0.545 253 866 332 628 829 603 505 348 4 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.545 253 866 332 628 829 603 505 348 4(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
0.545 253 866 332 628 829 603 505 348 4(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: 0.545 253 866 332 628 829 603 505 348 4.

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) 0.545 253 866 332 628 829 603 505 348 4 × 2 = 1 + 0.090 507 732 665 257 659 207 010 696 8;
  • 2) 0.090 507 732 665 257 659 207 010 696 8 × 2 = 0 + 0.181 015 465 330 515 318 414 021 393 6;
  • 3) 0.181 015 465 330 515 318 414 021 393 6 × 2 = 0 + 0.362 030 930 661 030 636 828 042 787 2;
  • 4) 0.362 030 930 661 030 636 828 042 787 2 × 2 = 0 + 0.724 061 861 322 061 273 656 085 574 4;
  • 5) 0.724 061 861 322 061 273 656 085 574 4 × 2 = 1 + 0.448 123 722 644 122 547 312 171 148 8;
  • 6) 0.448 123 722 644 122 547 312 171 148 8 × 2 = 0 + 0.896 247 445 288 245 094 624 342 297 6;
  • 7) 0.896 247 445 288 245 094 624 342 297 6 × 2 = 1 + 0.792 494 890 576 490 189 248 684 595 2;
  • 8) 0.792 494 890 576 490 189 248 684 595 2 × 2 = 1 + 0.584 989 781 152 980 378 497 369 190 4;
  • 9) 0.584 989 781 152 980 378 497 369 190 4 × 2 = 1 + 0.169 979 562 305 960 756 994 738 380 8;
  • 10) 0.169 979 562 305 960 756 994 738 380 8 × 2 = 0 + 0.339 959 124 611 921 513 989 476 761 6;
  • 11) 0.339 959 124 611 921 513 989 476 761 6 × 2 = 0 + 0.679 918 249 223 843 027 978 953 523 2;
  • 12) 0.679 918 249 223 843 027 978 953 523 2 × 2 = 1 + 0.359 836 498 447 686 055 957 907 046 4;
  • 13) 0.359 836 498 447 686 055 957 907 046 4 × 2 = 0 + 0.719 672 996 895 372 111 915 814 092 8;
  • 14) 0.719 672 996 895 372 111 915 814 092 8 × 2 = 1 + 0.439 345 993 790 744 223 831 628 185 6;
  • 15) 0.439 345 993 790 744 223 831 628 185 6 × 2 = 0 + 0.878 691 987 581 488 447 663 256 371 2;
  • 16) 0.878 691 987 581 488 447 663 256 371 2 × 2 = 1 + 0.757 383 975 162 976 895 326 512 742 4;
  • 17) 0.757 383 975 162 976 895 326 512 742 4 × 2 = 1 + 0.514 767 950 325 953 790 653 025 484 8;
  • 18) 0.514 767 950 325 953 790 653 025 484 8 × 2 = 1 + 0.029 535 900 651 907 581 306 050 969 6;
  • 19) 0.029 535 900 651 907 581 306 050 969 6 × 2 = 0 + 0.059 071 801 303 815 162 612 101 939 2;
  • 20) 0.059 071 801 303 815 162 612 101 939 2 × 2 = 0 + 0.118 143 602 607 630 325 224 203 878 4;
  • 21) 0.118 143 602 607 630 325 224 203 878 4 × 2 = 0 + 0.236 287 205 215 260 650 448 407 756 8;
  • 22) 0.236 287 205 215 260 650 448 407 756 8 × 2 = 0 + 0.472 574 410 430 521 300 896 815 513 6;
  • 23) 0.472 574 410 430 521 300 896 815 513 6 × 2 = 0 + 0.945 148 820 861 042 601 793 631 027 2;
  • 24) 0.945 148 820 861 042 601 793 631 027 2 × 2 = 1 + 0.890 297 641 722 085 203 587 262 054 4;
  • 25) 0.890 297 641 722 085 203 587 262 054 4 × 2 = 1 + 0.780 595 283 444 170 407 174 524 108 8;
  • 26) 0.780 595 283 444 170 407 174 524 108 8 × 2 = 1 + 0.561 190 566 888 340 814 349 048 217 6;
  • 27) 0.561 190 566 888 340 814 349 048 217 6 × 2 = 1 + 0.122 381 133 776 681 628 698 096 435 2;
  • 28) 0.122 381 133 776 681 628 698 096 435 2 × 2 = 0 + 0.244 762 267 553 363 257 396 192 870 4;
  • 29) 0.244 762 267 553 363 257 396 192 870 4 × 2 = 0 + 0.489 524 535 106 726 514 792 385 740 8;
  • 30) 0.489 524 535 106 726 514 792 385 740 8 × 2 = 0 + 0.979 049 070 213 453 029 584 771 481 6;
  • 31) 0.979 049 070 213 453 029 584 771 481 6 × 2 = 1 + 0.958 098 140 426 906 059 169 542 963 2;
  • 32) 0.958 098 140 426 906 059 169 542 963 2 × 2 = 1 + 0.916 196 280 853 812 118 339 085 926 4;
  • 33) 0.916 196 280 853 812 118 339 085 926 4 × 2 = 1 + 0.832 392 561 707 624 236 678 171 852 8;
  • 34) 0.832 392 561 707 624 236 678 171 852 8 × 2 = 1 + 0.664 785 123 415 248 473 356 343 705 6;
  • 35) 0.664 785 123 415 248 473 356 343 705 6 × 2 = 1 + 0.329 570 246 830 496 946 712 687 411 2;
  • 36) 0.329 570 246 830 496 946 712 687 411 2 × 2 = 0 + 0.659 140 493 660 993 893 425 374 822 4;
  • 37) 0.659 140 493 660 993 893 425 374 822 4 × 2 = 1 + 0.318 280 987 321 987 786 850 749 644 8;
  • 38) 0.318 280 987 321 987 786 850 749 644 8 × 2 = 0 + 0.636 561 974 643 975 573 701 499 289 6;
  • 39) 0.636 561 974 643 975 573 701 499 289 6 × 2 = 1 + 0.273 123 949 287 951 147 402 998 579 2;
  • 40) 0.273 123 949 287 951 147 402 998 579 2 × 2 = 0 + 0.546 247 898 575 902 294 805 997 158 4;
  • 41) 0.546 247 898 575 902 294 805 997 158 4 × 2 = 1 + 0.092 495 797 151 804 589 611 994 316 8;
  • 42) 0.092 495 797 151 804 589 611 994 316 8 × 2 = 0 + 0.184 991 594 303 609 179 223 988 633 6;
  • 43) 0.184 991 594 303 609 179 223 988 633 6 × 2 = 0 + 0.369 983 188 607 218 358 447 977 267 2;
  • 44) 0.369 983 188 607 218 358 447 977 267 2 × 2 = 0 + 0.739 966 377 214 436 716 895 954 534 4;
  • 45) 0.739 966 377 214 436 716 895 954 534 4 × 2 = 1 + 0.479 932 754 428 873 433 791 909 068 8;
  • 46) 0.479 932 754 428 873 433 791 909 068 8 × 2 = 0 + 0.959 865 508 857 746 867 583 818 137 6;
  • 47) 0.959 865 508 857 746 867 583 818 137 6 × 2 = 1 + 0.919 731 017 715 493 735 167 636 275 2;
  • 48) 0.919 731 017 715 493 735 167 636 275 2 × 2 = 1 + 0.839 462 035 430 987 470 335 272 550 4;
  • 49) 0.839 462 035 430 987 470 335 272 550 4 × 2 = 1 + 0.678 924 070 861 974 940 670 545 100 8;
  • 50) 0.678 924 070 861 974 940 670 545 100 8 × 2 = 1 + 0.357 848 141 723 949 881 341 090 201 6;
  • 51) 0.357 848 141 723 949 881 341 090 201 6 × 2 = 0 + 0.715 696 283 447 899 762 682 180 403 2;
  • 52) 0.715 696 283 447 899 762 682 180 403 2 × 2 = 1 + 0.431 392 566 895 799 525 364 360 806 4;
  • 53) 0.431 392 566 895 799 525 364 360 806 4 × 2 = 0 + 0.862 785 133 791 599 050 728 721 612 8;

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:


0.545 253 866 332 628 829 603 505 348 4(10) =

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

5. Positive number before normalization:

0.545 253 866 332 628 829 603 505 348 4(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:


0.545 253 866 332 628 829 603 505 348 4(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 0.545 253 866 332 628 829 603 505 348 4 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