2.444 089 209 850 062 616 169 452 667 06 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 2.444 089 209 850 062 616 169 452 667 06(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
2.444 089 209 850 062 616 169 452 667 06(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: 0.444 089 209 850 062 616 169 452 667 06.

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) 0.444 089 209 850 062 616 169 452 667 06 × 2 = 0 + 0.888 178 419 700 125 232 338 905 334 12;
  • 2) 0.888 178 419 700 125 232 338 905 334 12 × 2 = 1 + 0.776 356 839 400 250 464 677 810 668 24;
  • 3) 0.776 356 839 400 250 464 677 810 668 24 × 2 = 1 + 0.552 713 678 800 500 929 355 621 336 48;
  • 4) 0.552 713 678 800 500 929 355 621 336 48 × 2 = 1 + 0.105 427 357 601 001 858 711 242 672 96;
  • 5) 0.105 427 357 601 001 858 711 242 672 96 × 2 = 0 + 0.210 854 715 202 003 717 422 485 345 92;
  • 6) 0.210 854 715 202 003 717 422 485 345 92 × 2 = 0 + 0.421 709 430 404 007 434 844 970 691 84;
  • 7) 0.421 709 430 404 007 434 844 970 691 84 × 2 = 0 + 0.843 418 860 808 014 869 689 941 383 68;
  • 8) 0.843 418 860 808 014 869 689 941 383 68 × 2 = 1 + 0.686 837 721 616 029 739 379 882 767 36;
  • 9) 0.686 837 721 616 029 739 379 882 767 36 × 2 = 1 + 0.373 675 443 232 059 478 759 765 534 72;
  • 10) 0.373 675 443 232 059 478 759 765 534 72 × 2 = 0 + 0.747 350 886 464 118 957 519 531 069 44;
  • 11) 0.747 350 886 464 118 957 519 531 069 44 × 2 = 1 + 0.494 701 772 928 237 915 039 062 138 88;
  • 12) 0.494 701 772 928 237 915 039 062 138 88 × 2 = 0 + 0.989 403 545 856 475 830 078 124 277 76;
  • 13) 0.989 403 545 856 475 830 078 124 277 76 × 2 = 1 + 0.978 807 091 712 951 660 156 248 555 52;
  • 14) 0.978 807 091 712 951 660 156 248 555 52 × 2 = 1 + 0.957 614 183 425 903 320 312 497 111 04;
  • 15) 0.957 614 183 425 903 320 312 497 111 04 × 2 = 1 + 0.915 228 366 851 806 640 624 994 222 08;
  • 16) 0.915 228 366 851 806 640 624 994 222 08 × 2 = 1 + 0.830 456 733 703 613 281 249 988 444 16;
  • 17) 0.830 456 733 703 613 281 249 988 444 16 × 2 = 1 + 0.660 913 467 407 226 562 499 976 888 32;
  • 18) 0.660 913 467 407 226 562 499 976 888 32 × 2 = 1 + 0.321 826 934 814 453 124 999 953 776 64;
  • 19) 0.321 826 934 814 453 124 999 953 776 64 × 2 = 0 + 0.643 653 869 628 906 249 999 907 553 28;
  • 20) 0.643 653 869 628 906 249 999 907 553 28 × 2 = 1 + 0.287 307 739 257 812 499 999 815 106 56;
  • 21) 0.287 307 739 257 812 499 999 815 106 56 × 2 = 0 + 0.574 615 478 515 624 999 999 630 213 12;
  • 22) 0.574 615 478 515 624 999 999 630 213 12 × 2 = 1 + 0.149 230 957 031 249 999 999 260 426 24;
  • 23) 0.149 230 957 031 249 999 999 260 426 24 × 2 = 0 + 0.298 461 914 062 499 999 998 520 852 48;
  • 24) 0.298 461 914 062 499 999 998 520 852 48 × 2 = 0 + 0.596 923 828 124 999 999 997 041 704 96;
  • 25) 0.596 923 828 124 999 999 997 041 704 96 × 2 = 1 + 0.193 847 656 249 999 999 994 083 409 92;
  • 26) 0.193 847 656 249 999 999 994 083 409 92 × 2 = 0 + 0.387 695 312 499 999 999 988 166 819 84;
  • 27) 0.387 695 312 499 999 999 988 166 819 84 × 2 = 0 + 0.775 390 624 999 999 999 976 333 639 68;
  • 28) 0.775 390 624 999 999 999 976 333 639 68 × 2 = 1 + 0.550 781 249 999 999 999 952 667 279 36;
  • 29) 0.550 781 249 999 999 999 952 667 279 36 × 2 = 1 + 0.101 562 499 999 999 999 905 334 558 72;
  • 30) 0.101 562 499 999 999 999 905 334 558 72 × 2 = 0 + 0.203 124 999 999 999 999 810 669 117 44;
  • 31) 0.203 124 999 999 999 999 810 669 117 44 × 2 = 0 + 0.406 249 999 999 999 999 621 338 234 88;
  • 32) 0.406 249 999 999 999 999 621 338 234 88 × 2 = 0 + 0.812 499 999 999 999 999 242 676 469 76;
  • 33) 0.812 499 999 999 999 999 242 676 469 76 × 2 = 1 + 0.624 999 999 999 999 998 485 352 939 52;
  • 34) 0.624 999 999 999 999 998 485 352 939 52 × 2 = 1 + 0.249 999 999 999 999 996 970 705 879 04;
  • 35) 0.249 999 999 999 999 996 970 705 879 04 × 2 = 0 + 0.499 999 999 999 999 993 941 411 758 08;
  • 36) 0.499 999 999 999 999 993 941 411 758 08 × 2 = 0 + 0.999 999 999 999 999 987 882 823 516 16;
  • 37) 0.999 999 999 999 999 987 882 823 516 16 × 2 = 1 + 0.999 999 999 999 999 975 765 647 032 32;
  • 38) 0.999 999 999 999 999 975 765 647 032 32 × 2 = 1 + 0.999 999 999 999 999 951 531 294 064 64;
  • 39) 0.999 999 999 999 999 951 531 294 064 64 × 2 = 1 + 0.999 999 999 999 999 903 062 588 129 28;
  • 40) 0.999 999 999 999 999 903 062 588 129 28 × 2 = 1 + 0.999 999 999 999 999 806 125 176 258 56;
  • 41) 0.999 999 999 999 999 806 125 176 258 56 × 2 = 1 + 0.999 999 999 999 999 612 250 352 517 12;
  • 42) 0.999 999 999 999 999 612 250 352 517 12 × 2 = 1 + 0.999 999 999 999 999 224 500 705 034 24;
  • 43) 0.999 999 999 999 999 224 500 705 034 24 × 2 = 1 + 0.999 999 999 999 998 449 001 410 068 48;
  • 44) 0.999 999 999 999 998 449 001 410 068 48 × 2 = 1 + 0.999 999 999 999 996 898 002 820 136 96;
  • 45) 0.999 999 999 999 996 898 002 820 136 96 × 2 = 1 + 0.999 999 999 999 993 796 005 640 273 92;
  • 46) 0.999 999 999 999 993 796 005 640 273 92 × 2 = 1 + 0.999 999 999 999 987 592 011 280 547 84;
  • 47) 0.999 999 999 999 987 592 011 280 547 84 × 2 = 1 + 0.999 999 999 999 975 184 022 561 095 68;
  • 48) 0.999 999 999 999 975 184 022 561 095 68 × 2 = 1 + 0.999 999 999 999 950 368 045 122 191 36;
  • 49) 0.999 999 999 999 950 368 045 122 191 36 × 2 = 1 + 0.999 999 999 999 900 736 090 244 382 72;
  • 50) 0.999 999 999 999 900 736 090 244 382 72 × 2 = 1 + 0.999 999 999 999 801 472 180 488 765 44;
  • 51) 0.999 999 999 999 801 472 180 488 765 44 × 2 = 1 + 0.999 999 999 999 602 944 360 977 530 88;
  • 52) 0.999 999 999 999 602 944 360 977 530 88 × 2 = 1 + 0.999 999 999 999 205 888 721 955 061 76;
  • 53) 0.999 999 999 999 205 888 721 955 061 76 × 2 = 1 + 0.999 999 999 998 411 777 443 910 123 52;

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:


0.444 089 209 850 062 616 169 452 667 06(10) =

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

5. Positive number before normalization:

2.444 089 209 850 062 616 169 452 667 06(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:


2.444 089 209 850 062 616 169 452 667 06(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 2.444 089 209 850 062 616 169 452 667 06 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