90.899 999 999 999 991 473 487 171 16 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 90.899 999 999 999 991 473 487 171 16(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
90.899 999 999 999 991 473 487 171 16(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: 0.899 999 999 999 991 473 487 171 16.

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) 0.899 999 999 999 991 473 487 171 16 × 2 = 1 + 0.799 999 999 999 982 946 974 342 32;
  • 2) 0.799 999 999 999 982 946 974 342 32 × 2 = 1 + 0.599 999 999 999 965 893 948 684 64;
  • 3) 0.599 999 999 999 965 893 948 684 64 × 2 = 1 + 0.199 999 999 999 931 787 897 369 28;
  • 4) 0.199 999 999 999 931 787 897 369 28 × 2 = 0 + 0.399 999 999 999 863 575 794 738 56;
  • 5) 0.399 999 999 999 863 575 794 738 56 × 2 = 0 + 0.799 999 999 999 727 151 589 477 12;
  • 6) 0.799 999 999 999 727 151 589 477 12 × 2 = 1 + 0.599 999 999 999 454 303 178 954 24;
  • 7) 0.599 999 999 999 454 303 178 954 24 × 2 = 1 + 0.199 999 999 998 908 606 357 908 48;
  • 8) 0.199 999 999 998 908 606 357 908 48 × 2 = 0 + 0.399 999 999 997 817 212 715 816 96;
  • 9) 0.399 999 999 997 817 212 715 816 96 × 2 = 0 + 0.799 999 999 995 634 425 431 633 92;
  • 10) 0.799 999 999 995 634 425 431 633 92 × 2 = 1 + 0.599 999 999 991 268 850 863 267 84;
  • 11) 0.599 999 999 991 268 850 863 267 84 × 2 = 1 + 0.199 999 999 982 537 701 726 535 68;
  • 12) 0.199 999 999 982 537 701 726 535 68 × 2 = 0 + 0.399 999 999 965 075 403 453 071 36;
  • 13) 0.399 999 999 965 075 403 453 071 36 × 2 = 0 + 0.799 999 999 930 150 806 906 142 72;
  • 14) 0.799 999 999 930 150 806 906 142 72 × 2 = 1 + 0.599 999 999 860 301 613 812 285 44;
  • 15) 0.599 999 999 860 301 613 812 285 44 × 2 = 1 + 0.199 999 999 720 603 227 624 570 88;
  • 16) 0.199 999 999 720 603 227 624 570 88 × 2 = 0 + 0.399 999 999 441 206 455 249 141 76;
  • 17) 0.399 999 999 441 206 455 249 141 76 × 2 = 0 + 0.799 999 998 882 412 910 498 283 52;
  • 18) 0.799 999 998 882 412 910 498 283 52 × 2 = 1 + 0.599 999 997 764 825 820 996 567 04;
  • 19) 0.599 999 997 764 825 820 996 567 04 × 2 = 1 + 0.199 999 995 529 651 641 993 134 08;
  • 20) 0.199 999 995 529 651 641 993 134 08 × 2 = 0 + 0.399 999 991 059 303 283 986 268 16;
  • 21) 0.399 999 991 059 303 283 986 268 16 × 2 = 0 + 0.799 999 982 118 606 567 972 536 32;
  • 22) 0.799 999 982 118 606 567 972 536 32 × 2 = 1 + 0.599 999 964 237 213 135 945 072 64;
  • 23) 0.599 999 964 237 213 135 945 072 64 × 2 = 1 + 0.199 999 928 474 426 271 890 145 28;
  • 24) 0.199 999 928 474 426 271 890 145 28 × 2 = 0 + 0.399 999 856 948 852 543 780 290 56;
  • 25) 0.399 999 856 948 852 543 780 290 56 × 2 = 0 + 0.799 999 713 897 705 087 560 581 12;
  • 26) 0.799 999 713 897 705 087 560 581 12 × 2 = 1 + 0.599 999 427 795 410 175 121 162 24;
  • 27) 0.599 999 427 795 410 175 121 162 24 × 2 = 1 + 0.199 998 855 590 820 350 242 324 48;
  • 28) 0.199 998 855 590 820 350 242 324 48 × 2 = 0 + 0.399 997 711 181 640 700 484 648 96;
  • 29) 0.399 997 711 181 640 700 484 648 96 × 2 = 0 + 0.799 995 422 363 281 400 969 297 92;
  • 30) 0.799 995 422 363 281 400 969 297 92 × 2 = 1 + 0.599 990 844 726 562 801 938 595 84;
  • 31) 0.599 990 844 726 562 801 938 595 84 × 2 = 1 + 0.199 981 689 453 125 603 877 191 68;
  • 32) 0.199 981 689 453 125 603 877 191 68 × 2 = 0 + 0.399 963 378 906 251 207 754 383 36;
  • 33) 0.399 963 378 906 251 207 754 383 36 × 2 = 0 + 0.799 926 757 812 502 415 508 766 72;
  • 34) 0.799 926 757 812 502 415 508 766 72 × 2 = 1 + 0.599 853 515 625 004 831 017 533 44;
  • 35) 0.599 853 515 625 004 831 017 533 44 × 2 = 1 + 0.199 707 031 250 009 662 035 066 88;
  • 36) 0.199 707 031 250 009 662 035 066 88 × 2 = 0 + 0.399 414 062 500 019 324 070 133 76;
  • 37) 0.399 414 062 500 019 324 070 133 76 × 2 = 0 + 0.798 828 125 000 038 648 140 267 52;
  • 38) 0.798 828 125 000 038 648 140 267 52 × 2 = 1 + 0.597 656 250 000 077 296 280 535 04;
  • 39) 0.597 656 250 000 077 296 280 535 04 × 2 = 1 + 0.195 312 500 000 154 592 561 070 08;
  • 40) 0.195 312 500 000 154 592 561 070 08 × 2 = 0 + 0.390 625 000 000 309 185 122 140 16;
  • 41) 0.390 625 000 000 309 185 122 140 16 × 2 = 0 + 0.781 250 000 000 618 370 244 280 32;
  • 42) 0.781 250 000 000 618 370 244 280 32 × 2 = 1 + 0.562 500 000 001 236 740 488 560 64;
  • 43) 0.562 500 000 001 236 740 488 560 64 × 2 = 1 + 0.125 000 000 002 473 480 977 121 28;
  • 44) 0.125 000 000 002 473 480 977 121 28 × 2 = 0 + 0.250 000 000 004 946 961 954 242 56;
  • 45) 0.250 000 000 004 946 961 954 242 56 × 2 = 0 + 0.500 000 000 009 893 923 908 485 12;
  • 46) 0.500 000 000 009 893 923 908 485 12 × 2 = 1 + 0.000 000 000 019 787 847 816 970 24;
  • 47) 0.000 000 000 019 787 847 816 970 24 × 2 = 0 + 0.000 000 000 039 575 695 633 940 48;
  • 48) 0.000 000 000 039 575 695 633 940 48 × 2 = 0 + 0.000 000 000 079 151 391 267 880 96;
  • 49) 0.000 000 000 079 151 391 267 880 96 × 2 = 0 + 0.000 000 000 158 302 782 535 761 92;
  • 50) 0.000 000 000 158 302 782 535 761 92 × 2 = 0 + 0.000 000 000 316 605 565 071 523 84;
  • 51) 0.000 000 000 316 605 565 071 523 84 × 2 = 0 + 0.000 000 000 633 211 130 143 047 68;
  • 52) 0.000 000 000 633 211 130 143 047 68 × 2 = 0 + 0.000 000 001 266 422 260 286 095 36;
  • 53) 0.000 000 001 266 422 260 286 095 36 × 2 = 0 + 0.000 000 002 532 844 520 572 190 72;

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:


0.899 999 999 999 991 473 487 171 16(10) =

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

5. Positive number before normalization:

90.899 999 999 999 991 473 487 171 16(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:


90.899 999 999 999 991 473 487 171 16(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 90.899 999 999 999 991 473 487 171 16 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 90.899 999 999 999 991 473 487 171 34 to 64 bit double precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 10, 2026 [October]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: October - by our visitors

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