1.301 029 995 663 981 195 213 771 9 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.301 029 995 663 981 195 213 771 9(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
1.301 029 995 663 981 195 213 771 9(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: 1.
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;
  • 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.

1(10) =

1(2)


3. Convert to binary (base 2) the fractional part: 0.301 029 995 663 981 195 213 771 9.

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) 0.301 029 995 663 981 195 213 771 9 × 2 = 0 + 0.602 059 991 327 962 390 427 543 8;
  • 2) 0.602 059 991 327 962 390 427 543 8 × 2 = 1 + 0.204 119 982 655 924 780 855 087 6;
  • 3) 0.204 119 982 655 924 780 855 087 6 × 2 = 0 + 0.408 239 965 311 849 561 710 175 2;
  • 4) 0.408 239 965 311 849 561 710 175 2 × 2 = 0 + 0.816 479 930 623 699 123 420 350 4;
  • 5) 0.816 479 930 623 699 123 420 350 4 × 2 = 1 + 0.632 959 861 247 398 246 840 700 8;
  • 6) 0.632 959 861 247 398 246 840 700 8 × 2 = 1 + 0.265 919 722 494 796 493 681 401 6;
  • 7) 0.265 919 722 494 796 493 681 401 6 × 2 = 0 + 0.531 839 444 989 592 987 362 803 2;
  • 8) 0.531 839 444 989 592 987 362 803 2 × 2 = 1 + 0.063 678 889 979 185 974 725 606 4;
  • 9) 0.063 678 889 979 185 974 725 606 4 × 2 = 0 + 0.127 357 779 958 371 949 451 212 8;
  • 10) 0.127 357 779 958 371 949 451 212 8 × 2 = 0 + 0.254 715 559 916 743 898 902 425 6;
  • 11) 0.254 715 559 916 743 898 902 425 6 × 2 = 0 + 0.509 431 119 833 487 797 804 851 2;
  • 12) 0.509 431 119 833 487 797 804 851 2 × 2 = 1 + 0.018 862 239 666 975 595 609 702 4;
  • 13) 0.018 862 239 666 975 595 609 702 4 × 2 = 0 + 0.037 724 479 333 951 191 219 404 8;
  • 14) 0.037 724 479 333 951 191 219 404 8 × 2 = 0 + 0.075 448 958 667 902 382 438 809 6;
  • 15) 0.075 448 958 667 902 382 438 809 6 × 2 = 0 + 0.150 897 917 335 804 764 877 619 2;
  • 16) 0.150 897 917 335 804 764 877 619 2 × 2 = 0 + 0.301 795 834 671 609 529 755 238 4;
  • 17) 0.301 795 834 671 609 529 755 238 4 × 2 = 0 + 0.603 591 669 343 219 059 510 476 8;
  • 18) 0.603 591 669 343 219 059 510 476 8 × 2 = 1 + 0.207 183 338 686 438 119 020 953 6;
  • 19) 0.207 183 338 686 438 119 020 953 6 × 2 = 0 + 0.414 366 677 372 876 238 041 907 2;
  • 20) 0.414 366 677 372 876 238 041 907 2 × 2 = 0 + 0.828 733 354 745 752 476 083 814 4;
  • 21) 0.828 733 354 745 752 476 083 814 4 × 2 = 1 + 0.657 466 709 491 504 952 167 628 8;
  • 22) 0.657 466 709 491 504 952 167 628 8 × 2 = 1 + 0.314 933 418 983 009 904 335 257 6;
  • 23) 0.314 933 418 983 009 904 335 257 6 × 2 = 0 + 0.629 866 837 966 019 808 670 515 2;
  • 24) 0.629 866 837 966 019 808 670 515 2 × 2 = 1 + 0.259 733 675 932 039 617 341 030 4;
  • 25) 0.259 733 675 932 039 617 341 030 4 × 2 = 0 + 0.519 467 351 864 079 234 682 060 8;
  • 26) 0.519 467 351 864 079 234 682 060 8 × 2 = 1 + 0.038 934 703 728 158 469 364 121 6;
  • 27) 0.038 934 703 728 158 469 364 121 6 × 2 = 0 + 0.077 869 407 456 316 938 728 243 2;
  • 28) 0.077 869 407 456 316 938 728 243 2 × 2 = 0 + 0.155 738 814 912 633 877 456 486 4;
  • 29) 0.155 738 814 912 633 877 456 486 4 × 2 = 0 + 0.311 477 629 825 267 754 912 972 8;
  • 30) 0.311 477 629 825 267 754 912 972 8 × 2 = 0 + 0.622 955 259 650 535 509 825 945 6;
  • 31) 0.622 955 259 650 535 509 825 945 6 × 2 = 1 + 0.245 910 519 301 071 019 651 891 2;
  • 32) 0.245 910 519 301 071 019 651 891 2 × 2 = 0 + 0.491 821 038 602 142 039 303 782 4;
  • 33) 0.491 821 038 602 142 039 303 782 4 × 2 = 0 + 0.983 642 077 204 284 078 607 564 8;
  • 34) 0.983 642 077 204 284 078 607 564 8 × 2 = 1 + 0.967 284 154 408 568 157 215 129 6;
  • 35) 0.967 284 154 408 568 157 215 129 6 × 2 = 1 + 0.934 568 308 817 136 314 430 259 2;
  • 36) 0.934 568 308 817 136 314 430 259 2 × 2 = 1 + 0.869 136 617 634 272 628 860 518 4;
  • 37) 0.869 136 617 634 272 628 860 518 4 × 2 = 1 + 0.738 273 235 268 545 257 721 036 8;
  • 38) 0.738 273 235 268 545 257 721 036 8 × 2 = 1 + 0.476 546 470 537 090 515 442 073 6;
  • 39) 0.476 546 470 537 090 515 442 073 6 × 2 = 0 + 0.953 092 941 074 181 030 884 147 2;
  • 40) 0.953 092 941 074 181 030 884 147 2 × 2 = 1 + 0.906 185 882 148 362 061 768 294 4;
  • 41) 0.906 185 882 148 362 061 768 294 4 × 2 = 1 + 0.812 371 764 296 724 123 536 588 8;
  • 42) 0.812 371 764 296 724 123 536 588 8 × 2 = 1 + 0.624 743 528 593 448 247 073 177 6;
  • 43) 0.624 743 528 593 448 247 073 177 6 × 2 = 1 + 0.249 487 057 186 896 494 146 355 2;
  • 44) 0.249 487 057 186 896 494 146 355 2 × 2 = 0 + 0.498 974 114 373 792 988 292 710 4;
  • 45) 0.498 974 114 373 792 988 292 710 4 × 2 = 0 + 0.997 948 228 747 585 976 585 420 8;
  • 46) 0.997 948 228 747 585 976 585 420 8 × 2 = 1 + 0.995 896 457 495 171 953 170 841 6;
  • 47) 0.995 896 457 495 171 953 170 841 6 × 2 = 1 + 0.991 792 914 990 343 906 341 683 2;
  • 48) 0.991 792 914 990 343 906 341 683 2 × 2 = 1 + 0.983 585 829 980 687 812 683 366 4;
  • 49) 0.983 585 829 980 687 812 683 366 4 × 2 = 1 + 0.967 171 659 961 375 625 366 732 8;
  • 50) 0.967 171 659 961 375 625 366 732 8 × 2 = 1 + 0.934 343 319 922 751 250 733 465 6;
  • 51) 0.934 343 319 922 751 250 733 465 6 × 2 = 1 + 0.868 686 639 845 502 501 466 931 2;
  • 52) 0.868 686 639 845 502 501 466 931 2 × 2 = 1 + 0.737 373 279 691 005 002 933 862 4;
  • 53) 0.737 373 279 691 005 002 933 862 4 × 2 = 1 + 0.474 746 559 382 010 005 867 724 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:


0.301 029 995 663 981 195 213 771 9(10) =

0.0100 1101 0001 0000 0100 1101 0100 0010 0111 1101 1110 0111 1111 1(2)

5. Positive number before normalization:

1.301 029 995 663 981 195 213 771 9(10) =

1.0100 1101 0001 0000 0100 1101 0100 0010 0111 1101 1110 0111 1111 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 0 positions to the left, so that only one non zero digit remains to the left of it:


1.301 029 995 663 981 195 213 771 9(10) =

1.0100 1101 0001 0000 0100 1101 0100 0010 0111 1101 1110 0111 1111 1(2) =

1.0100 1101 0001 0000 0100 1101 0100 0010 0111 1101 1110 0111 1111 1(2) × 20


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): 0

Mantissa (not normalized):
1.0100 1101 0001 0000 0100 1101 0100 0010 0111 1101 1110 0111 1111 1


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

0 + 2(11-1) - 1 =

(0 + 1 023)(10) =

1 023(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;
  • 1 023 ÷ 2 = 511 + 1;
  • 511 ÷ 2 = 255 + 1;
  • 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) =

1023(10) =

011 1111 1111(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. 0100 1101 0001 0000 0100 1101 0100 0010 0111 1101 1110 0111 1111 1 =

0100 1101 0001 0000 0100 1101 0100 0010 0111 1101 1110 0111 1111


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 1111

Mantissa (52 bits) =
0100 1101 0001 0000 0100 1101 0100 0010 0111 1101 1110 0111 1111


Decimal number 1.301 029 995 663 981 195 213 771 9 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1111 - 0100 1101 0001 0000 0100 1101 0100 0010 0111 1101 1110 0111 1111

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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