0.545 253 866 332 628 829 603 504 88 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.545 253 866 332 628 829 603 504 88(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
0.545 253 866 332 628 829 603 504 88(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: 0.545 253 866 332 628 829 603 504 88.

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) 0.545 253 866 332 628 829 603 504 88 × 2 = 1 + 0.090 507 732 665 257 659 207 009 76;
  • 2) 0.090 507 732 665 257 659 207 009 76 × 2 = 0 + 0.181 015 465 330 515 318 414 019 52;
  • 3) 0.181 015 465 330 515 318 414 019 52 × 2 = 0 + 0.362 030 930 661 030 636 828 039 04;
  • 4) 0.362 030 930 661 030 636 828 039 04 × 2 = 0 + 0.724 061 861 322 061 273 656 078 08;
  • 5) 0.724 061 861 322 061 273 656 078 08 × 2 = 1 + 0.448 123 722 644 122 547 312 156 16;
  • 6) 0.448 123 722 644 122 547 312 156 16 × 2 = 0 + 0.896 247 445 288 245 094 624 312 32;
  • 7) 0.896 247 445 288 245 094 624 312 32 × 2 = 1 + 0.792 494 890 576 490 189 248 624 64;
  • 8) 0.792 494 890 576 490 189 248 624 64 × 2 = 1 + 0.584 989 781 152 980 378 497 249 28;
  • 9) 0.584 989 781 152 980 378 497 249 28 × 2 = 1 + 0.169 979 562 305 960 756 994 498 56;
  • 10) 0.169 979 562 305 960 756 994 498 56 × 2 = 0 + 0.339 959 124 611 921 513 988 997 12;
  • 11) 0.339 959 124 611 921 513 988 997 12 × 2 = 0 + 0.679 918 249 223 843 027 977 994 24;
  • 12) 0.679 918 249 223 843 027 977 994 24 × 2 = 1 + 0.359 836 498 447 686 055 955 988 48;
  • 13) 0.359 836 498 447 686 055 955 988 48 × 2 = 0 + 0.719 672 996 895 372 111 911 976 96;
  • 14) 0.719 672 996 895 372 111 911 976 96 × 2 = 1 + 0.439 345 993 790 744 223 823 953 92;
  • 15) 0.439 345 993 790 744 223 823 953 92 × 2 = 0 + 0.878 691 987 581 488 447 647 907 84;
  • 16) 0.878 691 987 581 488 447 647 907 84 × 2 = 1 + 0.757 383 975 162 976 895 295 815 68;
  • 17) 0.757 383 975 162 976 895 295 815 68 × 2 = 1 + 0.514 767 950 325 953 790 591 631 36;
  • 18) 0.514 767 950 325 953 790 591 631 36 × 2 = 1 + 0.029 535 900 651 907 581 183 262 72;
  • 19) 0.029 535 900 651 907 581 183 262 72 × 2 = 0 + 0.059 071 801 303 815 162 366 525 44;
  • 20) 0.059 071 801 303 815 162 366 525 44 × 2 = 0 + 0.118 143 602 607 630 324 733 050 88;
  • 21) 0.118 143 602 607 630 324 733 050 88 × 2 = 0 + 0.236 287 205 215 260 649 466 101 76;
  • 22) 0.236 287 205 215 260 649 466 101 76 × 2 = 0 + 0.472 574 410 430 521 298 932 203 52;
  • 23) 0.472 574 410 430 521 298 932 203 52 × 2 = 0 + 0.945 148 820 861 042 597 864 407 04;
  • 24) 0.945 148 820 861 042 597 864 407 04 × 2 = 1 + 0.890 297 641 722 085 195 728 814 08;
  • 25) 0.890 297 641 722 085 195 728 814 08 × 2 = 1 + 0.780 595 283 444 170 391 457 628 16;
  • 26) 0.780 595 283 444 170 391 457 628 16 × 2 = 1 + 0.561 190 566 888 340 782 915 256 32;
  • 27) 0.561 190 566 888 340 782 915 256 32 × 2 = 1 + 0.122 381 133 776 681 565 830 512 64;
  • 28) 0.122 381 133 776 681 565 830 512 64 × 2 = 0 + 0.244 762 267 553 363 131 661 025 28;
  • 29) 0.244 762 267 553 363 131 661 025 28 × 2 = 0 + 0.489 524 535 106 726 263 322 050 56;
  • 30) 0.489 524 535 106 726 263 322 050 56 × 2 = 0 + 0.979 049 070 213 452 526 644 101 12;
  • 31) 0.979 049 070 213 452 526 644 101 12 × 2 = 1 + 0.958 098 140 426 905 053 288 202 24;
  • 32) 0.958 098 140 426 905 053 288 202 24 × 2 = 1 + 0.916 196 280 853 810 106 576 404 48;
  • 33) 0.916 196 280 853 810 106 576 404 48 × 2 = 1 + 0.832 392 561 707 620 213 152 808 96;
  • 34) 0.832 392 561 707 620 213 152 808 96 × 2 = 1 + 0.664 785 123 415 240 426 305 617 92;
  • 35) 0.664 785 123 415 240 426 305 617 92 × 2 = 1 + 0.329 570 246 830 480 852 611 235 84;
  • 36) 0.329 570 246 830 480 852 611 235 84 × 2 = 0 + 0.659 140 493 660 961 705 222 471 68;
  • 37) 0.659 140 493 660 961 705 222 471 68 × 2 = 1 + 0.318 280 987 321 923 410 444 943 36;
  • 38) 0.318 280 987 321 923 410 444 943 36 × 2 = 0 + 0.636 561 974 643 846 820 889 886 72;
  • 39) 0.636 561 974 643 846 820 889 886 72 × 2 = 1 + 0.273 123 949 287 693 641 779 773 44;
  • 40) 0.273 123 949 287 693 641 779 773 44 × 2 = 0 + 0.546 247 898 575 387 283 559 546 88;
  • 41) 0.546 247 898 575 387 283 559 546 88 × 2 = 1 + 0.092 495 797 150 774 567 119 093 76;
  • 42) 0.092 495 797 150 774 567 119 093 76 × 2 = 0 + 0.184 991 594 301 549 134 238 187 52;
  • 43) 0.184 991 594 301 549 134 238 187 52 × 2 = 0 + 0.369 983 188 603 098 268 476 375 04;
  • 44) 0.369 983 188 603 098 268 476 375 04 × 2 = 0 + 0.739 966 377 206 196 536 952 750 08;
  • 45) 0.739 966 377 206 196 536 952 750 08 × 2 = 1 + 0.479 932 754 412 393 073 905 500 16;
  • 46) 0.479 932 754 412 393 073 905 500 16 × 2 = 0 + 0.959 865 508 824 786 147 811 000 32;
  • 47) 0.959 865 508 824 786 147 811 000 32 × 2 = 1 + 0.919 731 017 649 572 295 622 000 64;
  • 48) 0.919 731 017 649 572 295 622 000 64 × 2 = 1 + 0.839 462 035 299 144 591 244 001 28;
  • 49) 0.839 462 035 299 144 591 244 001 28 × 2 = 1 + 0.678 924 070 598 289 182 488 002 56;
  • 50) 0.678 924 070 598 289 182 488 002 56 × 2 = 1 + 0.357 848 141 196 578 364 976 005 12;
  • 51) 0.357 848 141 196 578 364 976 005 12 × 2 = 0 + 0.715 696 282 393 156 729 952 010 24;
  • 52) 0.715 696 282 393 156 729 952 010 24 × 2 = 1 + 0.431 392 564 786 313 459 904 020 48;
  • 53) 0.431 392 564 786 313 459 904 020 48 × 2 = 0 + 0.862 785 129 572 626 919 808 040 96;

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:


0.545 253 866 332 628 829 603 504 88(10) =

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

5. Positive number before normalization:

0.545 253 866 332 628 829 603 504 88(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:


0.545 253 866 332 628 829 603 504 88(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 0.545 253 866 332 628 829 603 504 88 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 0.545 253 866 332 628 829 603 504 3 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: Are You Using Communication by Design, or by Default? Gemini Notebook YouTube Video of 6.33 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