0.299 999 999 999 999 975 019 975 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.299 999 999 999 999 975 019 975(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
0.299 999 999 999 999 975 019 975(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: 0.299 999 999 999 999 975 019 975.

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) 0.299 999 999 999 999 975 019 975 × 2 = 0 + 0.599 999 999 999 999 950 039 95;
  • 2) 0.599 999 999 999 999 950 039 95 × 2 = 1 + 0.199 999 999 999 999 900 079 9;
  • 3) 0.199 999 999 999 999 900 079 9 × 2 = 0 + 0.399 999 999 999 999 800 159 8;
  • 4) 0.399 999 999 999 999 800 159 8 × 2 = 0 + 0.799 999 999 999 999 600 319 6;
  • 5) 0.799 999 999 999 999 600 319 6 × 2 = 1 + 0.599 999 999 999 999 200 639 2;
  • 6) 0.599 999 999 999 999 200 639 2 × 2 = 1 + 0.199 999 999 999 998 401 278 4;
  • 7) 0.199 999 999 999 998 401 278 4 × 2 = 0 + 0.399 999 999 999 996 802 556 8;
  • 8) 0.399 999 999 999 996 802 556 8 × 2 = 0 + 0.799 999 999 999 993 605 113 6;
  • 9) 0.799 999 999 999 993 605 113 6 × 2 = 1 + 0.599 999 999 999 987 210 227 2;
  • 10) 0.599 999 999 999 987 210 227 2 × 2 = 1 + 0.199 999 999 999 974 420 454 4;
  • 11) 0.199 999 999 999 974 420 454 4 × 2 = 0 + 0.399 999 999 999 948 840 908 8;
  • 12) 0.399 999 999 999 948 840 908 8 × 2 = 0 + 0.799 999 999 999 897 681 817 6;
  • 13) 0.799 999 999 999 897 681 817 6 × 2 = 1 + 0.599 999 999 999 795 363 635 2;
  • 14) 0.599 999 999 999 795 363 635 2 × 2 = 1 + 0.199 999 999 999 590 727 270 4;
  • 15) 0.199 999 999 999 590 727 270 4 × 2 = 0 + 0.399 999 999 999 181 454 540 8;
  • 16) 0.399 999 999 999 181 454 540 8 × 2 = 0 + 0.799 999 999 998 362 909 081 6;
  • 17) 0.799 999 999 998 362 909 081 6 × 2 = 1 + 0.599 999 999 996 725 818 163 2;
  • 18) 0.599 999 999 996 725 818 163 2 × 2 = 1 + 0.199 999 999 993 451 636 326 4;
  • 19) 0.199 999 999 993 451 636 326 4 × 2 = 0 + 0.399 999 999 986 903 272 652 8;
  • 20) 0.399 999 999 986 903 272 652 8 × 2 = 0 + 0.799 999 999 973 806 545 305 6;
  • 21) 0.799 999 999 973 806 545 305 6 × 2 = 1 + 0.599 999 999 947 613 090 611 2;
  • 22) 0.599 999 999 947 613 090 611 2 × 2 = 1 + 0.199 999 999 895 226 181 222 4;
  • 23) 0.199 999 999 895 226 181 222 4 × 2 = 0 + 0.399 999 999 790 452 362 444 8;
  • 24) 0.399 999 999 790 452 362 444 8 × 2 = 0 + 0.799 999 999 580 904 724 889 6;
  • 25) 0.799 999 999 580 904 724 889 6 × 2 = 1 + 0.599 999 999 161 809 449 779 2;
  • 26) 0.599 999 999 161 809 449 779 2 × 2 = 1 + 0.199 999 998 323 618 899 558 4;
  • 27) 0.199 999 998 323 618 899 558 4 × 2 = 0 + 0.399 999 996 647 237 799 116 8;
  • 28) 0.399 999 996 647 237 799 116 8 × 2 = 0 + 0.799 999 993 294 475 598 233 6;
  • 29) 0.799 999 993 294 475 598 233 6 × 2 = 1 + 0.599 999 986 588 951 196 467 2;
  • 30) 0.599 999 986 588 951 196 467 2 × 2 = 1 + 0.199 999 973 177 902 392 934 4;
  • 31) 0.199 999 973 177 902 392 934 4 × 2 = 0 + 0.399 999 946 355 804 785 868 8;
  • 32) 0.399 999 946 355 804 785 868 8 × 2 = 0 + 0.799 999 892 711 609 571 737 6;
  • 33) 0.799 999 892 711 609 571 737 6 × 2 = 1 + 0.599 999 785 423 219 143 475 2;
  • 34) 0.599 999 785 423 219 143 475 2 × 2 = 1 + 0.199 999 570 846 438 286 950 4;
  • 35) 0.199 999 570 846 438 286 950 4 × 2 = 0 + 0.399 999 141 692 876 573 900 8;
  • 36) 0.399 999 141 692 876 573 900 8 × 2 = 0 + 0.799 998 283 385 753 147 801 6;
  • 37) 0.799 998 283 385 753 147 801 6 × 2 = 1 + 0.599 996 566 771 506 295 603 2;
  • 38) 0.599 996 566 771 506 295 603 2 × 2 = 1 + 0.199 993 133 543 012 591 206 4;
  • 39) 0.199 993 133 543 012 591 206 4 × 2 = 0 + 0.399 986 267 086 025 182 412 8;
  • 40) 0.399 986 267 086 025 182 412 8 × 2 = 0 + 0.799 972 534 172 050 364 825 6;
  • 41) 0.799 972 534 172 050 364 825 6 × 2 = 1 + 0.599 945 068 344 100 729 651 2;
  • 42) 0.599 945 068 344 100 729 651 2 × 2 = 1 + 0.199 890 136 688 201 459 302 4;
  • 43) 0.199 890 136 688 201 459 302 4 × 2 = 0 + 0.399 780 273 376 402 918 604 8;
  • 44) 0.399 780 273 376 402 918 604 8 × 2 = 0 + 0.799 560 546 752 805 837 209 6;
  • 45) 0.799 560 546 752 805 837 209 6 × 2 = 1 + 0.599 121 093 505 611 674 419 2;
  • 46) 0.599 121 093 505 611 674 419 2 × 2 = 1 + 0.198 242 187 011 223 348 838 4;
  • 47) 0.198 242 187 011 223 348 838 4 × 2 = 0 + 0.396 484 374 022 446 697 676 8;
  • 48) 0.396 484 374 022 446 697 676 8 × 2 = 0 + 0.792 968 748 044 893 395 353 6;
  • 49) 0.792 968 748 044 893 395 353 6 × 2 = 1 + 0.585 937 496 089 786 790 707 2;
  • 50) 0.585 937 496 089 786 790 707 2 × 2 = 1 + 0.171 874 992 179 573 581 414 4;
  • 51) 0.171 874 992 179 573 581 414 4 × 2 = 0 + 0.343 749 984 359 147 162 828 8;
  • 52) 0.343 749 984 359 147 162 828 8 × 2 = 0 + 0.687 499 968 718 294 325 657 6;
  • 53) 0.687 499 968 718 294 325 657 6 × 2 = 1 + 0.374 999 937 436 588 651 315 2;
  • 54) 0.374 999 937 436 588 651 315 2 × 2 = 0 + 0.749 999 874 873 177 302 630 4;

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:


0.299 999 999 999 999 975 019 975(10) =

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

5. Positive number before normalization:

0.299 999 999 999 999 975 019 975(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:


0.299 999 999 999 999 975 019 975(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 0.299 999 999 999 999 975 019 975 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