90.899 999 999 999 991 473 487 169 92 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 90.899 999 999 999 991 473 487 169 92(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
90.899 999 999 999 991 473 487 169 92(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: 90.
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;
  • 90 ÷ 2 = 45 + 0;
  • 45 ÷ 2 = 22 + 1;
  • 22 ÷ 2 = 11 + 0;
  • 11 ÷ 2 = 5 + 1;
  • 5 ÷ 2 = 2 + 1;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

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.

90(10) =

101 1010(2)


3. Convert to binary (base 2) the fractional part: 0.899 999 999 999 991 473 487 169 92.

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) 0.899 999 999 999 991 473 487 169 92 × 2 = 1 + 0.799 999 999 999 982 946 974 339 84;
  • 2) 0.799 999 999 999 982 946 974 339 84 × 2 = 1 + 0.599 999 999 999 965 893 948 679 68;
  • 3) 0.599 999 999 999 965 893 948 679 68 × 2 = 1 + 0.199 999 999 999 931 787 897 359 36;
  • 4) 0.199 999 999 999 931 787 897 359 36 × 2 = 0 + 0.399 999 999 999 863 575 794 718 72;
  • 5) 0.399 999 999 999 863 575 794 718 72 × 2 = 0 + 0.799 999 999 999 727 151 589 437 44;
  • 6) 0.799 999 999 999 727 151 589 437 44 × 2 = 1 + 0.599 999 999 999 454 303 178 874 88;
  • 7) 0.599 999 999 999 454 303 178 874 88 × 2 = 1 + 0.199 999 999 998 908 606 357 749 76;
  • 8) 0.199 999 999 998 908 606 357 749 76 × 2 = 0 + 0.399 999 999 997 817 212 715 499 52;
  • 9) 0.399 999 999 997 817 212 715 499 52 × 2 = 0 + 0.799 999 999 995 634 425 430 999 04;
  • 10) 0.799 999 999 995 634 425 430 999 04 × 2 = 1 + 0.599 999 999 991 268 850 861 998 08;
  • 11) 0.599 999 999 991 268 850 861 998 08 × 2 = 1 + 0.199 999 999 982 537 701 723 996 16;
  • 12) 0.199 999 999 982 537 701 723 996 16 × 2 = 0 + 0.399 999 999 965 075 403 447 992 32;
  • 13) 0.399 999 999 965 075 403 447 992 32 × 2 = 0 + 0.799 999 999 930 150 806 895 984 64;
  • 14) 0.799 999 999 930 150 806 895 984 64 × 2 = 1 + 0.599 999 999 860 301 613 791 969 28;
  • 15) 0.599 999 999 860 301 613 791 969 28 × 2 = 1 + 0.199 999 999 720 603 227 583 938 56;
  • 16) 0.199 999 999 720 603 227 583 938 56 × 2 = 0 + 0.399 999 999 441 206 455 167 877 12;
  • 17) 0.399 999 999 441 206 455 167 877 12 × 2 = 0 + 0.799 999 998 882 412 910 335 754 24;
  • 18) 0.799 999 998 882 412 910 335 754 24 × 2 = 1 + 0.599 999 997 764 825 820 671 508 48;
  • 19) 0.599 999 997 764 825 820 671 508 48 × 2 = 1 + 0.199 999 995 529 651 641 343 016 96;
  • 20) 0.199 999 995 529 651 641 343 016 96 × 2 = 0 + 0.399 999 991 059 303 282 686 033 92;
  • 21) 0.399 999 991 059 303 282 686 033 92 × 2 = 0 + 0.799 999 982 118 606 565 372 067 84;
  • 22) 0.799 999 982 118 606 565 372 067 84 × 2 = 1 + 0.599 999 964 237 213 130 744 135 68;
  • 23) 0.599 999 964 237 213 130 744 135 68 × 2 = 1 + 0.199 999 928 474 426 261 488 271 36;
  • 24) 0.199 999 928 474 426 261 488 271 36 × 2 = 0 + 0.399 999 856 948 852 522 976 542 72;
  • 25) 0.399 999 856 948 852 522 976 542 72 × 2 = 0 + 0.799 999 713 897 705 045 953 085 44;
  • 26) 0.799 999 713 897 705 045 953 085 44 × 2 = 1 + 0.599 999 427 795 410 091 906 170 88;
  • 27) 0.599 999 427 795 410 091 906 170 88 × 2 = 1 + 0.199 998 855 590 820 183 812 341 76;
  • 28) 0.199 998 855 590 820 183 812 341 76 × 2 = 0 + 0.399 997 711 181 640 367 624 683 52;
  • 29) 0.399 997 711 181 640 367 624 683 52 × 2 = 0 + 0.799 995 422 363 280 735 249 367 04;
  • 30) 0.799 995 422 363 280 735 249 367 04 × 2 = 1 + 0.599 990 844 726 561 470 498 734 08;
  • 31) 0.599 990 844 726 561 470 498 734 08 × 2 = 1 + 0.199 981 689 453 122 940 997 468 16;
  • 32) 0.199 981 689 453 122 940 997 468 16 × 2 = 0 + 0.399 963 378 906 245 881 994 936 32;
  • 33) 0.399 963 378 906 245 881 994 936 32 × 2 = 0 + 0.799 926 757 812 491 763 989 872 64;
  • 34) 0.799 926 757 812 491 763 989 872 64 × 2 = 1 + 0.599 853 515 624 983 527 979 745 28;
  • 35) 0.599 853 515 624 983 527 979 745 28 × 2 = 1 + 0.199 707 031 249 967 055 959 490 56;
  • 36) 0.199 707 031 249 967 055 959 490 56 × 2 = 0 + 0.399 414 062 499 934 111 918 981 12;
  • 37) 0.399 414 062 499 934 111 918 981 12 × 2 = 0 + 0.798 828 124 999 868 223 837 962 24;
  • 38) 0.798 828 124 999 868 223 837 962 24 × 2 = 1 + 0.597 656 249 999 736 447 675 924 48;
  • 39) 0.597 656 249 999 736 447 675 924 48 × 2 = 1 + 0.195 312 499 999 472 895 351 848 96;
  • 40) 0.195 312 499 999 472 895 351 848 96 × 2 = 0 + 0.390 624 999 998 945 790 703 697 92;
  • 41) 0.390 624 999 998 945 790 703 697 92 × 2 = 0 + 0.781 249 999 997 891 581 407 395 84;
  • 42) 0.781 249 999 997 891 581 407 395 84 × 2 = 1 + 0.562 499 999 995 783 162 814 791 68;
  • 43) 0.562 499 999 995 783 162 814 791 68 × 2 = 1 + 0.124 999 999 991 566 325 629 583 36;
  • 44) 0.124 999 999 991 566 325 629 583 36 × 2 = 0 + 0.249 999 999 983 132 651 259 166 72;
  • 45) 0.249 999 999 983 132 651 259 166 72 × 2 = 0 + 0.499 999 999 966 265 302 518 333 44;
  • 46) 0.499 999 999 966 265 302 518 333 44 × 2 = 0 + 0.999 999 999 932 530 605 036 666 88;
  • 47) 0.999 999 999 932 530 605 036 666 88 × 2 = 1 + 0.999 999 999 865 061 210 073 333 76;
  • 48) 0.999 999 999 865 061 210 073 333 76 × 2 = 1 + 0.999 999 999 730 122 420 146 667 52;
  • 49) 0.999 999 999 730 122 420 146 667 52 × 2 = 1 + 0.999 999 999 460 244 840 293 335 04;
  • 50) 0.999 999 999 460 244 840 293 335 04 × 2 = 1 + 0.999 999 998 920 489 680 586 670 08;
  • 51) 0.999 999 998 920 489 680 586 670 08 × 2 = 1 + 0.999 999 997 840 979 361 173 340 16;
  • 52) 0.999 999 997 840 979 361 173 340 16 × 2 = 1 + 0.999 999 995 681 958 722 346 680 32;
  • 53) 0.999 999 995 681 958 722 346 680 32 × 2 = 1 + 0.999 999 991 363 917 444 693 360 64;

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:


0.899 999 999 999 991 473 487 169 92(10) =

0.1110 0110 0110 0110 0110 0110 0110 0110 0110 0110 0110 0011 1111 1(2)

5. Positive number before normalization:

90.899 999 999 999 991 473 487 169 92(10) =

101 1010.1110 0110 0110 0110 0110 0110 0110 0110 0110 0110 0110 0011 1111 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 6 positions to the left, so that only one non zero digit remains to the left of it:


90.899 999 999 999 991 473 487 169 92(10) =

101 1010.1110 0110 0110 0110 0110 0110 0110 0110 0110 0110 0110 0011 1111 1(2) =

101 1010.1110 0110 0110 0110 0110 0110 0110 0110 0110 0110 0110 0011 1111 1(2) × 20 =

1.0110 1011 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1000 1111 111(2) × 26


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): 6

Mantissa (not normalized):
1.0110 1011 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1000 1111 111


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

6 + 2(11-1) - 1 =

(6 + 1 023)(10) =

1 029(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;
  • 1 029 ÷ 2 = 514 + 1;
  • 514 ÷ 2 = 257 + 0;
  • 257 ÷ 2 = 128 + 1;
  • 128 ÷ 2 = 64 + 0;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 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) =

1029(10) =

100 0000 0101(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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =

1. 0110 1011 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1000 111 1111 =

0110 1011 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1000


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) =
100 0000 0101

Mantissa (52 bits) =
0110 1011 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1000


Decimal number 90.899 999 999 999 991 473 487 169 92 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0101 - 0110 1011 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1000

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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