0.545 253 866 332 628 829 603 505 353 1 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.545 253 866 332 628 829 603 505 353 1(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
0.545 253 866 332 628 829 603 505 353 1(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: 0.545 253 866 332 628 829 603 505 353 1.

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) 0.545 253 866 332 628 829 603 505 353 1 × 2 = 1 + 0.090 507 732 665 257 659 207 010 706 2;
  • 2) 0.090 507 732 665 257 659 207 010 706 2 × 2 = 0 + 0.181 015 465 330 515 318 414 021 412 4;
  • 3) 0.181 015 465 330 515 318 414 021 412 4 × 2 = 0 + 0.362 030 930 661 030 636 828 042 824 8;
  • 4) 0.362 030 930 661 030 636 828 042 824 8 × 2 = 0 + 0.724 061 861 322 061 273 656 085 649 6;
  • 5) 0.724 061 861 322 061 273 656 085 649 6 × 2 = 1 + 0.448 123 722 644 122 547 312 171 299 2;
  • 6) 0.448 123 722 644 122 547 312 171 299 2 × 2 = 0 + 0.896 247 445 288 245 094 624 342 598 4;
  • 7) 0.896 247 445 288 245 094 624 342 598 4 × 2 = 1 + 0.792 494 890 576 490 189 248 685 196 8;
  • 8) 0.792 494 890 576 490 189 248 685 196 8 × 2 = 1 + 0.584 989 781 152 980 378 497 370 393 6;
  • 9) 0.584 989 781 152 980 378 497 370 393 6 × 2 = 1 + 0.169 979 562 305 960 756 994 740 787 2;
  • 10) 0.169 979 562 305 960 756 994 740 787 2 × 2 = 0 + 0.339 959 124 611 921 513 989 481 574 4;
  • 11) 0.339 959 124 611 921 513 989 481 574 4 × 2 = 0 + 0.679 918 249 223 843 027 978 963 148 8;
  • 12) 0.679 918 249 223 843 027 978 963 148 8 × 2 = 1 + 0.359 836 498 447 686 055 957 926 297 6;
  • 13) 0.359 836 498 447 686 055 957 926 297 6 × 2 = 0 + 0.719 672 996 895 372 111 915 852 595 2;
  • 14) 0.719 672 996 895 372 111 915 852 595 2 × 2 = 1 + 0.439 345 993 790 744 223 831 705 190 4;
  • 15) 0.439 345 993 790 744 223 831 705 190 4 × 2 = 0 + 0.878 691 987 581 488 447 663 410 380 8;
  • 16) 0.878 691 987 581 488 447 663 410 380 8 × 2 = 1 + 0.757 383 975 162 976 895 326 820 761 6;
  • 17) 0.757 383 975 162 976 895 326 820 761 6 × 2 = 1 + 0.514 767 950 325 953 790 653 641 523 2;
  • 18) 0.514 767 950 325 953 790 653 641 523 2 × 2 = 1 + 0.029 535 900 651 907 581 307 283 046 4;
  • 19) 0.029 535 900 651 907 581 307 283 046 4 × 2 = 0 + 0.059 071 801 303 815 162 614 566 092 8;
  • 20) 0.059 071 801 303 815 162 614 566 092 8 × 2 = 0 + 0.118 143 602 607 630 325 229 132 185 6;
  • 21) 0.118 143 602 607 630 325 229 132 185 6 × 2 = 0 + 0.236 287 205 215 260 650 458 264 371 2;
  • 22) 0.236 287 205 215 260 650 458 264 371 2 × 2 = 0 + 0.472 574 410 430 521 300 916 528 742 4;
  • 23) 0.472 574 410 430 521 300 916 528 742 4 × 2 = 0 + 0.945 148 820 861 042 601 833 057 484 8;
  • 24) 0.945 148 820 861 042 601 833 057 484 8 × 2 = 1 + 0.890 297 641 722 085 203 666 114 969 6;
  • 25) 0.890 297 641 722 085 203 666 114 969 6 × 2 = 1 + 0.780 595 283 444 170 407 332 229 939 2;
  • 26) 0.780 595 283 444 170 407 332 229 939 2 × 2 = 1 + 0.561 190 566 888 340 814 664 459 878 4;
  • 27) 0.561 190 566 888 340 814 664 459 878 4 × 2 = 1 + 0.122 381 133 776 681 629 328 919 756 8;
  • 28) 0.122 381 133 776 681 629 328 919 756 8 × 2 = 0 + 0.244 762 267 553 363 258 657 839 513 6;
  • 29) 0.244 762 267 553 363 258 657 839 513 6 × 2 = 0 + 0.489 524 535 106 726 517 315 679 027 2;
  • 30) 0.489 524 535 106 726 517 315 679 027 2 × 2 = 0 + 0.979 049 070 213 453 034 631 358 054 4;
  • 31) 0.979 049 070 213 453 034 631 358 054 4 × 2 = 1 + 0.958 098 140 426 906 069 262 716 108 8;
  • 32) 0.958 098 140 426 906 069 262 716 108 8 × 2 = 1 + 0.916 196 280 853 812 138 525 432 217 6;
  • 33) 0.916 196 280 853 812 138 525 432 217 6 × 2 = 1 + 0.832 392 561 707 624 277 050 864 435 2;
  • 34) 0.832 392 561 707 624 277 050 864 435 2 × 2 = 1 + 0.664 785 123 415 248 554 101 728 870 4;
  • 35) 0.664 785 123 415 248 554 101 728 870 4 × 2 = 1 + 0.329 570 246 830 497 108 203 457 740 8;
  • 36) 0.329 570 246 830 497 108 203 457 740 8 × 2 = 0 + 0.659 140 493 660 994 216 406 915 481 6;
  • 37) 0.659 140 493 660 994 216 406 915 481 6 × 2 = 1 + 0.318 280 987 321 988 432 813 830 963 2;
  • 38) 0.318 280 987 321 988 432 813 830 963 2 × 2 = 0 + 0.636 561 974 643 976 865 627 661 926 4;
  • 39) 0.636 561 974 643 976 865 627 661 926 4 × 2 = 1 + 0.273 123 949 287 953 731 255 323 852 8;
  • 40) 0.273 123 949 287 953 731 255 323 852 8 × 2 = 0 + 0.546 247 898 575 907 462 510 647 705 6;
  • 41) 0.546 247 898 575 907 462 510 647 705 6 × 2 = 1 + 0.092 495 797 151 814 925 021 295 411 2;
  • 42) 0.092 495 797 151 814 925 021 295 411 2 × 2 = 0 + 0.184 991 594 303 629 850 042 590 822 4;
  • 43) 0.184 991 594 303 629 850 042 590 822 4 × 2 = 0 + 0.369 983 188 607 259 700 085 181 644 8;
  • 44) 0.369 983 188 607 259 700 085 181 644 8 × 2 = 0 + 0.739 966 377 214 519 400 170 363 289 6;
  • 45) 0.739 966 377 214 519 400 170 363 289 6 × 2 = 1 + 0.479 932 754 429 038 800 340 726 579 2;
  • 46) 0.479 932 754 429 038 800 340 726 579 2 × 2 = 0 + 0.959 865 508 858 077 600 681 453 158 4;
  • 47) 0.959 865 508 858 077 600 681 453 158 4 × 2 = 1 + 0.919 731 017 716 155 201 362 906 316 8;
  • 48) 0.919 731 017 716 155 201 362 906 316 8 × 2 = 1 + 0.839 462 035 432 310 402 725 812 633 6;
  • 49) 0.839 462 035 432 310 402 725 812 633 6 × 2 = 1 + 0.678 924 070 864 620 805 451 625 267 2;
  • 50) 0.678 924 070 864 620 805 451 625 267 2 × 2 = 1 + 0.357 848 141 729 241 610 903 250 534 4;
  • 51) 0.357 848 141 729 241 610 903 250 534 4 × 2 = 0 + 0.715 696 283 458 483 221 806 501 068 8;
  • 52) 0.715 696 283 458 483 221 806 501 068 8 × 2 = 1 + 0.431 392 566 916 966 443 613 002 137 6;
  • 53) 0.431 392 566 916 966 443 613 002 137 6 × 2 = 0 + 0.862 785 133 833 932 887 226 004 275 2;

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:


0.545 253 866 332 628 829 603 505 353 1(10) =

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

5. Positive number before normalization:

0.545 253 866 332 628 829 603 505 353 1(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:


0.545 253 866 332 628 829 603 505 353 1(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 0.545 253 866 332 628 829 603 505 353 1 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