0.142 857 142 857 142 857 146 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.142 857 142 857 142 857 146 3(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
0.142 857 142 857 142 857 146 3(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: 0.142 857 142 857 142 857 146 3.

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) 0.142 857 142 857 142 857 146 3 × 2 = 0 + 0.285 714 285 714 285 714 292 6;
  • 2) 0.285 714 285 714 285 714 292 6 × 2 = 0 + 0.571 428 571 428 571 428 585 2;
  • 3) 0.571 428 571 428 571 428 585 2 × 2 = 1 + 0.142 857 142 857 142 857 170 4;
  • 4) 0.142 857 142 857 142 857 170 4 × 2 = 0 + 0.285 714 285 714 285 714 340 8;
  • 5) 0.285 714 285 714 285 714 340 8 × 2 = 0 + 0.571 428 571 428 571 428 681 6;
  • 6) 0.571 428 571 428 571 428 681 6 × 2 = 1 + 0.142 857 142 857 142 857 363 2;
  • 7) 0.142 857 142 857 142 857 363 2 × 2 = 0 + 0.285 714 285 714 285 714 726 4;
  • 8) 0.285 714 285 714 285 714 726 4 × 2 = 0 + 0.571 428 571 428 571 429 452 8;
  • 9) 0.571 428 571 428 571 429 452 8 × 2 = 1 + 0.142 857 142 857 142 858 905 6;
  • 10) 0.142 857 142 857 142 858 905 6 × 2 = 0 + 0.285 714 285 714 285 717 811 2;
  • 11) 0.285 714 285 714 285 717 811 2 × 2 = 0 + 0.571 428 571 428 571 435 622 4;
  • 12) 0.571 428 571 428 571 435 622 4 × 2 = 1 + 0.142 857 142 857 142 871 244 8;
  • 13) 0.142 857 142 857 142 871 244 8 × 2 = 0 + 0.285 714 285 714 285 742 489 6;
  • 14) 0.285 714 285 714 285 742 489 6 × 2 = 0 + 0.571 428 571 428 571 484 979 2;
  • 15) 0.571 428 571 428 571 484 979 2 × 2 = 1 + 0.142 857 142 857 142 969 958 4;
  • 16) 0.142 857 142 857 142 969 958 4 × 2 = 0 + 0.285 714 285 714 285 939 916 8;
  • 17) 0.285 714 285 714 285 939 916 8 × 2 = 0 + 0.571 428 571 428 571 879 833 6;
  • 18) 0.571 428 571 428 571 879 833 6 × 2 = 1 + 0.142 857 142 857 143 759 667 2;
  • 19) 0.142 857 142 857 143 759 667 2 × 2 = 0 + 0.285 714 285 714 287 519 334 4;
  • 20) 0.285 714 285 714 287 519 334 4 × 2 = 0 + 0.571 428 571 428 575 038 668 8;
  • 21) 0.571 428 571 428 575 038 668 8 × 2 = 1 + 0.142 857 142 857 150 077 337 6;
  • 22) 0.142 857 142 857 150 077 337 6 × 2 = 0 + 0.285 714 285 714 300 154 675 2;
  • 23) 0.285 714 285 714 300 154 675 2 × 2 = 0 + 0.571 428 571 428 600 309 350 4;
  • 24) 0.571 428 571 428 600 309 350 4 × 2 = 1 + 0.142 857 142 857 200 618 700 8;
  • 25) 0.142 857 142 857 200 618 700 8 × 2 = 0 + 0.285 714 285 714 401 237 401 6;
  • 26) 0.285 714 285 714 401 237 401 6 × 2 = 0 + 0.571 428 571 428 802 474 803 2;
  • 27) 0.571 428 571 428 802 474 803 2 × 2 = 1 + 0.142 857 142 857 604 949 606 4;
  • 28) 0.142 857 142 857 604 949 606 4 × 2 = 0 + 0.285 714 285 715 209 899 212 8;
  • 29) 0.285 714 285 715 209 899 212 8 × 2 = 0 + 0.571 428 571 430 419 798 425 6;
  • 30) 0.571 428 571 430 419 798 425 6 × 2 = 1 + 0.142 857 142 860 839 596 851 2;
  • 31) 0.142 857 142 860 839 596 851 2 × 2 = 0 + 0.285 714 285 721 679 193 702 4;
  • 32) 0.285 714 285 721 679 193 702 4 × 2 = 0 + 0.571 428 571 443 358 387 404 8;
  • 33) 0.571 428 571 443 358 387 404 8 × 2 = 1 + 0.142 857 142 886 716 774 809 6;
  • 34) 0.142 857 142 886 716 774 809 6 × 2 = 0 + 0.285 714 285 773 433 549 619 2;
  • 35) 0.285 714 285 773 433 549 619 2 × 2 = 0 + 0.571 428 571 546 867 099 238 4;
  • 36) 0.571 428 571 546 867 099 238 4 × 2 = 1 + 0.142 857 143 093 734 198 476 8;
  • 37) 0.142 857 143 093 734 198 476 8 × 2 = 0 + 0.285 714 286 187 468 396 953 6;
  • 38) 0.285 714 286 187 468 396 953 6 × 2 = 0 + 0.571 428 572 374 936 793 907 2;
  • 39) 0.571 428 572 374 936 793 907 2 × 2 = 1 + 0.142 857 144 749 873 587 814 4;
  • 40) 0.142 857 144 749 873 587 814 4 × 2 = 0 + 0.285 714 289 499 747 175 628 8;
  • 41) 0.285 714 289 499 747 175 628 8 × 2 = 0 + 0.571 428 578 999 494 351 257 6;
  • 42) 0.571 428 578 999 494 351 257 6 × 2 = 1 + 0.142 857 157 998 988 702 515 2;
  • 43) 0.142 857 157 998 988 702 515 2 × 2 = 0 + 0.285 714 315 997 977 405 030 4;
  • 44) 0.285 714 315 997 977 405 030 4 × 2 = 0 + 0.571 428 631 995 954 810 060 8;
  • 45) 0.571 428 631 995 954 810 060 8 × 2 = 1 + 0.142 857 263 991 909 620 121 6;
  • 46) 0.142 857 263 991 909 620 121 6 × 2 = 0 + 0.285 714 527 983 819 240 243 2;
  • 47) 0.285 714 527 983 819 240 243 2 × 2 = 0 + 0.571 429 055 967 638 480 486 4;
  • 48) 0.571 429 055 967 638 480 486 4 × 2 = 1 + 0.142 858 111 935 276 960 972 8;
  • 49) 0.142 858 111 935 276 960 972 8 × 2 = 0 + 0.285 716 223 870 553 921 945 6;
  • 50) 0.285 716 223 870 553 921 945 6 × 2 = 0 + 0.571 432 447 741 107 843 891 2;
  • 51) 0.571 432 447 741 107 843 891 2 × 2 = 1 + 0.142 864 895 482 215 687 782 4;
  • 52) 0.142 864 895 482 215 687 782 4 × 2 = 0 + 0.285 729 790 964 431 375 564 8;
  • 53) 0.285 729 790 964 431 375 564 8 × 2 = 0 + 0.571 459 581 928 862 751 129 6;
  • 54) 0.571 459 581 928 862 751 129 6 × 2 = 1 + 0.142 919 163 857 725 502 259 2;
  • 55) 0.142 919 163 857 725 502 259 2 × 2 = 0 + 0.285 838 327 715 451 004 518 4;

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:


0.142 857 142 857 142 857 146 3(10) =

0.0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 010(2)

5. Positive number before normalization:

0.142 857 142 857 142 857 146 3(10) =

0.0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 010(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:


0.142 857 142 857 142 857 146 3(10) =

0.0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 010(2) =

0.0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 010(2) × 20 =

1.0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010(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):
1.0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010


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. 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 =

0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010


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) =
0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010


Decimal number 0.142 857 142 857 142 857 146 3 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1100 - 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010

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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