0.545 253 866 332 628 829 606 97 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.545 253 866 332 628 829 606 97(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
0.545 253 866 332 628 829 606 97(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: 0.545 253 866 332 628 829 606 97.

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) 0.545 253 866 332 628 829 606 97 × 2 = 1 + 0.090 507 732 665 257 659 213 94;
  • 2) 0.090 507 732 665 257 659 213 94 × 2 = 0 + 0.181 015 465 330 515 318 427 88;
  • 3) 0.181 015 465 330 515 318 427 88 × 2 = 0 + 0.362 030 930 661 030 636 855 76;
  • 4) 0.362 030 930 661 030 636 855 76 × 2 = 0 + 0.724 061 861 322 061 273 711 52;
  • 5) 0.724 061 861 322 061 273 711 52 × 2 = 1 + 0.448 123 722 644 122 547 423 04;
  • 6) 0.448 123 722 644 122 547 423 04 × 2 = 0 + 0.896 247 445 288 245 094 846 08;
  • 7) 0.896 247 445 288 245 094 846 08 × 2 = 1 + 0.792 494 890 576 490 189 692 16;
  • 8) 0.792 494 890 576 490 189 692 16 × 2 = 1 + 0.584 989 781 152 980 379 384 32;
  • 9) 0.584 989 781 152 980 379 384 32 × 2 = 1 + 0.169 979 562 305 960 758 768 64;
  • 10) 0.169 979 562 305 960 758 768 64 × 2 = 0 + 0.339 959 124 611 921 517 537 28;
  • 11) 0.339 959 124 611 921 517 537 28 × 2 = 0 + 0.679 918 249 223 843 035 074 56;
  • 12) 0.679 918 249 223 843 035 074 56 × 2 = 1 + 0.359 836 498 447 686 070 149 12;
  • 13) 0.359 836 498 447 686 070 149 12 × 2 = 0 + 0.719 672 996 895 372 140 298 24;
  • 14) 0.719 672 996 895 372 140 298 24 × 2 = 1 + 0.439 345 993 790 744 280 596 48;
  • 15) 0.439 345 993 790 744 280 596 48 × 2 = 0 + 0.878 691 987 581 488 561 192 96;
  • 16) 0.878 691 987 581 488 561 192 96 × 2 = 1 + 0.757 383 975 162 977 122 385 92;
  • 17) 0.757 383 975 162 977 122 385 92 × 2 = 1 + 0.514 767 950 325 954 244 771 84;
  • 18) 0.514 767 950 325 954 244 771 84 × 2 = 1 + 0.029 535 900 651 908 489 543 68;
  • 19) 0.029 535 900 651 908 489 543 68 × 2 = 0 + 0.059 071 801 303 816 979 087 36;
  • 20) 0.059 071 801 303 816 979 087 36 × 2 = 0 + 0.118 143 602 607 633 958 174 72;
  • 21) 0.118 143 602 607 633 958 174 72 × 2 = 0 + 0.236 287 205 215 267 916 349 44;
  • 22) 0.236 287 205 215 267 916 349 44 × 2 = 0 + 0.472 574 410 430 535 832 698 88;
  • 23) 0.472 574 410 430 535 832 698 88 × 2 = 0 + 0.945 148 820 861 071 665 397 76;
  • 24) 0.945 148 820 861 071 665 397 76 × 2 = 1 + 0.890 297 641 722 143 330 795 52;
  • 25) 0.890 297 641 722 143 330 795 52 × 2 = 1 + 0.780 595 283 444 286 661 591 04;
  • 26) 0.780 595 283 444 286 661 591 04 × 2 = 1 + 0.561 190 566 888 573 323 182 08;
  • 27) 0.561 190 566 888 573 323 182 08 × 2 = 1 + 0.122 381 133 777 146 646 364 16;
  • 28) 0.122 381 133 777 146 646 364 16 × 2 = 0 + 0.244 762 267 554 293 292 728 32;
  • 29) 0.244 762 267 554 293 292 728 32 × 2 = 0 + 0.489 524 535 108 586 585 456 64;
  • 30) 0.489 524 535 108 586 585 456 64 × 2 = 0 + 0.979 049 070 217 173 170 913 28;
  • 31) 0.979 049 070 217 173 170 913 28 × 2 = 1 + 0.958 098 140 434 346 341 826 56;
  • 32) 0.958 098 140 434 346 341 826 56 × 2 = 1 + 0.916 196 280 868 692 683 653 12;
  • 33) 0.916 196 280 868 692 683 653 12 × 2 = 1 + 0.832 392 561 737 385 367 306 24;
  • 34) 0.832 392 561 737 385 367 306 24 × 2 = 1 + 0.664 785 123 474 770 734 612 48;
  • 35) 0.664 785 123 474 770 734 612 48 × 2 = 1 + 0.329 570 246 949 541 469 224 96;
  • 36) 0.329 570 246 949 541 469 224 96 × 2 = 0 + 0.659 140 493 899 082 938 449 92;
  • 37) 0.659 140 493 899 082 938 449 92 × 2 = 1 + 0.318 280 987 798 165 876 899 84;
  • 38) 0.318 280 987 798 165 876 899 84 × 2 = 0 + 0.636 561 975 596 331 753 799 68;
  • 39) 0.636 561 975 596 331 753 799 68 × 2 = 1 + 0.273 123 951 192 663 507 599 36;
  • 40) 0.273 123 951 192 663 507 599 36 × 2 = 0 + 0.546 247 902 385 327 015 198 72;
  • 41) 0.546 247 902 385 327 015 198 72 × 2 = 1 + 0.092 495 804 770 654 030 397 44;
  • 42) 0.092 495 804 770 654 030 397 44 × 2 = 0 + 0.184 991 609 541 308 060 794 88;
  • 43) 0.184 991 609 541 308 060 794 88 × 2 = 0 + 0.369 983 219 082 616 121 589 76;
  • 44) 0.369 983 219 082 616 121 589 76 × 2 = 0 + 0.739 966 438 165 232 243 179 52;
  • 45) 0.739 966 438 165 232 243 179 52 × 2 = 1 + 0.479 932 876 330 464 486 359 04;
  • 46) 0.479 932 876 330 464 486 359 04 × 2 = 0 + 0.959 865 752 660 928 972 718 08;
  • 47) 0.959 865 752 660 928 972 718 08 × 2 = 1 + 0.919 731 505 321 857 945 436 16;
  • 48) 0.919 731 505 321 857 945 436 16 × 2 = 1 + 0.839 463 010 643 715 890 872 32;
  • 49) 0.839 463 010 643 715 890 872 32 × 2 = 1 + 0.678 926 021 287 431 781 744 64;
  • 50) 0.678 926 021 287 431 781 744 64 × 2 = 1 + 0.357 852 042 574 863 563 489 28;
  • 51) 0.357 852 042 574 863 563 489 28 × 2 = 0 + 0.715 704 085 149 727 126 978 56;
  • 52) 0.715 704 085 149 727 126 978 56 × 2 = 1 + 0.431 408 170 299 454 253 957 12;
  • 53) 0.431 408 170 299 454 253 957 12 × 2 = 0 + 0.862 816 340 598 908 507 914 24;

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:


0.545 253 866 332 628 829 606 97(10) =

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

5. Positive number before normalization:

0.545 253 866 332 628 829 606 97(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:


0.545 253 866 332 628 829 606 97(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 0.545 253 866 332 628 829 606 97 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