0.545 253 866 332 628 829 603 505 342 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.545 253 866 332 628 829 603 505 342(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
0.545 253 866 332 628 829 603 505 342(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: 0.545 253 866 332 628 829 603 505 342.

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) 0.545 253 866 332 628 829 603 505 342 × 2 = 1 + 0.090 507 732 665 257 659 207 010 684;
  • 2) 0.090 507 732 665 257 659 207 010 684 × 2 = 0 + 0.181 015 465 330 515 318 414 021 368;
  • 3) 0.181 015 465 330 515 318 414 021 368 × 2 = 0 + 0.362 030 930 661 030 636 828 042 736;
  • 4) 0.362 030 930 661 030 636 828 042 736 × 2 = 0 + 0.724 061 861 322 061 273 656 085 472;
  • 5) 0.724 061 861 322 061 273 656 085 472 × 2 = 1 + 0.448 123 722 644 122 547 312 170 944;
  • 6) 0.448 123 722 644 122 547 312 170 944 × 2 = 0 + 0.896 247 445 288 245 094 624 341 888;
  • 7) 0.896 247 445 288 245 094 624 341 888 × 2 = 1 + 0.792 494 890 576 490 189 248 683 776;
  • 8) 0.792 494 890 576 490 189 248 683 776 × 2 = 1 + 0.584 989 781 152 980 378 497 367 552;
  • 9) 0.584 989 781 152 980 378 497 367 552 × 2 = 1 + 0.169 979 562 305 960 756 994 735 104;
  • 10) 0.169 979 562 305 960 756 994 735 104 × 2 = 0 + 0.339 959 124 611 921 513 989 470 208;
  • 11) 0.339 959 124 611 921 513 989 470 208 × 2 = 0 + 0.679 918 249 223 843 027 978 940 416;
  • 12) 0.679 918 249 223 843 027 978 940 416 × 2 = 1 + 0.359 836 498 447 686 055 957 880 832;
  • 13) 0.359 836 498 447 686 055 957 880 832 × 2 = 0 + 0.719 672 996 895 372 111 915 761 664;
  • 14) 0.719 672 996 895 372 111 915 761 664 × 2 = 1 + 0.439 345 993 790 744 223 831 523 328;
  • 15) 0.439 345 993 790 744 223 831 523 328 × 2 = 0 + 0.878 691 987 581 488 447 663 046 656;
  • 16) 0.878 691 987 581 488 447 663 046 656 × 2 = 1 + 0.757 383 975 162 976 895 326 093 312;
  • 17) 0.757 383 975 162 976 895 326 093 312 × 2 = 1 + 0.514 767 950 325 953 790 652 186 624;
  • 18) 0.514 767 950 325 953 790 652 186 624 × 2 = 1 + 0.029 535 900 651 907 581 304 373 248;
  • 19) 0.029 535 900 651 907 581 304 373 248 × 2 = 0 + 0.059 071 801 303 815 162 608 746 496;
  • 20) 0.059 071 801 303 815 162 608 746 496 × 2 = 0 + 0.118 143 602 607 630 325 217 492 992;
  • 21) 0.118 143 602 607 630 325 217 492 992 × 2 = 0 + 0.236 287 205 215 260 650 434 985 984;
  • 22) 0.236 287 205 215 260 650 434 985 984 × 2 = 0 + 0.472 574 410 430 521 300 869 971 968;
  • 23) 0.472 574 410 430 521 300 869 971 968 × 2 = 0 + 0.945 148 820 861 042 601 739 943 936;
  • 24) 0.945 148 820 861 042 601 739 943 936 × 2 = 1 + 0.890 297 641 722 085 203 479 887 872;
  • 25) 0.890 297 641 722 085 203 479 887 872 × 2 = 1 + 0.780 595 283 444 170 406 959 775 744;
  • 26) 0.780 595 283 444 170 406 959 775 744 × 2 = 1 + 0.561 190 566 888 340 813 919 551 488;
  • 27) 0.561 190 566 888 340 813 919 551 488 × 2 = 1 + 0.122 381 133 776 681 627 839 102 976;
  • 28) 0.122 381 133 776 681 627 839 102 976 × 2 = 0 + 0.244 762 267 553 363 255 678 205 952;
  • 29) 0.244 762 267 553 363 255 678 205 952 × 2 = 0 + 0.489 524 535 106 726 511 356 411 904;
  • 30) 0.489 524 535 106 726 511 356 411 904 × 2 = 0 + 0.979 049 070 213 453 022 712 823 808;
  • 31) 0.979 049 070 213 453 022 712 823 808 × 2 = 1 + 0.958 098 140 426 906 045 425 647 616;
  • 32) 0.958 098 140 426 906 045 425 647 616 × 2 = 1 + 0.916 196 280 853 812 090 851 295 232;
  • 33) 0.916 196 280 853 812 090 851 295 232 × 2 = 1 + 0.832 392 561 707 624 181 702 590 464;
  • 34) 0.832 392 561 707 624 181 702 590 464 × 2 = 1 + 0.664 785 123 415 248 363 405 180 928;
  • 35) 0.664 785 123 415 248 363 405 180 928 × 2 = 1 + 0.329 570 246 830 496 726 810 361 856;
  • 36) 0.329 570 246 830 496 726 810 361 856 × 2 = 0 + 0.659 140 493 660 993 453 620 723 712;
  • 37) 0.659 140 493 660 993 453 620 723 712 × 2 = 1 + 0.318 280 987 321 986 907 241 447 424;
  • 38) 0.318 280 987 321 986 907 241 447 424 × 2 = 0 + 0.636 561 974 643 973 814 482 894 848;
  • 39) 0.636 561 974 643 973 814 482 894 848 × 2 = 1 + 0.273 123 949 287 947 628 965 789 696;
  • 40) 0.273 123 949 287 947 628 965 789 696 × 2 = 0 + 0.546 247 898 575 895 257 931 579 392;
  • 41) 0.546 247 898 575 895 257 931 579 392 × 2 = 1 + 0.092 495 797 151 790 515 863 158 784;
  • 42) 0.092 495 797 151 790 515 863 158 784 × 2 = 0 + 0.184 991 594 303 581 031 726 317 568;
  • 43) 0.184 991 594 303 581 031 726 317 568 × 2 = 0 + 0.369 983 188 607 162 063 452 635 136;
  • 44) 0.369 983 188 607 162 063 452 635 136 × 2 = 0 + 0.739 966 377 214 324 126 905 270 272;
  • 45) 0.739 966 377 214 324 126 905 270 272 × 2 = 1 + 0.479 932 754 428 648 253 810 540 544;
  • 46) 0.479 932 754 428 648 253 810 540 544 × 2 = 0 + 0.959 865 508 857 296 507 621 081 088;
  • 47) 0.959 865 508 857 296 507 621 081 088 × 2 = 1 + 0.919 731 017 714 593 015 242 162 176;
  • 48) 0.919 731 017 714 593 015 242 162 176 × 2 = 1 + 0.839 462 035 429 186 030 484 324 352;
  • 49) 0.839 462 035 429 186 030 484 324 352 × 2 = 1 + 0.678 924 070 858 372 060 968 648 704;
  • 50) 0.678 924 070 858 372 060 968 648 704 × 2 = 1 + 0.357 848 141 716 744 121 937 297 408;
  • 51) 0.357 848 141 716 744 121 937 297 408 × 2 = 0 + 0.715 696 283 433 488 243 874 594 816;
  • 52) 0.715 696 283 433 488 243 874 594 816 × 2 = 1 + 0.431 392 566 866 976 487 749 189 632;
  • 53) 0.431 392 566 866 976 487 749 189 632 × 2 = 0 + 0.862 785 133 733 952 975 498 379 264;

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:


0.545 253 866 332 628 829 603 505 342(10) =

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

5. Positive number before normalization:

0.545 253 866 332 628 829 603 505 342(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:


0.545 253 866 332 628 829 603 505 342(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 0.545 253 866 332 628 829 603 505 342 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