92.299 999 999 999 997 157 829 026 4 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 92.299 999 999 999 997 157 829 026 4(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
92.299 999 999 999 997 157 829 026 4(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: 92.
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;
  • 92 ÷ 2 = 46 + 0;
  • 46 ÷ 2 = 23 + 0;
  • 23 ÷ 2 = 11 + 1;
  • 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.

92(10) =

101 1100(2)


3. Convert to binary (base 2) the fractional part: 0.299 999 999 999 997 157 829 026 4.

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) 0.299 999 999 999 997 157 829 026 4 × 2 = 0 + 0.599 999 999 999 994 315 658 052 8;
  • 2) 0.599 999 999 999 994 315 658 052 8 × 2 = 1 + 0.199 999 999 999 988 631 316 105 6;
  • 3) 0.199 999 999 999 988 631 316 105 6 × 2 = 0 + 0.399 999 999 999 977 262 632 211 2;
  • 4) 0.399 999 999 999 977 262 632 211 2 × 2 = 0 + 0.799 999 999 999 954 525 264 422 4;
  • 5) 0.799 999 999 999 954 525 264 422 4 × 2 = 1 + 0.599 999 999 999 909 050 528 844 8;
  • 6) 0.599 999 999 999 909 050 528 844 8 × 2 = 1 + 0.199 999 999 999 818 101 057 689 6;
  • 7) 0.199 999 999 999 818 101 057 689 6 × 2 = 0 + 0.399 999 999 999 636 202 115 379 2;
  • 8) 0.399 999 999 999 636 202 115 379 2 × 2 = 0 + 0.799 999 999 999 272 404 230 758 4;
  • 9) 0.799 999 999 999 272 404 230 758 4 × 2 = 1 + 0.599 999 999 998 544 808 461 516 8;
  • 10) 0.599 999 999 998 544 808 461 516 8 × 2 = 1 + 0.199 999 999 997 089 616 923 033 6;
  • 11) 0.199 999 999 997 089 616 923 033 6 × 2 = 0 + 0.399 999 999 994 179 233 846 067 2;
  • 12) 0.399 999 999 994 179 233 846 067 2 × 2 = 0 + 0.799 999 999 988 358 467 692 134 4;
  • 13) 0.799 999 999 988 358 467 692 134 4 × 2 = 1 + 0.599 999 999 976 716 935 384 268 8;
  • 14) 0.599 999 999 976 716 935 384 268 8 × 2 = 1 + 0.199 999 999 953 433 870 768 537 6;
  • 15) 0.199 999 999 953 433 870 768 537 6 × 2 = 0 + 0.399 999 999 906 867 741 537 075 2;
  • 16) 0.399 999 999 906 867 741 537 075 2 × 2 = 0 + 0.799 999 999 813 735 483 074 150 4;
  • 17) 0.799 999 999 813 735 483 074 150 4 × 2 = 1 + 0.599 999 999 627 470 966 148 300 8;
  • 18) 0.599 999 999 627 470 966 148 300 8 × 2 = 1 + 0.199 999 999 254 941 932 296 601 6;
  • 19) 0.199 999 999 254 941 932 296 601 6 × 2 = 0 + 0.399 999 998 509 883 864 593 203 2;
  • 20) 0.399 999 998 509 883 864 593 203 2 × 2 = 0 + 0.799 999 997 019 767 729 186 406 4;
  • 21) 0.799 999 997 019 767 729 186 406 4 × 2 = 1 + 0.599 999 994 039 535 458 372 812 8;
  • 22) 0.599 999 994 039 535 458 372 812 8 × 2 = 1 + 0.199 999 988 079 070 916 745 625 6;
  • 23) 0.199 999 988 079 070 916 745 625 6 × 2 = 0 + 0.399 999 976 158 141 833 491 251 2;
  • 24) 0.399 999 976 158 141 833 491 251 2 × 2 = 0 + 0.799 999 952 316 283 666 982 502 4;
  • 25) 0.799 999 952 316 283 666 982 502 4 × 2 = 1 + 0.599 999 904 632 567 333 965 004 8;
  • 26) 0.599 999 904 632 567 333 965 004 8 × 2 = 1 + 0.199 999 809 265 134 667 930 009 6;
  • 27) 0.199 999 809 265 134 667 930 009 6 × 2 = 0 + 0.399 999 618 530 269 335 860 019 2;
  • 28) 0.399 999 618 530 269 335 860 019 2 × 2 = 0 + 0.799 999 237 060 538 671 720 038 4;
  • 29) 0.799 999 237 060 538 671 720 038 4 × 2 = 1 + 0.599 998 474 121 077 343 440 076 8;
  • 30) 0.599 998 474 121 077 343 440 076 8 × 2 = 1 + 0.199 996 948 242 154 686 880 153 6;
  • 31) 0.199 996 948 242 154 686 880 153 6 × 2 = 0 + 0.399 993 896 484 309 373 760 307 2;
  • 32) 0.399 993 896 484 309 373 760 307 2 × 2 = 0 + 0.799 987 792 968 618 747 520 614 4;
  • 33) 0.799 987 792 968 618 747 520 614 4 × 2 = 1 + 0.599 975 585 937 237 495 041 228 8;
  • 34) 0.599 975 585 937 237 495 041 228 8 × 2 = 1 + 0.199 951 171 874 474 990 082 457 6;
  • 35) 0.199 951 171 874 474 990 082 457 6 × 2 = 0 + 0.399 902 343 748 949 980 164 915 2;
  • 36) 0.399 902 343 748 949 980 164 915 2 × 2 = 0 + 0.799 804 687 497 899 960 329 830 4;
  • 37) 0.799 804 687 497 899 960 329 830 4 × 2 = 1 + 0.599 609 374 995 799 920 659 660 8;
  • 38) 0.599 609 374 995 799 920 659 660 8 × 2 = 1 + 0.199 218 749 991 599 841 319 321 6;
  • 39) 0.199 218 749 991 599 841 319 321 6 × 2 = 0 + 0.398 437 499 983 199 682 638 643 2;
  • 40) 0.398 437 499 983 199 682 638 643 2 × 2 = 0 + 0.796 874 999 966 399 365 277 286 4;
  • 41) 0.796 874 999 966 399 365 277 286 4 × 2 = 1 + 0.593 749 999 932 798 730 554 572 8;
  • 42) 0.593 749 999 932 798 730 554 572 8 × 2 = 1 + 0.187 499 999 865 597 461 109 145 6;
  • 43) 0.187 499 999 865 597 461 109 145 6 × 2 = 0 + 0.374 999 999 731 194 922 218 291 2;
  • 44) 0.374 999 999 731 194 922 218 291 2 × 2 = 0 + 0.749 999 999 462 389 844 436 582 4;
  • 45) 0.749 999 999 462 389 844 436 582 4 × 2 = 1 + 0.499 999 998 924 779 688 873 164 8;
  • 46) 0.499 999 998 924 779 688 873 164 8 × 2 = 0 + 0.999 999 997 849 559 377 746 329 6;
  • 47) 0.999 999 997 849 559 377 746 329 6 × 2 = 1 + 0.999 999 995 699 118 755 492 659 2;
  • 48) 0.999 999 995 699 118 755 492 659 2 × 2 = 1 + 0.999 999 991 398 237 510 985 318 4;
  • 49) 0.999 999 991 398 237 510 985 318 4 × 2 = 1 + 0.999 999 982 796 475 021 970 636 8;
  • 50) 0.999 999 982 796 475 021 970 636 8 × 2 = 1 + 0.999 999 965 592 950 043 941 273 6;
  • 51) 0.999 999 965 592 950 043 941 273 6 × 2 = 1 + 0.999 999 931 185 900 087 882 547 2;
  • 52) 0.999 999 931 185 900 087 882 547 2 × 2 = 1 + 0.999 999 862 371 800 175 765 094 4;
  • 53) 0.999 999 862 371 800 175 765 094 4 × 2 = 1 + 0.999 999 724 743 600 351 530 188 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:


0.299 999 999 999 997 157 829 026 4(10) =

0.0100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1011 1111 1(2)

5. Positive number before normalization:

92.299 999 999 999 997 157 829 026 4(10) =

101 1100.0100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1011 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:


92.299 999 999 999 997 157 829 026 4(10) =

101 1100.0100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1011 1111 1(2) =

101 1100.0100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1011 1111 1(2) × 20 =

1.0111 0001 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0010 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):
1.0111 0001 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0010 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. 0111 0001 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0010 111 1111 =

0111 0001 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0010


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) =
0111 0001 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0010


Decimal number 92.299 999 999 999 997 157 829 026 4 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0101 - 0111 0001 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0010

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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