293.309 999 999 999 945 430 317 893 624 305 725 05 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 293.309 999 999 999 945 430 317 893 624 305 725 05(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
293.309 999 999 999 945 430 317 893 624 305 725 05(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: 293.
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;
  • 293 ÷ 2 = 146 + 1;
  • 146 ÷ 2 = 73 + 0;
  • 73 ÷ 2 = 36 + 1;
  • 36 ÷ 2 = 18 + 0;
  • 18 ÷ 2 = 9 + 0;
  • 9 ÷ 2 = 4 + 1;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 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.

293(10) =

1 0010 0101(2)


3. Convert to binary (base 2) the fractional part: 0.309 999 999 999 945 430 317 893 624 305 725 05.

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) 0.309 999 999 999 945 430 317 893 624 305 725 05 × 2 = 0 + 0.619 999 999 999 890 860 635 787 248 611 450 1;
  • 2) 0.619 999 999 999 890 860 635 787 248 611 450 1 × 2 = 1 + 0.239 999 999 999 781 721 271 574 497 222 900 2;
  • 3) 0.239 999 999 999 781 721 271 574 497 222 900 2 × 2 = 0 + 0.479 999 999 999 563 442 543 148 994 445 800 4;
  • 4) 0.479 999 999 999 563 442 543 148 994 445 800 4 × 2 = 0 + 0.959 999 999 999 126 885 086 297 988 891 600 8;
  • 5) 0.959 999 999 999 126 885 086 297 988 891 600 8 × 2 = 1 + 0.919 999 999 998 253 770 172 595 977 783 201 6;
  • 6) 0.919 999 999 998 253 770 172 595 977 783 201 6 × 2 = 1 + 0.839 999 999 996 507 540 345 191 955 566 403 2;
  • 7) 0.839 999 999 996 507 540 345 191 955 566 403 2 × 2 = 1 + 0.679 999 999 993 015 080 690 383 911 132 806 4;
  • 8) 0.679 999 999 993 015 080 690 383 911 132 806 4 × 2 = 1 + 0.359 999 999 986 030 161 380 767 822 265 612 8;
  • 9) 0.359 999 999 986 030 161 380 767 822 265 612 8 × 2 = 0 + 0.719 999 999 972 060 322 761 535 644 531 225 6;
  • 10) 0.719 999 999 972 060 322 761 535 644 531 225 6 × 2 = 1 + 0.439 999 999 944 120 645 523 071 289 062 451 2;
  • 11) 0.439 999 999 944 120 645 523 071 289 062 451 2 × 2 = 0 + 0.879 999 999 888 241 291 046 142 578 124 902 4;
  • 12) 0.879 999 999 888 241 291 046 142 578 124 902 4 × 2 = 1 + 0.759 999 999 776 482 582 092 285 156 249 804 8;
  • 13) 0.759 999 999 776 482 582 092 285 156 249 804 8 × 2 = 1 + 0.519 999 999 552 965 164 184 570 312 499 609 6;
  • 14) 0.519 999 999 552 965 164 184 570 312 499 609 6 × 2 = 1 + 0.039 999 999 105 930 328 369 140 624 999 219 2;
  • 15) 0.039 999 999 105 930 328 369 140 624 999 219 2 × 2 = 0 + 0.079 999 998 211 860 656 738 281 249 998 438 4;
  • 16) 0.079 999 998 211 860 656 738 281 249 998 438 4 × 2 = 0 + 0.159 999 996 423 721 313 476 562 499 996 876 8;
  • 17) 0.159 999 996 423 721 313 476 562 499 996 876 8 × 2 = 0 + 0.319 999 992 847 442 626 953 124 999 993 753 6;
  • 18) 0.319 999 992 847 442 626 953 124 999 993 753 6 × 2 = 0 + 0.639 999 985 694 885 253 906 249 999 987 507 2;
  • 19) 0.639 999 985 694 885 253 906 249 999 987 507 2 × 2 = 1 + 0.279 999 971 389 770 507 812 499 999 975 014 4;
  • 20) 0.279 999 971 389 770 507 812 499 999 975 014 4 × 2 = 0 + 0.559 999 942 779 541 015 624 999 999 950 028 8;
  • 21) 0.559 999 942 779 541 015 624 999 999 950 028 8 × 2 = 1 + 0.119 999 885 559 082 031 249 999 999 900 057 6;
  • 22) 0.119 999 885 559 082 031 249 999 999 900 057 6 × 2 = 0 + 0.239 999 771 118 164 062 499 999 999 800 115 2;
  • 23) 0.239 999 771 118 164 062 499 999 999 800 115 2 × 2 = 0 + 0.479 999 542 236 328 124 999 999 999 600 230 4;
  • 24) 0.479 999 542 236 328 124 999 999 999 600 230 4 × 2 = 0 + 0.959 999 084 472 656 249 999 999 999 200 460 8;
  • 25) 0.959 999 084 472 656 249 999 999 999 200 460 8 × 2 = 1 + 0.919 998 168 945 312 499 999 999 998 400 921 6;
  • 26) 0.919 998 168 945 312 499 999 999 998 400 921 6 × 2 = 1 + 0.839 996 337 890 624 999 999 999 996 801 843 2;
  • 27) 0.839 996 337 890 624 999 999 999 996 801 843 2 × 2 = 1 + 0.679 992 675 781 249 999 999 999 993 603 686 4;
  • 28) 0.679 992 675 781 249 999 999 999 993 603 686 4 × 2 = 1 + 0.359 985 351 562 499 999 999 999 987 207 372 8;
  • 29) 0.359 985 351 562 499 999 999 999 987 207 372 8 × 2 = 0 + 0.719 970 703 124 999 999 999 999 974 414 745 6;
  • 30) 0.719 970 703 124 999 999 999 999 974 414 745 6 × 2 = 1 + 0.439 941 406 249 999 999 999 999 948 829 491 2;
  • 31) 0.439 941 406 249 999 999 999 999 948 829 491 2 × 2 = 0 + 0.879 882 812 499 999 999 999 999 897 658 982 4;
  • 32) 0.879 882 812 499 999 999 999 999 897 658 982 4 × 2 = 1 + 0.759 765 624 999 999 999 999 999 795 317 964 8;
  • 33) 0.759 765 624 999 999 999 999 999 795 317 964 8 × 2 = 1 + 0.519 531 249 999 999 999 999 999 590 635 929 6;
  • 34) 0.519 531 249 999 999 999 999 999 590 635 929 6 × 2 = 1 + 0.039 062 499 999 999 999 999 999 181 271 859 2;
  • 35) 0.039 062 499 999 999 999 999 999 181 271 859 2 × 2 = 0 + 0.078 124 999 999 999 999 999 998 362 543 718 4;
  • 36) 0.078 124 999 999 999 999 999 998 362 543 718 4 × 2 = 0 + 0.156 249 999 999 999 999 999 996 725 087 436 8;
  • 37) 0.156 249 999 999 999 999 999 996 725 087 436 8 × 2 = 0 + 0.312 499 999 999 999 999 999 993 450 174 873 6;
  • 38) 0.312 499 999 999 999 999 999 993 450 174 873 6 × 2 = 0 + 0.624 999 999 999 999 999 999 986 900 349 747 2;
  • 39) 0.624 999 999 999 999 999 999 986 900 349 747 2 × 2 = 1 + 0.249 999 999 999 999 999 999 973 800 699 494 4;
  • 40) 0.249 999 999 999 999 999 999 973 800 699 494 4 × 2 = 0 + 0.499 999 999 999 999 999 999 947 601 398 988 8;
  • 41) 0.499 999 999 999 999 999 999 947 601 398 988 8 × 2 = 0 + 0.999 999 999 999 999 999 999 895 202 797 977 6;
  • 42) 0.999 999 999 999 999 999 999 895 202 797 977 6 × 2 = 1 + 0.999 999 999 999 999 999 999 790 405 595 955 2;
  • 43) 0.999 999 999 999 999 999 999 790 405 595 955 2 × 2 = 1 + 0.999 999 999 999 999 999 999 580 811 191 910 4;
  • 44) 0.999 999 999 999 999 999 999 580 811 191 910 4 × 2 = 1 + 0.999 999 999 999 999 999 999 161 622 383 820 8;
  • 45) 0.999 999 999 999 999 999 999 161 622 383 820 8 × 2 = 1 + 0.999 999 999 999 999 999 998 323 244 767 641 6;
  • 46) 0.999 999 999 999 999 999 998 323 244 767 641 6 × 2 = 1 + 0.999 999 999 999 999 999 996 646 489 535 283 2;
  • 47) 0.999 999 999 999 999 999 996 646 489 535 283 2 × 2 = 1 + 0.999 999 999 999 999 999 993 292 979 070 566 4;
  • 48) 0.999 999 999 999 999 999 993 292 979 070 566 4 × 2 = 1 + 0.999 999 999 999 999 999 986 585 958 141 132 8;
  • 49) 0.999 999 999 999 999 999 986 585 958 141 132 8 × 2 = 1 + 0.999 999 999 999 999 999 973 171 916 282 265 6;
  • 50) 0.999 999 999 999 999 999 973 171 916 282 265 6 × 2 = 1 + 0.999 999 999 999 999 999 946 343 832 564 531 2;
  • 51) 0.999 999 999 999 999 999 946 343 832 564 531 2 × 2 = 1 + 0.999 999 999 999 999 999 892 687 665 129 062 4;
  • 52) 0.999 999 999 999 999 999 892 687 665 129 062 4 × 2 = 1 + 0.999 999 999 999 999 999 785 375 330 258 124 8;
  • 53) 0.999 999 999 999 999 999 785 375 330 258 124 8 × 2 = 1 + 0.999 999 999 999 999 999 570 750 660 516 249 6;

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:


0.309 999 999 999 945 430 317 893 624 305 725 05(10) =

0.0100 1111 0101 1100 0010 1000 1111 0101 1100 0010 0111 1111 1111 1(2)

5. Positive number before normalization:

293.309 999 999 999 945 430 317 893 624 305 725 05(10) =

1 0010 0101.0100 1111 0101 1100 0010 1000 1111 0101 1100 0010 0111 1111 1111 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 8 positions to the left, so that only one non zero digit remains to the left of it:


293.309 999 999 999 945 430 317 893 624 305 725 05(10) =

1 0010 0101.0100 1111 0101 1100 0010 1000 1111 0101 1100 0010 0111 1111 1111 1(2) =

1 0010 0101.0100 1111 0101 1100 0010 1000 1111 0101 1100 0010 0111 1111 1111 1(2) × 20 =

1.0010 0101 0100 1111 0101 1100 0010 1000 1111 0101 1100 0010 0111 1111 1111 1(2) × 28


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): 8

Mantissa (not normalized):
1.0010 0101 0100 1111 0101 1100 0010 1000 1111 0101 1100 0010 0111 1111 1111 1


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

8 + 2(11-1) - 1 =

(8 + 1 023)(10) =

1 031(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;
  • 1 031 ÷ 2 = 515 + 1;
  • 515 ÷ 2 = 257 + 1;
  • 257 ÷ 2 = 128 + 1;
  • 128 ÷ 2 = 64 + 0;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 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) =

1031(10) =

100 0000 0111(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. 0010 0101 0100 1111 0101 1100 0010 1000 1111 0101 1100 0010 0111 1 1111 1111 =

0010 0101 0100 1111 0101 1100 0010 1000 1111 0101 1100 0010 0111


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) =
100 0000 0111

Mantissa (52 bits) =
0010 0101 0100 1111 0101 1100 0010 1000 1111 0101 1100 0010 0111


Decimal number 293.309 999 999 999 945 430 317 893 624 305 725 05 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0111 - 0010 0101 0100 1111 0101 1100 0010 1000 1111 0101 1100 0010 0111

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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