0.234 234 234 234 234 234 268 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.234 234 234 234 234 234 268(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
0.234 234 234 234 234 234 268(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: 0.234 234 234 234 234 234 268.

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) 0.234 234 234 234 234 234 268 × 2 = 0 + 0.468 468 468 468 468 468 536;
  • 2) 0.468 468 468 468 468 468 536 × 2 = 0 + 0.936 936 936 936 936 937 072;
  • 3) 0.936 936 936 936 936 937 072 × 2 = 1 + 0.873 873 873 873 873 874 144;
  • 4) 0.873 873 873 873 873 874 144 × 2 = 1 + 0.747 747 747 747 747 748 288;
  • 5) 0.747 747 747 747 747 748 288 × 2 = 1 + 0.495 495 495 495 495 496 576;
  • 6) 0.495 495 495 495 495 496 576 × 2 = 0 + 0.990 990 990 990 990 993 152;
  • 7) 0.990 990 990 990 990 993 152 × 2 = 1 + 0.981 981 981 981 981 986 304;
  • 8) 0.981 981 981 981 981 986 304 × 2 = 1 + 0.963 963 963 963 963 972 608;
  • 9) 0.963 963 963 963 963 972 608 × 2 = 1 + 0.927 927 927 927 927 945 216;
  • 10) 0.927 927 927 927 927 945 216 × 2 = 1 + 0.855 855 855 855 855 890 432;
  • 11) 0.855 855 855 855 855 890 432 × 2 = 1 + 0.711 711 711 711 711 780 864;
  • 12) 0.711 711 711 711 711 780 864 × 2 = 1 + 0.423 423 423 423 423 561 728;
  • 13) 0.423 423 423 423 423 561 728 × 2 = 0 + 0.846 846 846 846 847 123 456;
  • 14) 0.846 846 846 846 847 123 456 × 2 = 1 + 0.693 693 693 693 694 246 912;
  • 15) 0.693 693 693 693 694 246 912 × 2 = 1 + 0.387 387 387 387 388 493 824;
  • 16) 0.387 387 387 387 388 493 824 × 2 = 0 + 0.774 774 774 774 776 987 648;
  • 17) 0.774 774 774 774 776 987 648 × 2 = 1 + 0.549 549 549 549 553 975 296;
  • 18) 0.549 549 549 549 553 975 296 × 2 = 1 + 0.099 099 099 099 107 950 592;
  • 19) 0.099 099 099 099 107 950 592 × 2 = 0 + 0.198 198 198 198 215 901 184;
  • 20) 0.198 198 198 198 215 901 184 × 2 = 0 + 0.396 396 396 396 431 802 368;
  • 21) 0.396 396 396 396 431 802 368 × 2 = 0 + 0.792 792 792 792 863 604 736;
  • 22) 0.792 792 792 792 863 604 736 × 2 = 1 + 0.585 585 585 585 727 209 472;
  • 23) 0.585 585 585 585 727 209 472 × 2 = 1 + 0.171 171 171 171 454 418 944;
  • 24) 0.171 171 171 171 454 418 944 × 2 = 0 + 0.342 342 342 342 908 837 888;
  • 25) 0.342 342 342 342 908 837 888 × 2 = 0 + 0.684 684 684 685 817 675 776;
  • 26) 0.684 684 684 685 817 675 776 × 2 = 1 + 0.369 369 369 371 635 351 552;
  • 27) 0.369 369 369 371 635 351 552 × 2 = 0 + 0.738 738 738 743 270 703 104;
  • 28) 0.738 738 738 743 270 703 104 × 2 = 1 + 0.477 477 477 486 541 406 208;
  • 29) 0.477 477 477 486 541 406 208 × 2 = 0 + 0.954 954 954 973 082 812 416;
  • 30) 0.954 954 954 973 082 812 416 × 2 = 1 + 0.909 909 909 946 165 624 832;
  • 31) 0.909 909 909 946 165 624 832 × 2 = 1 + 0.819 819 819 892 331 249 664;
  • 32) 0.819 819 819 892 331 249 664 × 2 = 1 + 0.639 639 639 784 662 499 328;
  • 33) 0.639 639 639 784 662 499 328 × 2 = 1 + 0.279 279 279 569 324 998 656;
  • 34) 0.279 279 279 569 324 998 656 × 2 = 0 + 0.558 558 559 138 649 997 312;
  • 35) 0.558 558 559 138 649 997 312 × 2 = 1 + 0.117 117 118 277 299 994 624;
  • 36) 0.117 117 118 277 299 994 624 × 2 = 0 + 0.234 234 236 554 599 989 248;
  • 37) 0.234 234 236 554 599 989 248 × 2 = 0 + 0.468 468 473 109 199 978 496;
  • 38) 0.468 468 473 109 199 978 496 × 2 = 0 + 0.936 936 946 218 399 956 992;
  • 39) 0.936 936 946 218 399 956 992 × 2 = 1 + 0.873 873 892 436 799 913 984;
  • 40) 0.873 873 892 436 799 913 984 × 2 = 1 + 0.747 747 784 873 599 827 968;
  • 41) 0.747 747 784 873 599 827 968 × 2 = 1 + 0.495 495 569 747 199 655 936;
  • 42) 0.495 495 569 747 199 655 936 × 2 = 0 + 0.990 991 139 494 399 311 872;
  • 43) 0.990 991 139 494 399 311 872 × 2 = 1 + 0.981 982 278 988 798 623 744;
  • 44) 0.981 982 278 988 798 623 744 × 2 = 1 + 0.963 964 557 977 597 247 488;
  • 45) 0.963 964 557 977 597 247 488 × 2 = 1 + 0.927 929 115 955 194 494 976;
  • 46) 0.927 929 115 955 194 494 976 × 2 = 1 + 0.855 858 231 910 388 989 952;
  • 47) 0.855 858 231 910 388 989 952 × 2 = 1 + 0.711 716 463 820 777 979 904;
  • 48) 0.711 716 463 820 777 979 904 × 2 = 1 + 0.423 432 927 641 555 959 808;
  • 49) 0.423 432 927 641 555 959 808 × 2 = 0 + 0.846 865 855 283 111 919 616;
  • 50) 0.846 865 855 283 111 919 616 × 2 = 1 + 0.693 731 710 566 223 839 232;
  • 51) 0.693 731 710 566 223 839 232 × 2 = 1 + 0.387 463 421 132 447 678 464;
  • 52) 0.387 463 421 132 447 678 464 × 2 = 0 + 0.774 926 842 264 895 356 928;
  • 53) 0.774 926 842 264 895 356 928 × 2 = 1 + 0.549 853 684 529 790 713 856;
  • 54) 0.549 853 684 529 790 713 856 × 2 = 1 + 0.099 707 369 059 581 427 712;
  • 55) 0.099 707 369 059 581 427 712 × 2 = 0 + 0.199 414 738 119 162 855 424;

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:


0.234 234 234 234 234 234 268(10) =

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

5. Positive number before normalization:

0.234 234 234 234 234 234 268(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:


0.234 234 234 234 234 234 268(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 0.234 234 234 234 234 234 268 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