38.811 999 999 999 997 612 576 407 846 063 372 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 38.811 999 999 999 997 612 576 407 846 063 372(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
38.811 999 999 999 997 612 576 407 846 063 372(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: 38.
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;
  • 38 ÷ 2 = 19 + 0;
  • 19 ÷ 2 = 9 + 1;
  • 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.

38(10) =

10 0110(2)


3. Convert to binary (base 2) the fractional part: 0.811 999 999 999 997 612 576 407 846 063 372.

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) 0.811 999 999 999 997 612 576 407 846 063 372 × 2 = 1 + 0.623 999 999 999 995 225 152 815 692 126 744;
  • 2) 0.623 999 999 999 995 225 152 815 692 126 744 × 2 = 1 + 0.247 999 999 999 990 450 305 631 384 253 488;
  • 3) 0.247 999 999 999 990 450 305 631 384 253 488 × 2 = 0 + 0.495 999 999 999 980 900 611 262 768 506 976;
  • 4) 0.495 999 999 999 980 900 611 262 768 506 976 × 2 = 0 + 0.991 999 999 999 961 801 222 525 537 013 952;
  • 5) 0.991 999 999 999 961 801 222 525 537 013 952 × 2 = 1 + 0.983 999 999 999 923 602 445 051 074 027 904;
  • 6) 0.983 999 999 999 923 602 445 051 074 027 904 × 2 = 1 + 0.967 999 999 999 847 204 890 102 148 055 808;
  • 7) 0.967 999 999 999 847 204 890 102 148 055 808 × 2 = 1 + 0.935 999 999 999 694 409 780 204 296 111 616;
  • 8) 0.935 999 999 999 694 409 780 204 296 111 616 × 2 = 1 + 0.871 999 999 999 388 819 560 408 592 223 232;
  • 9) 0.871 999 999 999 388 819 560 408 592 223 232 × 2 = 1 + 0.743 999 999 998 777 639 120 817 184 446 464;
  • 10) 0.743 999 999 998 777 639 120 817 184 446 464 × 2 = 1 + 0.487 999 999 997 555 278 241 634 368 892 928;
  • 11) 0.487 999 999 997 555 278 241 634 368 892 928 × 2 = 0 + 0.975 999 999 995 110 556 483 268 737 785 856;
  • 12) 0.975 999 999 995 110 556 483 268 737 785 856 × 2 = 1 + 0.951 999 999 990 221 112 966 537 475 571 712;
  • 13) 0.951 999 999 990 221 112 966 537 475 571 712 × 2 = 1 + 0.903 999 999 980 442 225 933 074 951 143 424;
  • 14) 0.903 999 999 980 442 225 933 074 951 143 424 × 2 = 1 + 0.807 999 999 960 884 451 866 149 902 286 848;
  • 15) 0.807 999 999 960 884 451 866 149 902 286 848 × 2 = 1 + 0.615 999 999 921 768 903 732 299 804 573 696;
  • 16) 0.615 999 999 921 768 903 732 299 804 573 696 × 2 = 1 + 0.231 999 999 843 537 807 464 599 609 147 392;
  • 17) 0.231 999 999 843 537 807 464 599 609 147 392 × 2 = 0 + 0.463 999 999 687 075 614 929 199 218 294 784;
  • 18) 0.463 999 999 687 075 614 929 199 218 294 784 × 2 = 0 + 0.927 999 999 374 151 229 858 398 436 589 568;
  • 19) 0.927 999 999 374 151 229 858 398 436 589 568 × 2 = 1 + 0.855 999 998 748 302 459 716 796 873 179 136;
  • 20) 0.855 999 998 748 302 459 716 796 873 179 136 × 2 = 1 + 0.711 999 997 496 604 919 433 593 746 358 272;
  • 21) 0.711 999 997 496 604 919 433 593 746 358 272 × 2 = 1 + 0.423 999 994 993 209 838 867 187 492 716 544;
  • 22) 0.423 999 994 993 209 838 867 187 492 716 544 × 2 = 0 + 0.847 999 989 986 419 677 734 374 985 433 088;
  • 23) 0.847 999 989 986 419 677 734 374 985 433 088 × 2 = 1 + 0.695 999 979 972 839 355 468 749 970 866 176;
  • 24) 0.695 999 979 972 839 355 468 749 970 866 176 × 2 = 1 + 0.391 999 959 945 678 710 937 499 941 732 352;
  • 25) 0.391 999 959 945 678 710 937 499 941 732 352 × 2 = 0 + 0.783 999 919 891 357 421 874 999 883 464 704;
  • 26) 0.783 999 919 891 357 421 874 999 883 464 704 × 2 = 1 + 0.567 999 839 782 714 843 749 999 766 929 408;
  • 27) 0.567 999 839 782 714 843 749 999 766 929 408 × 2 = 1 + 0.135 999 679 565 429 687 499 999 533 858 816;
  • 28) 0.135 999 679 565 429 687 499 999 533 858 816 × 2 = 0 + 0.271 999 359 130 859 374 999 999 067 717 632;
  • 29) 0.271 999 359 130 859 374 999 999 067 717 632 × 2 = 0 + 0.543 998 718 261 718 749 999 998 135 435 264;
  • 30) 0.543 998 718 261 718 749 999 998 135 435 264 × 2 = 1 + 0.087 997 436 523 437 499 999 996 270 870 528;
  • 31) 0.087 997 436 523 437 499 999 996 270 870 528 × 2 = 0 + 0.175 994 873 046 874 999 999 992 541 741 056;
  • 32) 0.175 994 873 046 874 999 999 992 541 741 056 × 2 = 0 + 0.351 989 746 093 749 999 999 985 083 482 112;
  • 33) 0.351 989 746 093 749 999 999 985 083 482 112 × 2 = 0 + 0.703 979 492 187 499 999 999 970 166 964 224;
  • 34) 0.703 979 492 187 499 999 999 970 166 964 224 × 2 = 1 + 0.407 958 984 374 999 999 999 940 333 928 448;
  • 35) 0.407 958 984 374 999 999 999 940 333 928 448 × 2 = 0 + 0.815 917 968 749 999 999 999 880 667 856 896;
  • 36) 0.815 917 968 749 999 999 999 880 667 856 896 × 2 = 1 + 0.631 835 937 499 999 999 999 761 335 713 792;
  • 37) 0.631 835 937 499 999 999 999 761 335 713 792 × 2 = 1 + 0.263 671 874 999 999 999 999 522 671 427 584;
  • 38) 0.263 671 874 999 999 999 999 522 671 427 584 × 2 = 0 + 0.527 343 749 999 999 999 999 045 342 855 168;
  • 39) 0.527 343 749 999 999 999 999 045 342 855 168 × 2 = 1 + 0.054 687 499 999 999 999 998 090 685 710 336;
  • 40) 0.054 687 499 999 999 999 998 090 685 710 336 × 2 = 0 + 0.109 374 999 999 999 999 996 181 371 420 672;
  • 41) 0.109 374 999 999 999 999 996 181 371 420 672 × 2 = 0 + 0.218 749 999 999 999 999 992 362 742 841 344;
  • 42) 0.218 749 999 999 999 999 992 362 742 841 344 × 2 = 0 + 0.437 499 999 999 999 999 984 725 485 682 688;
  • 43) 0.437 499 999 999 999 999 984 725 485 682 688 × 2 = 0 + 0.874 999 999 999 999 999 969 450 971 365 376;
  • 44) 0.874 999 999 999 999 999 969 450 971 365 376 × 2 = 1 + 0.749 999 999 999 999 999 938 901 942 730 752;
  • 45) 0.749 999 999 999 999 999 938 901 942 730 752 × 2 = 1 + 0.499 999 999 999 999 999 877 803 885 461 504;
  • 46) 0.499 999 999 999 999 999 877 803 885 461 504 × 2 = 0 + 0.999 999 999 999 999 999 755 607 770 923 008;
  • 47) 0.999 999 999 999 999 999 755 607 770 923 008 × 2 = 1 + 0.999 999 999 999 999 999 511 215 541 846 016;
  • 48) 0.999 999 999 999 999 999 511 215 541 846 016 × 2 = 1 + 0.999 999 999 999 999 999 022 431 083 692 032;
  • 49) 0.999 999 999 999 999 999 022 431 083 692 032 × 2 = 1 + 0.999 999 999 999 999 998 044 862 167 384 064;
  • 50) 0.999 999 999 999 999 998 044 862 167 384 064 × 2 = 1 + 0.999 999 999 999 999 996 089 724 334 768 128;
  • 51) 0.999 999 999 999 999 996 089 724 334 768 128 × 2 = 1 + 0.999 999 999 999 999 992 179 448 669 536 256;
  • 52) 0.999 999 999 999 999 992 179 448 669 536 256 × 2 = 1 + 0.999 999 999 999 999 984 358 897 339 072 512;
  • 53) 0.999 999 999 999 999 984 358 897 339 072 512 × 2 = 1 + 0.999 999 999 999 999 968 717 794 678 145 024;

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:


0.811 999 999 999 997 612 576 407 846 063 372(10) =

0.1100 1111 1101 1111 0011 1011 0110 0100 0101 1010 0001 1011 1111 1(2)

5. Positive number before normalization:

38.811 999 999 999 997 612 576 407 846 063 372(10) =

10 0110.1100 1111 1101 1111 0011 1011 0110 0100 0101 1010 0001 1011 1111 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 5 positions to the left, so that only one non zero digit remains to the left of it:


38.811 999 999 999 997 612 576 407 846 063 372(10) =

10 0110.1100 1111 1101 1111 0011 1011 0110 0100 0101 1010 0001 1011 1111 1(2) =

10 0110.1100 1111 1101 1111 0011 1011 0110 0100 0101 1010 0001 1011 1111 1(2) × 20 =

1.0011 0110 0111 1110 1111 1001 1101 1011 0010 0010 1101 0000 1101 1111 11(2) × 25


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): 5

Mantissa (not normalized):
1.0011 0110 0111 1110 1111 1001 1101 1011 0010 0010 1101 0000 1101 1111 11


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

5 + 2(11-1) - 1 =

(5 + 1 023)(10) =

1 028(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;
  • 1 028 ÷ 2 = 514 + 0;
  • 514 ÷ 2 = 257 + 0;
  • 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) =

1028(10) =

100 0000 0100(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 0110 0111 1110 1111 1001 1101 1011 0010 0010 1101 0000 1101 11 1111 =

0011 0110 0111 1110 1111 1001 1101 1011 0010 0010 1101 0000 1101


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 0100

Mantissa (52 bits) =
0011 0110 0111 1110 1111 1001 1101 1011 0010 0010 1101 0000 1101


Decimal number 38.811 999 999 999 997 612 576 407 846 063 372 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0100 - 0011 0110 0111 1110 1111 1001 1101 1011 0010 0010 1101 0000 1101

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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