0.299 999 999 999 999 975 017 55 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.299 999 999 999 999 975 017 55(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
0.299 999 999 999 999 975 017 55(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: 0.299 999 999 999 999 975 017 55.

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) 0.299 999 999 999 999 975 017 55 × 2 = 0 + 0.599 999 999 999 999 950 035 1;
  • 2) 0.599 999 999 999 999 950 035 1 × 2 = 1 + 0.199 999 999 999 999 900 070 2;
  • 3) 0.199 999 999 999 999 900 070 2 × 2 = 0 + 0.399 999 999 999 999 800 140 4;
  • 4) 0.399 999 999 999 999 800 140 4 × 2 = 0 + 0.799 999 999 999 999 600 280 8;
  • 5) 0.799 999 999 999 999 600 280 8 × 2 = 1 + 0.599 999 999 999 999 200 561 6;
  • 6) 0.599 999 999 999 999 200 561 6 × 2 = 1 + 0.199 999 999 999 998 401 123 2;
  • 7) 0.199 999 999 999 998 401 123 2 × 2 = 0 + 0.399 999 999 999 996 802 246 4;
  • 8) 0.399 999 999 999 996 802 246 4 × 2 = 0 + 0.799 999 999 999 993 604 492 8;
  • 9) 0.799 999 999 999 993 604 492 8 × 2 = 1 + 0.599 999 999 999 987 208 985 6;
  • 10) 0.599 999 999 999 987 208 985 6 × 2 = 1 + 0.199 999 999 999 974 417 971 2;
  • 11) 0.199 999 999 999 974 417 971 2 × 2 = 0 + 0.399 999 999 999 948 835 942 4;
  • 12) 0.399 999 999 999 948 835 942 4 × 2 = 0 + 0.799 999 999 999 897 671 884 8;
  • 13) 0.799 999 999 999 897 671 884 8 × 2 = 1 + 0.599 999 999 999 795 343 769 6;
  • 14) 0.599 999 999 999 795 343 769 6 × 2 = 1 + 0.199 999 999 999 590 687 539 2;
  • 15) 0.199 999 999 999 590 687 539 2 × 2 = 0 + 0.399 999 999 999 181 375 078 4;
  • 16) 0.399 999 999 999 181 375 078 4 × 2 = 0 + 0.799 999 999 998 362 750 156 8;
  • 17) 0.799 999 999 998 362 750 156 8 × 2 = 1 + 0.599 999 999 996 725 500 313 6;
  • 18) 0.599 999 999 996 725 500 313 6 × 2 = 1 + 0.199 999 999 993 451 000 627 2;
  • 19) 0.199 999 999 993 451 000 627 2 × 2 = 0 + 0.399 999 999 986 902 001 254 4;
  • 20) 0.399 999 999 986 902 001 254 4 × 2 = 0 + 0.799 999 999 973 804 002 508 8;
  • 21) 0.799 999 999 973 804 002 508 8 × 2 = 1 + 0.599 999 999 947 608 005 017 6;
  • 22) 0.599 999 999 947 608 005 017 6 × 2 = 1 + 0.199 999 999 895 216 010 035 2;
  • 23) 0.199 999 999 895 216 010 035 2 × 2 = 0 + 0.399 999 999 790 432 020 070 4;
  • 24) 0.399 999 999 790 432 020 070 4 × 2 = 0 + 0.799 999 999 580 864 040 140 8;
  • 25) 0.799 999 999 580 864 040 140 8 × 2 = 1 + 0.599 999 999 161 728 080 281 6;
  • 26) 0.599 999 999 161 728 080 281 6 × 2 = 1 + 0.199 999 998 323 456 160 563 2;
  • 27) 0.199 999 998 323 456 160 563 2 × 2 = 0 + 0.399 999 996 646 912 321 126 4;
  • 28) 0.399 999 996 646 912 321 126 4 × 2 = 0 + 0.799 999 993 293 824 642 252 8;
  • 29) 0.799 999 993 293 824 642 252 8 × 2 = 1 + 0.599 999 986 587 649 284 505 6;
  • 30) 0.599 999 986 587 649 284 505 6 × 2 = 1 + 0.199 999 973 175 298 569 011 2;
  • 31) 0.199 999 973 175 298 569 011 2 × 2 = 0 + 0.399 999 946 350 597 138 022 4;
  • 32) 0.399 999 946 350 597 138 022 4 × 2 = 0 + 0.799 999 892 701 194 276 044 8;
  • 33) 0.799 999 892 701 194 276 044 8 × 2 = 1 + 0.599 999 785 402 388 552 089 6;
  • 34) 0.599 999 785 402 388 552 089 6 × 2 = 1 + 0.199 999 570 804 777 104 179 2;
  • 35) 0.199 999 570 804 777 104 179 2 × 2 = 0 + 0.399 999 141 609 554 208 358 4;
  • 36) 0.399 999 141 609 554 208 358 4 × 2 = 0 + 0.799 998 283 219 108 416 716 8;
  • 37) 0.799 998 283 219 108 416 716 8 × 2 = 1 + 0.599 996 566 438 216 833 433 6;
  • 38) 0.599 996 566 438 216 833 433 6 × 2 = 1 + 0.199 993 132 876 433 666 867 2;
  • 39) 0.199 993 132 876 433 666 867 2 × 2 = 0 + 0.399 986 265 752 867 333 734 4;
  • 40) 0.399 986 265 752 867 333 734 4 × 2 = 0 + 0.799 972 531 505 734 667 468 8;
  • 41) 0.799 972 531 505 734 667 468 8 × 2 = 1 + 0.599 945 063 011 469 334 937 6;
  • 42) 0.599 945 063 011 469 334 937 6 × 2 = 1 + 0.199 890 126 022 938 669 875 2;
  • 43) 0.199 890 126 022 938 669 875 2 × 2 = 0 + 0.399 780 252 045 877 339 750 4;
  • 44) 0.399 780 252 045 877 339 750 4 × 2 = 0 + 0.799 560 504 091 754 679 500 8;
  • 45) 0.799 560 504 091 754 679 500 8 × 2 = 1 + 0.599 121 008 183 509 359 001 6;
  • 46) 0.599 121 008 183 509 359 001 6 × 2 = 1 + 0.198 242 016 367 018 718 003 2;
  • 47) 0.198 242 016 367 018 718 003 2 × 2 = 0 + 0.396 484 032 734 037 436 006 4;
  • 48) 0.396 484 032 734 037 436 006 4 × 2 = 0 + 0.792 968 065 468 074 872 012 8;
  • 49) 0.792 968 065 468 074 872 012 8 × 2 = 1 + 0.585 936 130 936 149 744 025 6;
  • 50) 0.585 936 130 936 149 744 025 6 × 2 = 1 + 0.171 872 261 872 299 488 051 2;
  • 51) 0.171 872 261 872 299 488 051 2 × 2 = 0 + 0.343 744 523 744 598 976 102 4;
  • 52) 0.343 744 523 744 598 976 102 4 × 2 = 0 + 0.687 489 047 489 197 952 204 8;
  • 53) 0.687 489 047 489 197 952 204 8 × 2 = 1 + 0.374 978 094 978 395 904 409 6;
  • 54) 0.374 978 094 978 395 904 409 6 × 2 = 0 + 0.749 956 189 956 791 808 819 2;

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:


0.299 999 999 999 999 975 017 55(10) =

0.0100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 10(2)

5. Positive number before normalization:

0.299 999 999 999 999 975 017 55(10) =

0.0100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 10(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 2 positions to the right, so that only one non zero digit remains to the left of it:


0.299 999 999 999 999 975 017 55(10) =

0.0100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 10(2) =

0.0100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 10(2) × 20 =

1.0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0010(2) × 2-2


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): -2

Mantissa (not normalized):
1.0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0010


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

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

(-2 + 1 023)(10) =

1 021(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;
  • 1 021 ÷ 2 = 510 + 1;
  • 510 ÷ 2 = 255 + 0;
  • 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) =

1021(10) =

011 1111 1101(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. 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0010 =

0011 0011 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) =
011 1111 1101

Mantissa (52 bits) =
0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0010


Decimal number 0.299 999 999 999 999 975 017 55 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1101 - 0011 0011 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