0.545 253 866 332 628 829 603 506 47 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.545 253 866 332 628 829 603 506 47(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
0.545 253 866 332 628 829 603 506 47(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: 0.545 253 866 332 628 829 603 506 47.

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) 0.545 253 866 332 628 829 603 506 47 × 2 = 1 + 0.090 507 732 665 257 659 207 012 94;
  • 2) 0.090 507 732 665 257 659 207 012 94 × 2 = 0 + 0.181 015 465 330 515 318 414 025 88;
  • 3) 0.181 015 465 330 515 318 414 025 88 × 2 = 0 + 0.362 030 930 661 030 636 828 051 76;
  • 4) 0.362 030 930 661 030 636 828 051 76 × 2 = 0 + 0.724 061 861 322 061 273 656 103 52;
  • 5) 0.724 061 861 322 061 273 656 103 52 × 2 = 1 + 0.448 123 722 644 122 547 312 207 04;
  • 6) 0.448 123 722 644 122 547 312 207 04 × 2 = 0 + 0.896 247 445 288 245 094 624 414 08;
  • 7) 0.896 247 445 288 245 094 624 414 08 × 2 = 1 + 0.792 494 890 576 490 189 248 828 16;
  • 8) 0.792 494 890 576 490 189 248 828 16 × 2 = 1 + 0.584 989 781 152 980 378 497 656 32;
  • 9) 0.584 989 781 152 980 378 497 656 32 × 2 = 1 + 0.169 979 562 305 960 756 995 312 64;
  • 10) 0.169 979 562 305 960 756 995 312 64 × 2 = 0 + 0.339 959 124 611 921 513 990 625 28;
  • 11) 0.339 959 124 611 921 513 990 625 28 × 2 = 0 + 0.679 918 249 223 843 027 981 250 56;
  • 12) 0.679 918 249 223 843 027 981 250 56 × 2 = 1 + 0.359 836 498 447 686 055 962 501 12;
  • 13) 0.359 836 498 447 686 055 962 501 12 × 2 = 0 + 0.719 672 996 895 372 111 925 002 24;
  • 14) 0.719 672 996 895 372 111 925 002 24 × 2 = 1 + 0.439 345 993 790 744 223 850 004 48;
  • 15) 0.439 345 993 790 744 223 850 004 48 × 2 = 0 + 0.878 691 987 581 488 447 700 008 96;
  • 16) 0.878 691 987 581 488 447 700 008 96 × 2 = 1 + 0.757 383 975 162 976 895 400 017 92;
  • 17) 0.757 383 975 162 976 895 400 017 92 × 2 = 1 + 0.514 767 950 325 953 790 800 035 84;
  • 18) 0.514 767 950 325 953 790 800 035 84 × 2 = 1 + 0.029 535 900 651 907 581 600 071 68;
  • 19) 0.029 535 900 651 907 581 600 071 68 × 2 = 0 + 0.059 071 801 303 815 163 200 143 36;
  • 20) 0.059 071 801 303 815 163 200 143 36 × 2 = 0 + 0.118 143 602 607 630 326 400 286 72;
  • 21) 0.118 143 602 607 630 326 400 286 72 × 2 = 0 + 0.236 287 205 215 260 652 800 573 44;
  • 22) 0.236 287 205 215 260 652 800 573 44 × 2 = 0 + 0.472 574 410 430 521 305 601 146 88;
  • 23) 0.472 574 410 430 521 305 601 146 88 × 2 = 0 + 0.945 148 820 861 042 611 202 293 76;
  • 24) 0.945 148 820 861 042 611 202 293 76 × 2 = 1 + 0.890 297 641 722 085 222 404 587 52;
  • 25) 0.890 297 641 722 085 222 404 587 52 × 2 = 1 + 0.780 595 283 444 170 444 809 175 04;
  • 26) 0.780 595 283 444 170 444 809 175 04 × 2 = 1 + 0.561 190 566 888 340 889 618 350 08;
  • 27) 0.561 190 566 888 340 889 618 350 08 × 2 = 1 + 0.122 381 133 776 681 779 236 700 16;
  • 28) 0.122 381 133 776 681 779 236 700 16 × 2 = 0 + 0.244 762 267 553 363 558 473 400 32;
  • 29) 0.244 762 267 553 363 558 473 400 32 × 2 = 0 + 0.489 524 535 106 727 116 946 800 64;
  • 30) 0.489 524 535 106 727 116 946 800 64 × 2 = 0 + 0.979 049 070 213 454 233 893 601 28;
  • 31) 0.979 049 070 213 454 233 893 601 28 × 2 = 1 + 0.958 098 140 426 908 467 787 202 56;
  • 32) 0.958 098 140 426 908 467 787 202 56 × 2 = 1 + 0.916 196 280 853 816 935 574 405 12;
  • 33) 0.916 196 280 853 816 935 574 405 12 × 2 = 1 + 0.832 392 561 707 633 871 148 810 24;
  • 34) 0.832 392 561 707 633 871 148 810 24 × 2 = 1 + 0.664 785 123 415 267 742 297 620 48;
  • 35) 0.664 785 123 415 267 742 297 620 48 × 2 = 1 + 0.329 570 246 830 535 484 595 240 96;
  • 36) 0.329 570 246 830 535 484 595 240 96 × 2 = 0 + 0.659 140 493 661 070 969 190 481 92;
  • 37) 0.659 140 493 661 070 969 190 481 92 × 2 = 1 + 0.318 280 987 322 141 938 380 963 84;
  • 38) 0.318 280 987 322 141 938 380 963 84 × 2 = 0 + 0.636 561 974 644 283 876 761 927 68;
  • 39) 0.636 561 974 644 283 876 761 927 68 × 2 = 1 + 0.273 123 949 288 567 753 523 855 36;
  • 40) 0.273 123 949 288 567 753 523 855 36 × 2 = 0 + 0.546 247 898 577 135 507 047 710 72;
  • 41) 0.546 247 898 577 135 507 047 710 72 × 2 = 1 + 0.092 495 797 154 271 014 095 421 44;
  • 42) 0.092 495 797 154 271 014 095 421 44 × 2 = 0 + 0.184 991 594 308 542 028 190 842 88;
  • 43) 0.184 991 594 308 542 028 190 842 88 × 2 = 0 + 0.369 983 188 617 084 056 381 685 76;
  • 44) 0.369 983 188 617 084 056 381 685 76 × 2 = 0 + 0.739 966 377 234 168 112 763 371 52;
  • 45) 0.739 966 377 234 168 112 763 371 52 × 2 = 1 + 0.479 932 754 468 336 225 526 743 04;
  • 46) 0.479 932 754 468 336 225 526 743 04 × 2 = 0 + 0.959 865 508 936 672 451 053 486 08;
  • 47) 0.959 865 508 936 672 451 053 486 08 × 2 = 1 + 0.919 731 017 873 344 902 106 972 16;
  • 48) 0.919 731 017 873 344 902 106 972 16 × 2 = 1 + 0.839 462 035 746 689 804 213 944 32;
  • 49) 0.839 462 035 746 689 804 213 944 32 × 2 = 1 + 0.678 924 071 493 379 608 427 888 64;
  • 50) 0.678 924 071 493 379 608 427 888 64 × 2 = 1 + 0.357 848 142 986 759 216 855 777 28;
  • 51) 0.357 848 142 986 759 216 855 777 28 × 2 = 0 + 0.715 696 285 973 518 433 711 554 56;
  • 52) 0.715 696 285 973 518 433 711 554 56 × 2 = 1 + 0.431 392 571 947 036 867 423 109 12;
  • 53) 0.431 392 571 947 036 867 423 109 12 × 2 = 0 + 0.862 785 143 894 073 734 846 218 24;

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:


0.545 253 866 332 628 829 603 506 47(10) =

0.1000 1011 1001 0101 1100 0001 1110 0011 1110 1010 1000 1011 1101 0(2)

5. Positive number before normalization:

0.545 253 866 332 628 829 603 506 47(10) =

0.1000 1011 1001 0101 1100 0001 1110 0011 1110 1010 1000 1011 1101 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:


0.545 253 866 332 628 829 603 506 47(10) =

0.1000 1011 1001 0101 1100 0001 1110 0011 1110 1010 1000 1011 1101 0(2) =

0.1000 1011 1001 0101 1100 0001 1110 0011 1110 1010 1000 1011 1101 0(2) × 20 =

1.0001 0111 0010 1011 1000 0011 1100 0111 1101 0101 0001 0111 1010(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):
1.0001 0111 0010 1011 1000 0011 1100 0111 1101 0101 0001 0111 1010


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. 0001 0111 0010 1011 1000 0011 1100 0111 1101 0101 0001 0111 1010 =

0001 0111 0010 1011 1000 0011 1100 0111 1101 0101 0001 0111 1010


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) =
0001 0111 0010 1011 1000 0011 1100 0111 1101 0101 0001 0111 1010


Decimal number 0.545 253 866 332 628 829 603 506 47 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1110 - 0001 0111 0010 1011 1000 0011 1100 0111 1101 0101 0001 0111 1010

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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