90.899 999 999 999 991 473 487 170 9 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 90.899 999 999 999 991 473 487 170 9(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
90.899 999 999 999 991 473 487 170 9(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: 0.899 999 999 999 991 473 487 170 9.

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) 0.899 999 999 999 991 473 487 170 9 × 2 = 1 + 0.799 999 999 999 982 946 974 341 8;
  • 2) 0.799 999 999 999 982 946 974 341 8 × 2 = 1 + 0.599 999 999 999 965 893 948 683 6;
  • 3) 0.599 999 999 999 965 893 948 683 6 × 2 = 1 + 0.199 999 999 999 931 787 897 367 2;
  • 4) 0.199 999 999 999 931 787 897 367 2 × 2 = 0 + 0.399 999 999 999 863 575 794 734 4;
  • 5) 0.399 999 999 999 863 575 794 734 4 × 2 = 0 + 0.799 999 999 999 727 151 589 468 8;
  • 6) 0.799 999 999 999 727 151 589 468 8 × 2 = 1 + 0.599 999 999 999 454 303 178 937 6;
  • 7) 0.599 999 999 999 454 303 178 937 6 × 2 = 1 + 0.199 999 999 998 908 606 357 875 2;
  • 8) 0.199 999 999 998 908 606 357 875 2 × 2 = 0 + 0.399 999 999 997 817 212 715 750 4;
  • 9) 0.399 999 999 997 817 212 715 750 4 × 2 = 0 + 0.799 999 999 995 634 425 431 500 8;
  • 10) 0.799 999 999 995 634 425 431 500 8 × 2 = 1 + 0.599 999 999 991 268 850 863 001 6;
  • 11) 0.599 999 999 991 268 850 863 001 6 × 2 = 1 + 0.199 999 999 982 537 701 726 003 2;
  • 12) 0.199 999 999 982 537 701 726 003 2 × 2 = 0 + 0.399 999 999 965 075 403 452 006 4;
  • 13) 0.399 999 999 965 075 403 452 006 4 × 2 = 0 + 0.799 999 999 930 150 806 904 012 8;
  • 14) 0.799 999 999 930 150 806 904 012 8 × 2 = 1 + 0.599 999 999 860 301 613 808 025 6;
  • 15) 0.599 999 999 860 301 613 808 025 6 × 2 = 1 + 0.199 999 999 720 603 227 616 051 2;
  • 16) 0.199 999 999 720 603 227 616 051 2 × 2 = 0 + 0.399 999 999 441 206 455 232 102 4;
  • 17) 0.399 999 999 441 206 455 232 102 4 × 2 = 0 + 0.799 999 998 882 412 910 464 204 8;
  • 18) 0.799 999 998 882 412 910 464 204 8 × 2 = 1 + 0.599 999 997 764 825 820 928 409 6;
  • 19) 0.599 999 997 764 825 820 928 409 6 × 2 = 1 + 0.199 999 995 529 651 641 856 819 2;
  • 20) 0.199 999 995 529 651 641 856 819 2 × 2 = 0 + 0.399 999 991 059 303 283 713 638 4;
  • 21) 0.399 999 991 059 303 283 713 638 4 × 2 = 0 + 0.799 999 982 118 606 567 427 276 8;
  • 22) 0.799 999 982 118 606 567 427 276 8 × 2 = 1 + 0.599 999 964 237 213 134 854 553 6;
  • 23) 0.599 999 964 237 213 134 854 553 6 × 2 = 1 + 0.199 999 928 474 426 269 709 107 2;
  • 24) 0.199 999 928 474 426 269 709 107 2 × 2 = 0 + 0.399 999 856 948 852 539 418 214 4;
  • 25) 0.399 999 856 948 852 539 418 214 4 × 2 = 0 + 0.799 999 713 897 705 078 836 428 8;
  • 26) 0.799 999 713 897 705 078 836 428 8 × 2 = 1 + 0.599 999 427 795 410 157 672 857 6;
  • 27) 0.599 999 427 795 410 157 672 857 6 × 2 = 1 + 0.199 998 855 590 820 315 345 715 2;
  • 28) 0.199 998 855 590 820 315 345 715 2 × 2 = 0 + 0.399 997 711 181 640 630 691 430 4;
  • 29) 0.399 997 711 181 640 630 691 430 4 × 2 = 0 + 0.799 995 422 363 281 261 382 860 8;
  • 30) 0.799 995 422 363 281 261 382 860 8 × 2 = 1 + 0.599 990 844 726 562 522 765 721 6;
  • 31) 0.599 990 844 726 562 522 765 721 6 × 2 = 1 + 0.199 981 689 453 125 045 531 443 2;
  • 32) 0.199 981 689 453 125 045 531 443 2 × 2 = 0 + 0.399 963 378 906 250 091 062 886 4;
  • 33) 0.399 963 378 906 250 091 062 886 4 × 2 = 0 + 0.799 926 757 812 500 182 125 772 8;
  • 34) 0.799 926 757 812 500 182 125 772 8 × 2 = 1 + 0.599 853 515 625 000 364 251 545 6;
  • 35) 0.599 853 515 625 000 364 251 545 6 × 2 = 1 + 0.199 707 031 250 000 728 503 091 2;
  • 36) 0.199 707 031 250 000 728 503 091 2 × 2 = 0 + 0.399 414 062 500 001 457 006 182 4;
  • 37) 0.399 414 062 500 001 457 006 182 4 × 2 = 0 + 0.798 828 125 000 002 914 012 364 8;
  • 38) 0.798 828 125 000 002 914 012 364 8 × 2 = 1 + 0.597 656 250 000 005 828 024 729 6;
  • 39) 0.597 656 250 000 005 828 024 729 6 × 2 = 1 + 0.195 312 500 000 011 656 049 459 2;
  • 40) 0.195 312 500 000 011 656 049 459 2 × 2 = 0 + 0.390 625 000 000 023 312 098 918 4;
  • 41) 0.390 625 000 000 023 312 098 918 4 × 2 = 0 + 0.781 250 000 000 046 624 197 836 8;
  • 42) 0.781 250 000 000 046 624 197 836 8 × 2 = 1 + 0.562 500 000 000 093 248 395 673 6;
  • 43) 0.562 500 000 000 093 248 395 673 6 × 2 = 1 + 0.125 000 000 000 186 496 791 347 2;
  • 44) 0.125 000 000 000 186 496 791 347 2 × 2 = 0 + 0.250 000 000 000 372 993 582 694 4;
  • 45) 0.250 000 000 000 372 993 582 694 4 × 2 = 0 + 0.500 000 000 000 745 987 165 388 8;
  • 46) 0.500 000 000 000 745 987 165 388 8 × 2 = 1 + 0.000 000 000 001 491 974 330 777 6;
  • 47) 0.000 000 000 001 491 974 330 777 6 × 2 = 0 + 0.000 000 000 002 983 948 661 555 2;
  • 48) 0.000 000 000 002 983 948 661 555 2 × 2 = 0 + 0.000 000 000 005 967 897 323 110 4;
  • 49) 0.000 000 000 005 967 897 323 110 4 × 2 = 0 + 0.000 000 000 011 935 794 646 220 8;
  • 50) 0.000 000 000 011 935 794 646 220 8 × 2 = 0 + 0.000 000 000 023 871 589 292 441 6;
  • 51) 0.000 000 000 023 871 589 292 441 6 × 2 = 0 + 0.000 000 000 047 743 178 584 883 2;
  • 52) 0.000 000 000 047 743 178 584 883 2 × 2 = 0 + 0.000 000 000 095 486 357 169 766 4;
  • 53) 0.000 000 000 095 486 357 169 766 4 × 2 = 0 + 0.000 000 000 190 972 714 339 532 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:


0.899 999 999 999 991 473 487 170 9(10) =

0.1110 0110 0110 0110 0110 0110 0110 0110 0110 0110 0110 0100 0000 0(2)

5. Positive number before normalization:

90.899 999 999 999 991 473 487 170 9(10) =

101 1010.1110 0110 0110 0110 0110 0110 0110 0110 0110 0110 0110 0100 0000 0(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:


90.899 999 999 999 991 473 487 170 9(10) =

101 1010.1110 0110 0110 0110 0110 0110 0110 0110 0110 0110 0110 0100 0000 0(2) =

101 1010.1110 0110 0110 0110 0110 0110 0110 0110 0110 0110 0110 0100 0000 0(2) × 20 =

1.0110 1011 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 0000 000(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 1001 0000 000


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 1001 000 0000 =

0110 1011 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001


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 1001


Decimal number 90.899 999 999 999 991 473 487 170 9 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 1001

» Convert decimal 90.899 999 999 999 991 473 487 175 to 64 bit double precision IEEE 754 binary floating point representation standard

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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