 832 × 2 = 1 + 0.279 998 779 731 071 729 664;
  • 35) 0.279 998 779 731 071 729 664 × 2 = 0 + 0.559 997 559 462 143 459 328;
  • 36) 0.559 997 559 462 143 459 328 × 2 = 1 + 0.119 995 118 924 286 918 656;
  • 37) 0.119 995 118 924 286 918 656 × 2 = 0 + 0.239 990 237 848 573 837 312;
  • 38) 0.239 990 237 848 573 837 312 × 2 = 0 + 0.479 980 475 697 147 674 624;
  • 39) 0.479 980 475 697 147 674 624 × 2 = 0 + 0.959 960 951 394 295 349 248;
  • 40) 0.959 960 951 394 295 349 248 × 2 = 1 + 0.919 921 902 788 590 698 496;
  • 41) 0.919 921 902 788 590 698 496 × 2 = 1 + 0.839 843 805 577 181 396 992;
  • 42) 0.839 843 805 577 181 396 992 × 2 = 1 + 0.679 687 611 154 362 793 984;
  • 43) 0.679 687 611 154 362 793 984 × 2 = 1 + 0.359 375 222 308 725 587 968;
  • 44) 0.359 375 222 308 725 587 968 × 2 = 0 + 0.718 750 444 617 451 175 936;
  • 45) 0.718 750 444 617 451 175 936 × 2 = 1 + 0.437 500 889 234 902 351 872;
  • 46) 0.437 500 889 234 902 351 872 × 2 = 0 + 0.875 001 778 469 804 703 744;
  • 47) 0.875 001 778 469 804 703 744 × 2 = 1 + 0.750 003 556 939 609 407 488;
  • 48) 0.750 003 556 939 609 407 488 × 2 = 1 + 0.500 007 113 879 218 814 976;
  • 49) 0.500 007 113 879 218 814 976 × 2 = 1 + 0.000 014 227 758 437 629 952;
  • 50) 0.000 014 227 758 437 629 952 × 2 = 0 + 0.000 028 455 516 875 259 904;
  • 51) 0.000 028 455 516 875 259 904 × 2 = 0 + 0.000 056 911 033 750 519 808;
  • 52) 0.000 056 911 033 750 519 808 × 2 = 0 + 0.000 113 822 067 501 039 616;
  • 53) 0.000 113 822 067 501 039 616 × 2 = 0 + 0.000 227 644 135 002 079 232;

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:


0.919 999 999 999 999 928 971(10) =

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

5. Positive number before normalization:

5.919 999 999 999 999 928 971(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:


5.919 999 999 999 999 928 971(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 5.919 999 999 999 999 928 971 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