0.234 234 234 234 234 234 328 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.234 234 234 234 234 234 328(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
0.234 234 234 234 234 234 328(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: 0.
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;
  • 0 ÷ 2 = 0 + 0;

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.

0(10) =

0(2)


3. Convert to binary (base 2) the fractional part: 0.234 234 234 234 234 234 328.

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) 0.234 234 234 234 234 234 328 × 2 = 0 + 0.468 468 468 468 468 468 656;
  • 2) 0.468 468 468 468 468 468 656 × 2 = 0 + 0.936 936 936 936 936 937 312;
  • 3) 0.936 936 936 936 936 937 312 × 2 = 1 + 0.873 873 873 873 873 874 624;
  • 4) 0.873 873 873 873 873 874 624 × 2 = 1 + 0.747 747 747 747 747 749 248;
  • 5) 0.747 747 747 747 747 749 248 × 2 = 1 + 0.495 495 495 495 495 498 496;
  • 6) 0.495 495 495 495 495 498 496 × 2 = 0 + 0.990 990 990 990 990 996 992;
  • 7) 0.990 990 990 990 990 996 992 × 2 = 1 + 0.981 981 981 981 981 993 984;
  • 8) 0.981 981 981 981 981 993 984 × 2 = 1 + 0.963 963 963 963 963 987 968;
  • 9) 0.963 963 963 963 963 987 968 × 2 = 1 + 0.927 927 927 927 927 975 936;
  • 10) 0.927 927 927 927 927 975 936 × 2 = 1 + 0.855 855 855 855 855 951 872;
  • 11) 0.855 855 855 855 855 951 872 × 2 = 1 + 0.711 711 711 711 711 903 744;
  • 12) 0.711 711 711 711 711 903 744 × 2 = 1 + 0.423 423 423 423 423 807 488;
  • 13) 0.423 423 423 423 423 807 488 × 2 = 0 + 0.846 846 846 846 847 614 976;
  • 14) 0.846 846 846 846 847 614 976 × 2 = 1 + 0.693 693 693 693 695 229 952;
  • 15) 0.693 693 693 693 695 229 952 × 2 = 1 + 0.387 387 387 387 390 459 904;
  • 16) 0.387 387 387 387 390 459 904 × 2 = 0 + 0.774 774 774 774 780 919 808;
  • 17) 0.774 774 774 774 780 919 808 × 2 = 1 + 0.549 549 549 549 561 839 616;
  • 18) 0.549 549 549 549 561 839 616 × 2 = 1 + 0.099 099 099 099 123 679 232;
  • 19) 0.099 099 099 099 123 679 232 × 2 = 0 + 0.198 198 198 198 247 358 464;
  • 20) 0.198 198 198 198 247 358 464 × 2 = 0 + 0.396 396 396 396 494 716 928;
  • 21) 0.396 396 396 396 494 716 928 × 2 = 0 + 0.792 792 792 792 989 433 856;
  • 22) 0.792 792 792 792 989 433 856 × 2 = 1 + 0.585 585 585 585 978 867 712;
  • 23) 0.585 585 585 585 978 867 712 × 2 = 1 + 0.171 171 171 171 957 735 424;
  • 24) 0.171 171 171 171 957 735 424 × 2 = 0 + 0.342 342 342 343 915 470 848;
  • 25) 0.342 342 342 343 915 470 848 × 2 = 0 + 0.684 684 684 687 830 941 696;
  • 26) 0.684 684 684 687 830 941 696 × 2 = 1 + 0.369 369 369 375 661 883 392;
  • 27) 0.369 369 369 375 661 883 392 × 2 = 0 + 0.738 738 738 751 323 766 784;
  • 28) 0.738 738 738 751 323 766 784 × 2 = 1 + 0.477 477 477 502 647 533 568;
  • 29) 0.477 477 477 502 647 533 568 × 2 = 0 + 0.954 954 955 005 295 067 136;
  • 30) 0.954 954 955 005 295 067 136 × 2 = 1 + 0.909 909 910 010 590 134 272;
  • 31) 0.909 909 910 010 590 134 272 × 2 = 1 + 0.819 819 820 021 180 268 544;
  • 32) 0.819 819 820 021 180 268 544 × 2 = 1 + 0.639 639 640 042 360 537 088;
  • 33) 0.639 639 640 042 360 537 088 × 2 = 1 + 0.279 279 280 084 721 074 176;
  • 34) 0.279 279 280 084 721 074 176 × 2 = 0 + 0.558 558 560 169 442 148 352;
  • 35) 0.558 558 560 169 442 148 352 × 2 = 1 + 0.117 117 120 338 884 296 704;
  • 36) 0.117 117 120 338 884 296 704 × 2 = 0 + 0.234 234 240 677 768 593 408;
  • 37) 0.234 234 240 677 768 593 408 × 2 = 0 + 0.468 468 481 355 537 186 816;
  • 38) 0.468 468 481 355 537 186 816 × 2 = 0 + 0.936 936 962 711 074 373 632;
  • 39) 0.936 936 962 711 074 373 632 × 2 = 1 + 0.873 873 925 422 148 747 264;
  • 40) 0.873 873 925 422 148 747 264 × 2 = 1 + 0.747 747 850 844 297 494 528;
  • 41) 0.747 747 850 844 297 494 528 × 2 = 1 + 0.495 495 701 688 594 989 056;
  • 42) 0.495 495 701 688 594 989 056 × 2 = 0 + 0.990 991 403 377 189 978 112;
  • 43) 0.990 991 403 377 189 978 112 × 2 = 1 + 0.981 982 806 754 379 956 224;
  • 44) 0.981 982 806 754 379 956 224 × 2 = 1 + 0.963 965 613 508 759 912 448;
  • 45) 0.963 965 613 508 759 912 448 × 2 = 1 + 0.927 931 227 017 519 824 896;
  • 46) 0.927 931 227 017 519 824 896 × 2 = 1 + 0.855 862 454 035 039 649 792;
  • 47) 0.855 862 454 035 039 649 792 × 2 = 1 + 0.711 724 908 070 079 299 584;
  • 48) 0.711 724 908 070 079 299 584 × 2 = 1 + 0.423 449 816 140 158 599 168;
  • 49) 0.423 449 816 140 158 599 168 × 2 = 0 + 0.846 899 632 280 317 198 336;
  • 50) 0.846 899 632 280 317 198 336 × 2 = 1 + 0.693 799 264 560 634 396 672;
  • 51) 0.693 799 264 560 634 396 672 × 2 = 1 + 0.387 598 529 121 268 793 344;
  • 52) 0.387 598 529 121 268 793 344 × 2 = 0 + 0.775 197 058 242 537 586 688;
  • 53) 0.775 197 058 242 537 586 688 × 2 = 1 + 0.550 394 116 485 075 173 376;
  • 54) 0.550 394 116 485 075 173 376 × 2 = 1 + 0.100 788 232 970 150 346 752;
  • 55) 0.100 788 232 970 150 346 752 × 2 = 0 + 0.201 576 465 940 300 693 504;

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:


0.234 234 234 234 234 234 328(10) =

0.0011 1011 1111 0110 1100 0110 0101 0111 1010 0011 1011 1111 0110 110(2)

5. Positive number before normalization:

0.234 234 234 234 234 234 328(10) =

0.0011 1011 1111 0110 1100 0110 0101 0111 1010 0011 1011 1111 0110 110(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 3 positions to the right, so that only one non zero digit remains to the left of it:


0.234 234 234 234 234 234 328(10) =

0.0011 1011 1111 0110 1100 0110 0101 0111 1010 0011 1011 1111 0110 110(2) =

0.0011 1011 1111 0110 1100 0110 0101 0111 1010 0011 1011 1111 0110 110(2) × 20 =

1.1101 1111 1011 0110 0011 0010 1011 1101 0001 1101 1111 1011 0110(2) × 2-3


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): -3

Mantissa (not normalized):
1.1101 1111 1011 0110 0011 0010 1011 1101 0001 1101 1111 1011 0110


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-3 + 2(11-1) - 1 =

(-3 + 1 023)(10) =

1 020(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;
  • 1 020 ÷ 2 = 510 + 0;
  • 510 ÷ 2 = 255 + 0;
  • 255 ÷ 2 = 127 + 1;
  • 127 ÷ 2 = 63 + 1;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 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) =

1020(10) =

011 1111 1100(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, only if necessary (not the case here).


Mantissa (normalized) =

1. 1101 1111 1011 0110 0011 0010 1011 1101 0001 1101 1111 1011 0110 =

1101 1111 1011 0110 0011 0010 1011 1101 0001 1101 1111 1011 0110


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) =
011 1111 1100

Mantissa (52 bits) =
1101 1111 1011 0110 0011 0010 1011 1101 0001 1101 1111 1011 0110


Decimal number 0.234 234 234 234 234 234 328 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1100 - 1101 1111 1011 0110 0011 0010 1011 1101 0001 1101 1111 1011 0110

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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