1.301 029 995 663 981 321 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.301 029 995 663 981 321(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
1.301 029 995 663 981 321(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: 0.301 029 995 663 981 321.

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) 0.301 029 995 663 981 321 × 2 = 0 + 0.602 059 991 327 962 642;
  • 2) 0.602 059 991 327 962 642 × 2 = 1 + 0.204 119 982 655 925 284;
  • 3) 0.204 119 982 655 925 284 × 2 = 0 + 0.408 239 965 311 850 568;
  • 4) 0.408 239 965 311 850 568 × 2 = 0 + 0.816 479 930 623 701 136;
  • 5) 0.816 479 930 623 701 136 × 2 = 1 + 0.632 959 861 247 402 272;
  • 6) 0.632 959 861 247 402 272 × 2 = 1 + 0.265 919 722 494 804 544;
  • 7) 0.265 919 722 494 804 544 × 2 = 0 + 0.531 839 444 989 609 088;
  • 8) 0.531 839 444 989 609 088 × 2 = 1 + 0.063 678 889 979 218 176;
  • 9) 0.063 678 889 979 218 176 × 2 = 0 + 0.127 357 779 958 436 352;
  • 10) 0.127 357 779 958 436 352 × 2 = 0 + 0.254 715 559 916 872 704;
  • 11) 0.254 715 559 916 872 704 × 2 = 0 + 0.509 431 119 833 745 408;
  • 12) 0.509 431 119 833 745 408 × 2 = 1 + 0.018 862 239 667 490 816;
  • 13) 0.018 862 239 667 490 816 × 2 = 0 + 0.037 724 479 334 981 632;
  • 14) 0.037 724 479 334 981 632 × 2 = 0 + 0.075 448 958 669 963 264;
  • 15) 0.075 448 958 669 963 264 × 2 = 0 + 0.150 897 917 339 926 528;
  • 16) 0.150 897 917 339 926 528 × 2 = 0 + 0.301 795 834 679 853 056;
  • 17) 0.301 795 834 679 853 056 × 2 = 0 + 0.603 591 669 359 706 112;
  • 18) 0.603 591 669 359 706 112 × 2 = 1 + 0.207 183 338 719 412 224;
  • 19) 0.207 183 338 719 412 224 × 2 = 0 + 0.414 366 677 438 824 448;
  • 20) 0.414 366 677 438 824 448 × 2 = 0 + 0.828 733 354 877 648 896;
  • 21) 0.828 733 354 877 648 896 × 2 = 1 + 0.657 466 709 755 297 792;
  • 22) 0.657 466 709 755 297 792 × 2 = 1 + 0.314 933 419 510 595 584;
  • 23) 0.314 933 419 510 595 584 × 2 = 0 + 0.629 866 839 021 191 168;
  • 24) 0.629 866 839 021 191 168 × 2 = 1 + 0.259 733 678 042 382 336;
  • 25) 0.259 733 678 042 382 336 × 2 = 0 + 0.519 467 356 084 764 672;
  • 26) 0.519 467 356 084 764 672 × 2 = 1 + 0.038 934 712 169 529 344;
  • 27) 0.038 934 712 169 529 344 × 2 = 0 + 0.077 869 424 339 058 688;
  • 28) 0.077 869 424 339 058 688 × 2 = 0 + 0.155 738 848 678 117 376;
  • 29) 0.155 738 848 678 117 376 × 2 = 0 + 0.311 477 697 356 234 752;
  • 30) 0.311 477 697 356 234 752 × 2 = 0 + 0.622 955 394 712 469 504;
  • 31) 0.622 955 394 712 469 504 × 2 = 1 + 0.245 910 789 424 939 008;
  • 32) 0.245 910 789 424 939 008 × 2 = 0 + 0.491 821 578 849 878 016;
  • 33) 0.491 821 578 849 878 016 × 2 = 0 + 0.983 643 157 699 756 032;
  • 34) 0.983 643 157 699 756 032 × 2 = 1 + 0.967 286 315 399 512 064;
  • 35) 0.967 286 315 399 512 064 × 2 = 1 + 0.934 572 630 799 024 128;
  • 36) 0.934 572 630 799 024 128 × 2 = 1 + 0.869 145 261 598 048 256;
  • 37) 0.869 145 261 598 048 256 × 2 = 1 + 0.738 290 523 196 096 512;
  • 38) 0.738 290 523 196 096 512 × 2 = 1 + 0.476 581 046 392 193 024;
  • 39) 0.476 581 046 392 193 024 × 2 = 0 + 0.953 162 092 784 386 048;
  • 40) 0.953 162 092 784 386 048 × 2 = 1 + 0.906 324 185 568 772 096;
  • 41) 0.906 324 185 568 772 096 × 2 = 1 + 0.812 648 371 137 544 192;
  • 42) 0.812 648 371 137 544 192 × 2 = 1 + 0.625 296 742 275 088 384;
  • 43) 0.625 296 742 275 088 384 × 2 = 1 + 0.250 593 484 550 176 768;
  • 44) 0.250 593 484 550 176 768 × 2 = 0 + 0.501 186 969 100 353 536;
  • 45) 0.501 186 969 100 353 536 × 2 = 1 + 0.002 373 938 200 707 072;
  • 46) 0.002 373 938 200 707 072 × 2 = 0 + 0.004 747 876 401 414 144;
  • 47) 0.004 747 876 401 414 144 × 2 = 0 + 0.009 495 752 802 828 288;
  • 48) 0.009 495 752 802 828 288 × 2 = 0 + 0.018 991 505 605 656 576;
  • 49) 0.018 991 505 605 656 576 × 2 = 0 + 0.037 983 011 211 313 152;
  • 50) 0.037 983 011 211 313 152 × 2 = 0 + 0.075 966 022 422 626 304;
  • 51) 0.075 966 022 422 626 304 × 2 = 0 + 0.151 932 044 845 252 608;
  • 52) 0.151 932 044 845 252 608 × 2 = 0 + 0.303 864 089 690 505 216;
  • 53) 0.303 864 089 690 505 216 × 2 = 0 + 0.607 728 179 381 010 432;

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:


0.301 029 995 663 981 321(10) =

0.0100 1101 0001 0000 0100 1101 0100 0010 0111 1101 1110 1000 0000 0(2)

5. Positive number before normalization:

1.301 029 995 663 981 321(10) =

1.0100 1101 0001 0000 0100 1101 0100 0010 0111 1101 1110 1000 0000 0(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:


1.301 029 995 663 981 321(10) =

1.0100 1101 0001 0000 0100 1101 0100 0010 0111 1101 1110 1000 0000 0(2) =

1.0100 1101 0001 0000 0100 1101 0100 0010 0111 1101 1110 1000 0000 0(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 1000 0000 0


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 1000 0000 0 =

0100 1101 0001 0000 0100 1101 0100 0010 0111 1101 1110 1000 0000


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 1000 0000


Decimal number 1.301 029 995 663 981 321 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 1000 0000

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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