2.444 089 209 850 062 616 169 452 666 9 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 2.444 089 209 850 062 616 169 452 666 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
2.444 089 209 850 062 616 169 452 666 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: 2.
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;
  • 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.

2(10) =

10(2)


3. Convert to binary (base 2) the fractional part: 0.444 089 209 850 062 616 169 452 666 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) 0.444 089 209 850 062 616 169 452 666 9 × 2 = 0 + 0.888 178 419 700 125 232 338 905 333 8;
  • 2) 0.888 178 419 700 125 232 338 905 333 8 × 2 = 1 + 0.776 356 839 400 250 464 677 810 667 6;
  • 3) 0.776 356 839 400 250 464 677 810 667 6 × 2 = 1 + 0.552 713 678 800 500 929 355 621 335 2;
  • 4) 0.552 713 678 800 500 929 355 621 335 2 × 2 = 1 + 0.105 427 357 601 001 858 711 242 670 4;
  • 5) 0.105 427 357 601 001 858 711 242 670 4 × 2 = 0 + 0.210 854 715 202 003 717 422 485 340 8;
  • 6) 0.210 854 715 202 003 717 422 485 340 8 × 2 = 0 + 0.421 709 430 404 007 434 844 970 681 6;
  • 7) 0.421 709 430 404 007 434 844 970 681 6 × 2 = 0 + 0.843 418 860 808 014 869 689 941 363 2;
  • 8) 0.843 418 860 808 014 869 689 941 363 2 × 2 = 1 + 0.686 837 721 616 029 739 379 882 726 4;
  • 9) 0.686 837 721 616 029 739 379 882 726 4 × 2 = 1 + 0.373 675 443 232 059 478 759 765 452 8;
  • 10) 0.373 675 443 232 059 478 759 765 452 8 × 2 = 0 + 0.747 350 886 464 118 957 519 530 905 6;
  • 11) 0.747 350 886 464 118 957 519 530 905 6 × 2 = 1 + 0.494 701 772 928 237 915 039 061 811 2;
  • 12) 0.494 701 772 928 237 915 039 061 811 2 × 2 = 0 + 0.989 403 545 856 475 830 078 123 622 4;
  • 13) 0.989 403 545 856 475 830 078 123 622 4 × 2 = 1 + 0.978 807 091 712 951 660 156 247 244 8;
  • 14) 0.978 807 091 712 951 660 156 247 244 8 × 2 = 1 + 0.957 614 183 425 903 320 312 494 489 6;
  • 15) 0.957 614 183 425 903 320 312 494 489 6 × 2 = 1 + 0.915 228 366 851 806 640 624 988 979 2;
  • 16) 0.915 228 366 851 806 640 624 988 979 2 × 2 = 1 + 0.830 456 733 703 613 281 249 977 958 4;
  • 17) 0.830 456 733 703 613 281 249 977 958 4 × 2 = 1 + 0.660 913 467 407 226 562 499 955 916 8;
  • 18) 0.660 913 467 407 226 562 499 955 916 8 × 2 = 1 + 0.321 826 934 814 453 124 999 911 833 6;
  • 19) 0.321 826 934 814 453 124 999 911 833 6 × 2 = 0 + 0.643 653 869 628 906 249 999 823 667 2;
  • 20) 0.643 653 869 628 906 249 999 823 667 2 × 2 = 1 + 0.287 307 739 257 812 499 999 647 334 4;
  • 21) 0.287 307 739 257 812 499 999 647 334 4 × 2 = 0 + 0.574 615 478 515 624 999 999 294 668 8;
  • 22) 0.574 615 478 515 624 999 999 294 668 8 × 2 = 1 + 0.149 230 957 031 249 999 998 589 337 6;
  • 23) 0.149 230 957 031 249 999 998 589 337 6 × 2 = 0 + 0.298 461 914 062 499 999 997 178 675 2;
  • 24) 0.298 461 914 062 499 999 997 178 675 2 × 2 = 0 + 0.596 923 828 124 999 999 994 357 350 4;
  • 25) 0.596 923 828 124 999 999 994 357 350 4 × 2 = 1 + 0.193 847 656 249 999 999 988 714 700 8;
  • 26) 0.193 847 656 249 999 999 988 714 700 8 × 2 = 0 + 0.387 695 312 499 999 999 977 429 401 6;
  • 27) 0.387 695 312 499 999 999 977 429 401 6 × 2 = 0 + 0.775 390 624 999 999 999 954 858 803 2;
  • 28) 0.775 390 624 999 999 999 954 858 803 2 × 2 = 1 + 0.550 781 249 999 999 999 909 717 606 4;
  • 29) 0.550 781 249 999 999 999 909 717 606 4 × 2 = 1 + 0.101 562 499 999 999 999 819 435 212 8;
  • 30) 0.101 562 499 999 999 999 819 435 212 8 × 2 = 0 + 0.203 124 999 999 999 999 638 870 425 6;
  • 31) 0.203 124 999 999 999 999 638 870 425 6 × 2 = 0 + 0.406 249 999 999 999 999 277 740 851 2;
  • 32) 0.406 249 999 999 999 999 277 740 851 2 × 2 = 0 + 0.812 499 999 999 999 998 555 481 702 4;
  • 33) 0.812 499 999 999 999 998 555 481 702 4 × 2 = 1 + 0.624 999 999 999 999 997 110 963 404 8;
  • 34) 0.624 999 999 999 999 997 110 963 404 8 × 2 = 1 + 0.249 999 999 999 999 994 221 926 809 6;
  • 35) 0.249 999 999 999 999 994 221 926 809 6 × 2 = 0 + 0.499 999 999 999 999 988 443 853 619 2;
  • 36) 0.499 999 999 999 999 988 443 853 619 2 × 2 = 0 + 0.999 999 999 999 999 976 887 707 238 4;
  • 37) 0.999 999 999 999 999 976 887 707 238 4 × 2 = 1 + 0.999 999 999 999 999 953 775 414 476 8;
  • 38) 0.999 999 999 999 999 953 775 414 476 8 × 2 = 1 + 0.999 999 999 999 999 907 550 828 953 6;
  • 39) 0.999 999 999 999 999 907 550 828 953 6 × 2 = 1 + 0.999 999 999 999 999 815 101 657 907 2;
  • 40) 0.999 999 999 999 999 815 101 657 907 2 × 2 = 1 + 0.999 999 999 999 999 630 203 315 814 4;
  • 41) 0.999 999 999 999 999 630 203 315 814 4 × 2 = 1 + 0.999 999 999 999 999 260 406 631 628 8;
  • 42) 0.999 999 999 999 999 260 406 631 628 8 × 2 = 1 + 0.999 999 999 999 998 520 813 263 257 6;
  • 43) 0.999 999 999 999 998 520 813 263 257 6 × 2 = 1 + 0.999 999 999 999 997 041 626 526 515 2;
  • 44) 0.999 999 999 999 997 041 626 526 515 2 × 2 = 1 + 0.999 999 999 999 994 083 253 053 030 4;
  • 45) 0.999 999 999 999 994 083 253 053 030 4 × 2 = 1 + 0.999 999 999 999 988 166 506 106 060 8;
  • 46) 0.999 999 999 999 988 166 506 106 060 8 × 2 = 1 + 0.999 999 999 999 976 333 012 212 121 6;
  • 47) 0.999 999 999 999 976 333 012 212 121 6 × 2 = 1 + 0.999 999 999 999 952 666 024 424 243 2;
  • 48) 0.999 999 999 999 952 666 024 424 243 2 × 2 = 1 + 0.999 999 999 999 905 332 048 848 486 4;
  • 49) 0.999 999 999 999 905 332 048 848 486 4 × 2 = 1 + 0.999 999 999 999 810 664 097 696 972 8;
  • 50) 0.999 999 999 999 810 664 097 696 972 8 × 2 = 1 + 0.999 999 999 999 621 328 195 393 945 6;
  • 51) 0.999 999 999 999 621 328 195 393 945 6 × 2 = 1 + 0.999 999 999 999 242 656 390 787 891 2;
  • 52) 0.999 999 999 999 242 656 390 787 891 2 × 2 = 1 + 0.999 999 999 998 485 312 781 575 782 4;
  • 53) 0.999 999 999 998 485 312 781 575 782 4 × 2 = 1 + 0.999 999 999 996 970 625 563 151 564 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:


0.444 089 209 850 062 616 169 452 666 9(10) =

0.0111 0001 1010 1111 1101 0100 1001 1000 1100 1111 1111 1111 1111 1(2)

5. Positive number before normalization:

2.444 089 209 850 062 616 169 452 666 9(10) =

10.0111 0001 1010 1111 1101 0100 1001 1000 1100 1111 1111 1111 1111 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 1 positions to the left, so that only one non zero digit remains to the left of it:


2.444 089 209 850 062 616 169 452 666 9(10) =

10.0111 0001 1010 1111 1101 0100 1001 1000 1100 1111 1111 1111 1111 1(2) =

10.0111 0001 1010 1111 1101 0100 1001 1000 1100 1111 1111 1111 1111 1(2) × 20 =

1.0011 1000 1101 0111 1110 1010 0100 1100 0110 0111 1111 1111 1111 11(2) × 21


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): 1

Mantissa (not normalized):
1.0011 1000 1101 0111 1110 1010 0100 1100 0110 0111 1111 1111 1111 11


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

1 + 2(11-1) - 1 =

(1 + 1 023)(10) =

1 024(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;
  • 1 024 ÷ 2 = 512 + 0;
  • 512 ÷ 2 = 256 + 0;
  • 256 ÷ 2 = 128 + 0;
  • 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) =

1024(10) =

100 0000 0000(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. 0011 1000 1101 0111 1110 1010 0100 1100 0110 0111 1111 1111 1111 11 =

0011 1000 1101 0111 1110 1010 0100 1100 0110 0111 1111 1111 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) =
100 0000 0000

Mantissa (52 bits) =
0011 1000 1101 0111 1110 1010 0100 1100 0110 0111 1111 1111 1111


Decimal number 2.444 089 209 850 062 616 169 452 666 9 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0000 - 0011 1000 1101 0111 1110 1010 0100 1100 0110 0111 1111 1111 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