2.444 089 209 850 062 616 169 452 666 903 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 2.444 089 209 850 062 616 169 452 666 903(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
2.444 089 209 850 062 616 169 452 666 903(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: 0.444 089 209 850 062 616 169 452 666 903.

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) 0.444 089 209 850 062 616 169 452 666 903 × 2 = 0 + 0.888 178 419 700 125 232 338 905 333 806;
  • 2) 0.888 178 419 700 125 232 338 905 333 806 × 2 = 1 + 0.776 356 839 400 250 464 677 810 667 612;
  • 3) 0.776 356 839 400 250 464 677 810 667 612 × 2 = 1 + 0.552 713 678 800 500 929 355 621 335 224;
  • 4) 0.552 713 678 800 500 929 355 621 335 224 × 2 = 1 + 0.105 427 357 601 001 858 711 242 670 448;
  • 5) 0.105 427 357 601 001 858 711 242 670 448 × 2 = 0 + 0.210 854 715 202 003 717 422 485 340 896;
  • 6) 0.210 854 715 202 003 717 422 485 340 896 × 2 = 0 + 0.421 709 430 404 007 434 844 970 681 792;
  • 7) 0.421 709 430 404 007 434 844 970 681 792 × 2 = 0 + 0.843 418 860 808 014 869 689 941 363 584;
  • 8) 0.843 418 860 808 014 869 689 941 363 584 × 2 = 1 + 0.686 837 721 616 029 739 379 882 727 168;
  • 9) 0.686 837 721 616 029 739 379 882 727 168 × 2 = 1 + 0.373 675 443 232 059 478 759 765 454 336;
  • 10) 0.373 675 443 232 059 478 759 765 454 336 × 2 = 0 + 0.747 350 886 464 118 957 519 530 908 672;
  • 11) 0.747 350 886 464 118 957 519 530 908 672 × 2 = 1 + 0.494 701 772 928 237 915 039 061 817 344;
  • 12) 0.494 701 772 928 237 915 039 061 817 344 × 2 = 0 + 0.989 403 545 856 475 830 078 123 634 688;
  • 13) 0.989 403 545 856 475 830 078 123 634 688 × 2 = 1 + 0.978 807 091 712 951 660 156 247 269 376;
  • 14) 0.978 807 091 712 951 660 156 247 269 376 × 2 = 1 + 0.957 614 183 425 903 320 312 494 538 752;
  • 15) 0.957 614 183 425 903 320 312 494 538 752 × 2 = 1 + 0.915 228 366 851 806 640 624 989 077 504;
  • 16) 0.915 228 366 851 806 640 624 989 077 504 × 2 = 1 + 0.830 456 733 703 613 281 249 978 155 008;
  • 17) 0.830 456 733 703 613 281 249 978 155 008 × 2 = 1 + 0.660 913 467 407 226 562 499 956 310 016;
  • 18) 0.660 913 467 407 226 562 499 956 310 016 × 2 = 1 + 0.321 826 934 814 453 124 999 912 620 032;
  • 19) 0.321 826 934 814 453 124 999 912 620 032 × 2 = 0 + 0.643 653 869 628 906 249 999 825 240 064;
  • 20) 0.643 653 869 628 906 249 999 825 240 064 × 2 = 1 + 0.287 307 739 257 812 499 999 650 480 128;
  • 21) 0.287 307 739 257 812 499 999 650 480 128 × 2 = 0 + 0.574 615 478 515 624 999 999 300 960 256;
  • 22) 0.574 615 478 515 624 999 999 300 960 256 × 2 = 1 + 0.149 230 957 031 249 999 998 601 920 512;
  • 23) 0.149 230 957 031 249 999 998 601 920 512 × 2 = 0 + 0.298 461 914 062 499 999 997 203 841 024;
  • 24) 0.298 461 914 062 499 999 997 203 841 024 × 2 = 0 + 0.596 923 828 124 999 999 994 407 682 048;
  • 25) 0.596 923 828 124 999 999 994 407 682 048 × 2 = 1 + 0.193 847 656 249 999 999 988 815 364 096;
  • 26) 0.193 847 656 249 999 999 988 815 364 096 × 2 = 0 + 0.387 695 312 499 999 999 977 630 728 192;
  • 27) 0.387 695 312 499 999 999 977 630 728 192 × 2 = 0 + 0.775 390 624 999 999 999 955 261 456 384;
  • 28) 0.775 390 624 999 999 999 955 261 456 384 × 2 = 1 + 0.550 781 249 999 999 999 910 522 912 768;
  • 29) 0.550 781 249 999 999 999 910 522 912 768 × 2 = 1 + 0.101 562 499 999 999 999 821 045 825 536;
  • 30) 0.101 562 499 999 999 999 821 045 825 536 × 2 = 0 + 0.203 124 999 999 999 999 642 091 651 072;
  • 31) 0.203 124 999 999 999 999 642 091 651 072 × 2 = 0 + 0.406 249 999 999 999 999 284 183 302 144;
  • 32) 0.406 249 999 999 999 999 284 183 302 144 × 2 = 0 + 0.812 499 999 999 999 998 568 366 604 288;
  • 33) 0.812 499 999 999 999 998 568 366 604 288 × 2 = 1 + 0.624 999 999 999 999 997 136 733 208 576;
  • 34) 0.624 999 999 999 999 997 136 733 208 576 × 2 = 1 + 0.249 999 999 999 999 994 273 466 417 152;
  • 35) 0.249 999 999 999 999 994 273 466 417 152 × 2 = 0 + 0.499 999 999 999 999 988 546 932 834 304;
  • 36) 0.499 999 999 999 999 988 546 932 834 304 × 2 = 0 + 0.999 999 999 999 999 977 093 865 668 608;
  • 37) 0.999 999 999 999 999 977 093 865 668 608 × 2 = 1 + 0.999 999 999 999 999 954 187 731 337 216;
  • 38) 0.999 999 999 999 999 954 187 731 337 216 × 2 = 1 + 0.999 999 999 999 999 908 375 462 674 432;
  • 39) 0.999 999 999 999 999 908 375 462 674 432 × 2 = 1 + 0.999 999 999 999 999 816 750 925 348 864;
  • 40) 0.999 999 999 999 999 816 750 925 348 864 × 2 = 1 + 0.999 999 999 999 999 633 501 850 697 728;
  • 41) 0.999 999 999 999 999 633 501 850 697 728 × 2 = 1 + 0.999 999 999 999 999 267 003 701 395 456;
  • 42) 0.999 999 999 999 999 267 003 701 395 456 × 2 = 1 + 0.999 999 999 999 998 534 007 402 790 912;
  • 43) 0.999 999 999 999 998 534 007 402 790 912 × 2 = 1 + 0.999 999 999 999 997 068 014 805 581 824;
  • 44) 0.999 999 999 999 997 068 014 805 581 824 × 2 = 1 + 0.999 999 999 999 994 136 029 611 163 648;
  • 45) 0.999 999 999 999 994 136 029 611 163 648 × 2 = 1 + 0.999 999 999 999 988 272 059 222 327 296;
  • 46) 0.999 999 999 999 988 272 059 222 327 296 × 2 = 1 + 0.999 999 999 999 976 544 118 444 654 592;
  • 47) 0.999 999 999 999 976 544 118 444 654 592 × 2 = 1 + 0.999 999 999 999 953 088 236 889 309 184;
  • 48) 0.999 999 999 999 953 088 236 889 309 184 × 2 = 1 + 0.999 999 999 999 906 176 473 778 618 368;
  • 49) 0.999 999 999 999 906 176 473 778 618 368 × 2 = 1 + 0.999 999 999 999 812 352 947 557 236 736;
  • 50) 0.999 999 999 999 812 352 947 557 236 736 × 2 = 1 + 0.999 999 999 999 624 705 895 114 473 472;
  • 51) 0.999 999 999 999 624 705 895 114 473 472 × 2 = 1 + 0.999 999 999 999 249 411 790 228 946 944;
  • 52) 0.999 999 999 999 249 411 790 228 946 944 × 2 = 1 + 0.999 999 999 998 498 823 580 457 893 888;
  • 53) 0.999 999 999 998 498 823 580 457 893 888 × 2 = 1 + 0.999 999 999 996 997 647 160 915 787 776;

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:


0.444 089 209 850 062 616 169 452 666 903(10) =

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

5. Positive number before normalization:

2.444 089 209 850 062 616 169 452 666 903(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:


2.444 089 209 850 062 616 169 452 666 903(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 2.444 089 209 850 062 616 169 452 666 903 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