0.545 253 866 332 628 829 603 505 09 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.545 253 866 332 628 829 603 505 09(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 09(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 09.

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 09 × 2 = 1 + 0.090 507 732 665 257 659 207 010 18;
  • 2) 0.090 507 732 665 257 659 207 010 18 × 2 = 0 + 0.181 015 465 330 515 318 414 020 36;
  • 3) 0.181 015 465 330 515 318 414 020 36 × 2 = 0 + 0.362 030 930 661 030 636 828 040 72;
  • 4) 0.362 030 930 661 030 636 828 040 72 × 2 = 0 + 0.724 061 861 322 061 273 656 081 44;
  • 5) 0.724 061 861 322 061 273 656 081 44 × 2 = 1 + 0.448 123 722 644 122 547 312 162 88;
  • 6) 0.448 123 722 644 122 547 312 162 88 × 2 = 0 + 0.896 247 445 288 245 094 624 325 76;
  • 7) 0.896 247 445 288 245 094 624 325 76 × 2 = 1 + 0.792 494 890 576 490 189 248 651 52;
  • 8) 0.792 494 890 576 490 189 248 651 52 × 2 = 1 + 0.584 989 781 152 980 378 497 303 04;
  • 9) 0.584 989 781 152 980 378 497 303 04 × 2 = 1 + 0.169 979 562 305 960 756 994 606 08;
  • 10) 0.169 979 562 305 960 756 994 606 08 × 2 = 0 + 0.339 959 124 611 921 513 989 212 16;
  • 11) 0.339 959 124 611 921 513 989 212 16 × 2 = 0 + 0.679 918 249 223 843 027 978 424 32;
  • 12) 0.679 918 249 223 843 027 978 424 32 × 2 = 1 + 0.359 836 498 447 686 055 956 848 64;
  • 13) 0.359 836 498 447 686 055 956 848 64 × 2 = 0 + 0.719 672 996 895 372 111 913 697 28;
  • 14) 0.719 672 996 895 372 111 913 697 28 × 2 = 1 + 0.439 345 993 790 744 223 827 394 56;
  • 15) 0.439 345 993 790 744 223 827 394 56 × 2 = 0 + 0.878 691 987 581 488 447 654 789 12;
  • 16) 0.878 691 987 581 488 447 654 789 12 × 2 = 1 + 0.757 383 975 162 976 895 309 578 24;
  • 17) 0.757 383 975 162 976 895 309 578 24 × 2 = 1 + 0.514 767 950 325 953 790 619 156 48;
  • 18) 0.514 767 950 325 953 790 619 156 48 × 2 = 1 + 0.029 535 900 651 907 581 238 312 96;
  • 19) 0.029 535 900 651 907 581 238 312 96 × 2 = 0 + 0.059 071 801 303 815 162 476 625 92;
  • 20) 0.059 071 801 303 815 162 476 625 92 × 2 = 0 + 0.118 143 602 607 630 324 953 251 84;
  • 21) 0.118 143 602 607 630 324 953 251 84 × 2 = 0 + 0.236 287 205 215 260 649 906 503 68;
  • 22) 0.236 287 205 215 260 649 906 503 68 × 2 = 0 + 0.472 574 410 430 521 299 813 007 36;
  • 23) 0.472 574 410 430 521 299 813 007 36 × 2 = 0 + 0.945 148 820 861 042 599 626 014 72;
  • 24) 0.945 148 820 861 042 599 626 014 72 × 2 = 1 + 0.890 297 641 722 085 199 252 029 44;
  • 25) 0.890 297 641 722 085 199 252 029 44 × 2 = 1 + 0.780 595 283 444 170 398 504 058 88;
  • 26) 0.780 595 283 444 170 398 504 058 88 × 2 = 1 + 0.561 190 566 888 340 797 008 117 76;
  • 27) 0.561 190 566 888 340 797 008 117 76 × 2 = 1 + 0.122 381 133 776 681 594 016 235 52;
  • 28) 0.122 381 133 776 681 594 016 235 52 × 2 = 0 + 0.244 762 267 553 363 188 032 471 04;
  • 29) 0.244 762 267 553 363 188 032 471 04 × 2 = 0 + 0.489 524 535 106 726 376 064 942 08;
  • 30) 0.489 524 535 106 726 376 064 942 08 × 2 = 0 + 0.979 049 070 213 452 752 129 884 16;
  • 31) 0.979 049 070 213 452 752 129 884 16 × 2 = 1 + 0.958 098 140 426 905 504 259 768 32;
  • 32) 0.958 098 140 426 905 504 259 768 32 × 2 = 1 + 0.916 196 280 853 811 008 519 536 64;
  • 33) 0.916 196 280 853 811 008 519 536 64 × 2 = 1 + 0.832 392 561 707 622 017 039 073 28;
  • 34) 0.832 392 561 707 622 017 039 073 28 × 2 = 1 + 0.664 785 123 415 244 034 078 146 56;
  • 35) 0.664 785 123 415 244 034 078 146 56 × 2 = 1 + 0.329 570 246 830 488 068 156 293 12;
  • 36) 0.329 570 246 830 488 068 156 293 12 × 2 = 0 + 0.659 140 493 660 976 136 312 586 24;
  • 37) 0.659 140 493 660 976 136 312 586 24 × 2 = 1 + 0.318 280 987 321 952 272 625 172 48;
  • 38) 0.318 280 987 321 952 272 625 172 48 × 2 = 0 + 0.636 561 974 643 904 545 250 344 96;
  • 39) 0.636 561 974 643 904 545 250 344 96 × 2 = 1 + 0.273 123 949 287 809 090 500 689 92;
  • 40) 0.273 123 949 287 809 090 500 689 92 × 2 = 0 + 0.546 247 898 575 618 181 001 379 84;
  • 41) 0.546 247 898 575 618 181 001 379 84 × 2 = 1 + 0.092 495 797 151 236 362 002 759 68;
  • 42) 0.092 495 797 151 236 362 002 759 68 × 2 = 0 + 0.184 991 594 302 472 724 005 519 36;
  • 43) 0.184 991 594 302 472 724 005 519 36 × 2 = 0 + 0.369 983 188 604 945 448 011 038 72;
  • 44) 0.369 983 188 604 945 448 011 038 72 × 2 = 0 + 0.739 966 377 209 890 896 022 077 44;
  • 45) 0.739 966 377 209 890 896 022 077 44 × 2 = 1 + 0.479 932 754 419 781 792 044 154 88;
  • 46) 0.479 932 754 419 781 792 044 154 88 × 2 = 0 + 0.959 865 508 839 563 584 088 309 76;
  • 47) 0.959 865 508 839 563 584 088 309 76 × 2 = 1 + 0.919 731 017 679 127 168 176 619 52;
  • 48) 0.919 731 017 679 127 168 176 619 52 × 2 = 1 + 0.839 462 035 358 254 336 353 239 04;
  • 49) 0.839 462 035 358 254 336 353 239 04 × 2 = 1 + 0.678 924 070 716 508 672 706 478 08;
  • 50) 0.678 924 070 716 508 672 706 478 08 × 2 = 1 + 0.357 848 141 433 017 345 412 956 16;
  • 51) 0.357 848 141 433 017 345 412 956 16 × 2 = 0 + 0.715 696 282 866 034 690 825 912 32;
  • 52) 0.715 696 282 866 034 690 825 912 32 × 2 = 1 + 0.431 392 565 732 069 381 651 824 64;
  • 53) 0.431 392 565 732 069 381 651 824 64 × 2 = 0 + 0.862 785 131 464 138 763 303 649 28;

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 09(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 09(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 09(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 09 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

Share

» Convert decimal 0.545 253 866 332 628 829 603 505 56 to 64 bit double precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 08, 2026 [August]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: August - by our visitors

» Improve your social skills: Why Do Colleagues From Other Teams Have Better Results? Gemini Notebook YouTube Video of 11.31 minutes

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