0.545 253 866 332 628 829 603 505 344 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.545 253 866 332 628 829 603 505 344(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
0.545 253 866 332 628 829 603 505 344(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: 0.545 253 866 332 628 829 603 505 344.

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) 0.545 253 866 332 628 829 603 505 344 × 2 = 1 + 0.090 507 732 665 257 659 207 010 688;
  • 2) 0.090 507 732 665 257 659 207 010 688 × 2 = 0 + 0.181 015 465 330 515 318 414 021 376;
  • 3) 0.181 015 465 330 515 318 414 021 376 × 2 = 0 + 0.362 030 930 661 030 636 828 042 752;
  • 4) 0.362 030 930 661 030 636 828 042 752 × 2 = 0 + 0.724 061 861 322 061 273 656 085 504;
  • 5) 0.724 061 861 322 061 273 656 085 504 × 2 = 1 + 0.448 123 722 644 122 547 312 171 008;
  • 6) 0.448 123 722 644 122 547 312 171 008 × 2 = 0 + 0.896 247 445 288 245 094 624 342 016;
  • 7) 0.896 247 445 288 245 094 624 342 016 × 2 = 1 + 0.792 494 890 576 490 189 248 684 032;
  • 8) 0.792 494 890 576 490 189 248 684 032 × 2 = 1 + 0.584 989 781 152 980 378 497 368 064;
  • 9) 0.584 989 781 152 980 378 497 368 064 × 2 = 1 + 0.169 979 562 305 960 756 994 736 128;
  • 10) 0.169 979 562 305 960 756 994 736 128 × 2 = 0 + 0.339 959 124 611 921 513 989 472 256;
  • 11) 0.339 959 124 611 921 513 989 472 256 × 2 = 0 + 0.679 918 249 223 843 027 978 944 512;
  • 12) 0.679 918 249 223 843 027 978 944 512 × 2 = 1 + 0.359 836 498 447 686 055 957 889 024;
  • 13) 0.359 836 498 447 686 055 957 889 024 × 2 = 0 + 0.719 672 996 895 372 111 915 778 048;
  • 14) 0.719 672 996 895 372 111 915 778 048 × 2 = 1 + 0.439 345 993 790 744 223 831 556 096;
  • 15) 0.439 345 993 790 744 223 831 556 096 × 2 = 0 + 0.878 691 987 581 488 447 663 112 192;
  • 16) 0.878 691 987 581 488 447 663 112 192 × 2 = 1 + 0.757 383 975 162 976 895 326 224 384;
  • 17) 0.757 383 975 162 976 895 326 224 384 × 2 = 1 + 0.514 767 950 325 953 790 652 448 768;
  • 18) 0.514 767 950 325 953 790 652 448 768 × 2 = 1 + 0.029 535 900 651 907 581 304 897 536;
  • 19) 0.029 535 900 651 907 581 304 897 536 × 2 = 0 + 0.059 071 801 303 815 162 609 795 072;
  • 20) 0.059 071 801 303 815 162 609 795 072 × 2 = 0 + 0.118 143 602 607 630 325 219 590 144;
  • 21) 0.118 143 602 607 630 325 219 590 144 × 2 = 0 + 0.236 287 205 215 260 650 439 180 288;
  • 22) 0.236 287 205 215 260 650 439 180 288 × 2 = 0 + 0.472 574 410 430 521 300 878 360 576;
  • 23) 0.472 574 410 430 521 300 878 360 576 × 2 = 0 + 0.945 148 820 861 042 601 756 721 152;
  • 24) 0.945 148 820 861 042 601 756 721 152 × 2 = 1 + 0.890 297 641 722 085 203 513 442 304;
  • 25) 0.890 297 641 722 085 203 513 442 304 × 2 = 1 + 0.780 595 283 444 170 407 026 884 608;
  • 26) 0.780 595 283 444 170 407 026 884 608 × 2 = 1 + 0.561 190 566 888 340 814 053 769 216;
  • 27) 0.561 190 566 888 340 814 053 769 216 × 2 = 1 + 0.122 381 133 776 681 628 107 538 432;
  • 28) 0.122 381 133 776 681 628 107 538 432 × 2 = 0 + 0.244 762 267 553 363 256 215 076 864;
  • 29) 0.244 762 267 553 363 256 215 076 864 × 2 = 0 + 0.489 524 535 106 726 512 430 153 728;
  • 30) 0.489 524 535 106 726 512 430 153 728 × 2 = 0 + 0.979 049 070 213 453 024 860 307 456;
  • 31) 0.979 049 070 213 453 024 860 307 456 × 2 = 1 + 0.958 098 140 426 906 049 720 614 912;
  • 32) 0.958 098 140 426 906 049 720 614 912 × 2 = 1 + 0.916 196 280 853 812 099 441 229 824;
  • 33) 0.916 196 280 853 812 099 441 229 824 × 2 = 1 + 0.832 392 561 707 624 198 882 459 648;
  • 34) 0.832 392 561 707 624 198 882 459 648 × 2 = 1 + 0.664 785 123 415 248 397 764 919 296;
  • 35) 0.664 785 123 415 248 397 764 919 296 × 2 = 1 + 0.329 570 246 830 496 795 529 838 592;
  • 36) 0.329 570 246 830 496 795 529 838 592 × 2 = 0 + 0.659 140 493 660 993 591 059 677 184;
  • 37) 0.659 140 493 660 993 591 059 677 184 × 2 = 1 + 0.318 280 987 321 987 182 119 354 368;
  • 38) 0.318 280 987 321 987 182 119 354 368 × 2 = 0 + 0.636 561 974 643 974 364 238 708 736;
  • 39) 0.636 561 974 643 974 364 238 708 736 × 2 = 1 + 0.273 123 949 287 948 728 477 417 472;
  • 40) 0.273 123 949 287 948 728 477 417 472 × 2 = 0 + 0.546 247 898 575 897 456 954 834 944;
  • 41) 0.546 247 898 575 897 456 954 834 944 × 2 = 1 + 0.092 495 797 151 794 913 909 669 888;
  • 42) 0.092 495 797 151 794 913 909 669 888 × 2 = 0 + 0.184 991 594 303 589 827 819 339 776;
  • 43) 0.184 991 594 303 589 827 819 339 776 × 2 = 0 + 0.369 983 188 607 179 655 638 679 552;
  • 44) 0.369 983 188 607 179 655 638 679 552 × 2 = 0 + 0.739 966 377 214 359 311 277 359 104;
  • 45) 0.739 966 377 214 359 311 277 359 104 × 2 = 1 + 0.479 932 754 428 718 622 554 718 208;
  • 46) 0.479 932 754 428 718 622 554 718 208 × 2 = 0 + 0.959 865 508 857 437 245 109 436 416;
  • 47) 0.959 865 508 857 437 245 109 436 416 × 2 = 1 + 0.919 731 017 714 874 490 218 872 832;
  • 48) 0.919 731 017 714 874 490 218 872 832 × 2 = 1 + 0.839 462 035 429 748 980 437 745 664;
  • 49) 0.839 462 035 429 748 980 437 745 664 × 2 = 1 + 0.678 924 070 859 497 960 875 491 328;
  • 50) 0.678 924 070 859 497 960 875 491 328 × 2 = 1 + 0.357 848 141 718 995 921 750 982 656;
  • 51) 0.357 848 141 718 995 921 750 982 656 × 2 = 0 + 0.715 696 283 437 991 843 501 965 312;
  • 52) 0.715 696 283 437 991 843 501 965 312 × 2 = 1 + 0.431 392 566 875 983 687 003 930 624;
  • 53) 0.431 392 566 875 983 687 003 930 624 × 2 = 0 + 0.862 785 133 751 967 374 007 861 248;

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:


0.545 253 866 332 628 829 603 505 344(10) =

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

5. Positive number before normalization:

0.545 253 866 332 628 829 603 505 344(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:


0.545 253 866 332 628 829 603 505 344(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 0.545 253 866 332 628 829 603 505 344 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