0.299 999 999 999 999 975 018 41 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.299 999 999 999 999 975 018 41(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
0.299 999 999 999 999 975 018 41(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: 0.299 999 999 999 999 975 018 41.

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) 0.299 999 999 999 999 975 018 41 × 2 = 0 + 0.599 999 999 999 999 950 036 82;
  • 2) 0.599 999 999 999 999 950 036 82 × 2 = 1 + 0.199 999 999 999 999 900 073 64;
  • 3) 0.199 999 999 999 999 900 073 64 × 2 = 0 + 0.399 999 999 999 999 800 147 28;
  • 4) 0.399 999 999 999 999 800 147 28 × 2 = 0 + 0.799 999 999 999 999 600 294 56;
  • 5) 0.799 999 999 999 999 600 294 56 × 2 = 1 + 0.599 999 999 999 999 200 589 12;
  • 6) 0.599 999 999 999 999 200 589 12 × 2 = 1 + 0.199 999 999 999 998 401 178 24;
  • 7) 0.199 999 999 999 998 401 178 24 × 2 = 0 + 0.399 999 999 999 996 802 356 48;
  • 8) 0.399 999 999 999 996 802 356 48 × 2 = 0 + 0.799 999 999 999 993 604 712 96;
  • 9) 0.799 999 999 999 993 604 712 96 × 2 = 1 + 0.599 999 999 999 987 209 425 92;
  • 10) 0.599 999 999 999 987 209 425 92 × 2 = 1 + 0.199 999 999 999 974 418 851 84;
  • 11) 0.199 999 999 999 974 418 851 84 × 2 = 0 + 0.399 999 999 999 948 837 703 68;
  • 12) 0.399 999 999 999 948 837 703 68 × 2 = 0 + 0.799 999 999 999 897 675 407 36;
  • 13) 0.799 999 999 999 897 675 407 36 × 2 = 1 + 0.599 999 999 999 795 350 814 72;
  • 14) 0.599 999 999 999 795 350 814 72 × 2 = 1 + 0.199 999 999 999 590 701 629 44;
  • 15) 0.199 999 999 999 590 701 629 44 × 2 = 0 + 0.399 999 999 999 181 403 258 88;
  • 16) 0.399 999 999 999 181 403 258 88 × 2 = 0 + 0.799 999 999 998 362 806 517 76;
  • 17) 0.799 999 999 998 362 806 517 76 × 2 = 1 + 0.599 999 999 996 725 613 035 52;
  • 18) 0.599 999 999 996 725 613 035 52 × 2 = 1 + 0.199 999 999 993 451 226 071 04;
  • 19) 0.199 999 999 993 451 226 071 04 × 2 = 0 + 0.399 999 999 986 902 452 142 08;
  • 20) 0.399 999 999 986 902 452 142 08 × 2 = 0 + 0.799 999 999 973 804 904 284 16;
  • 21) 0.799 999 999 973 804 904 284 16 × 2 = 1 + 0.599 999 999 947 609 808 568 32;
  • 22) 0.599 999 999 947 609 808 568 32 × 2 = 1 + 0.199 999 999 895 219 617 136 64;
  • 23) 0.199 999 999 895 219 617 136 64 × 2 = 0 + 0.399 999 999 790 439 234 273 28;
  • 24) 0.399 999 999 790 439 234 273 28 × 2 = 0 + 0.799 999 999 580 878 468 546 56;
  • 25) 0.799 999 999 580 878 468 546 56 × 2 = 1 + 0.599 999 999 161 756 937 093 12;
  • 26) 0.599 999 999 161 756 937 093 12 × 2 = 1 + 0.199 999 998 323 513 874 186 24;
  • 27) 0.199 999 998 323 513 874 186 24 × 2 = 0 + 0.399 999 996 647 027 748 372 48;
  • 28) 0.399 999 996 647 027 748 372 48 × 2 = 0 + 0.799 999 993 294 055 496 744 96;
  • 29) 0.799 999 993 294 055 496 744 96 × 2 = 1 + 0.599 999 986 588 110 993 489 92;
  • 30) 0.599 999 986 588 110 993 489 92 × 2 = 1 + 0.199 999 973 176 221 986 979 84;
  • 31) 0.199 999 973 176 221 986 979 84 × 2 = 0 + 0.399 999 946 352 443 973 959 68;
  • 32) 0.399 999 946 352 443 973 959 68 × 2 = 0 + 0.799 999 892 704 887 947 919 36;
  • 33) 0.799 999 892 704 887 947 919 36 × 2 = 1 + 0.599 999 785 409 775 895 838 72;
  • 34) 0.599 999 785 409 775 895 838 72 × 2 = 1 + 0.199 999 570 819 551 791 677 44;
  • 35) 0.199 999 570 819 551 791 677 44 × 2 = 0 + 0.399 999 141 639 103 583 354 88;
  • 36) 0.399 999 141 639 103 583 354 88 × 2 = 0 + 0.799 998 283 278 207 166 709 76;
  • 37) 0.799 998 283 278 207 166 709 76 × 2 = 1 + 0.599 996 566 556 414 333 419 52;
  • 38) 0.599 996 566 556 414 333 419 52 × 2 = 1 + 0.199 993 133 112 828 666 839 04;
  • 39) 0.199 993 133 112 828 666 839 04 × 2 = 0 + 0.399 986 266 225 657 333 678 08;
  • 40) 0.399 986 266 225 657 333 678 08 × 2 = 0 + 0.799 972 532 451 314 667 356 16;
  • 41) 0.799 972 532 451 314 667 356 16 × 2 = 1 + 0.599 945 064 902 629 334 712 32;
  • 42) 0.599 945 064 902 629 334 712 32 × 2 = 1 + 0.199 890 129 805 258 669 424 64;
  • 43) 0.199 890 129 805 258 669 424 64 × 2 = 0 + 0.399 780 259 610 517 338 849 28;
  • 44) 0.399 780 259 610 517 338 849 28 × 2 = 0 + 0.799 560 519 221 034 677 698 56;
  • 45) 0.799 560 519 221 034 677 698 56 × 2 = 1 + 0.599 121 038 442 069 355 397 12;
  • 46) 0.599 121 038 442 069 355 397 12 × 2 = 1 + 0.198 242 076 884 138 710 794 24;
  • 47) 0.198 242 076 884 138 710 794 24 × 2 = 0 + 0.396 484 153 768 277 421 588 48;
  • 48) 0.396 484 153 768 277 421 588 48 × 2 = 0 + 0.792 968 307 536 554 843 176 96;
  • 49) 0.792 968 307 536 554 843 176 96 × 2 = 1 + 0.585 936 615 073 109 686 353 92;
  • 50) 0.585 936 615 073 109 686 353 92 × 2 = 1 + 0.171 873 230 146 219 372 707 84;
  • 51) 0.171 873 230 146 219 372 707 84 × 2 = 0 + 0.343 746 460 292 438 745 415 68;
  • 52) 0.343 746 460 292 438 745 415 68 × 2 = 0 + 0.687 492 920 584 877 490 831 36;
  • 53) 0.687 492 920 584 877 490 831 36 × 2 = 1 + 0.374 985 841 169 754 981 662 72;
  • 54) 0.374 985 841 169 754 981 662 72 × 2 = 0 + 0.749 971 682 339 509 963 325 44;

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:


0.299 999 999 999 999 975 018 41(10) =

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

5. Positive number before normalization:

0.299 999 999 999 999 975 018 41(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:


0.299 999 999 999 999 975 018 41(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 0.299 999 999 999 999 975 018 41 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