0.142 857 142 857 146 9 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.142 857 142 857 146 9(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
0.142 857 142 857 146 9(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: 0.142 857 142 857 146 9.

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) 0.142 857 142 857 146 9 × 2 = 0 + 0.285 714 285 714 293 8;
  • 2) 0.285 714 285 714 293 8 × 2 = 0 + 0.571 428 571 428 587 6;
  • 3) 0.571 428 571 428 587 6 × 2 = 1 + 0.142 857 142 857 175 2;
  • 4) 0.142 857 142 857 175 2 × 2 = 0 + 0.285 714 285 714 350 4;
  • 5) 0.285 714 285 714 350 4 × 2 = 0 + 0.571 428 571 428 700 8;
  • 6) 0.571 428 571 428 700 8 × 2 = 1 + 0.142 857 142 857 401 6;
  • 7) 0.142 857 142 857 401 6 × 2 = 0 + 0.285 714 285 714 803 2;
  • 8) 0.285 714 285 714 803 2 × 2 = 0 + 0.571 428 571 429 606 4;
  • 9) 0.571 428 571 429 606 4 × 2 = 1 + 0.142 857 142 859 212 8;
  • 10) 0.142 857 142 859 212 8 × 2 = 0 + 0.285 714 285 718 425 6;
  • 11) 0.285 714 285 718 425 6 × 2 = 0 + 0.571 428 571 436 851 2;
  • 12) 0.571 428 571 436 851 2 × 2 = 1 + 0.142 857 142 873 702 4;
  • 13) 0.142 857 142 873 702 4 × 2 = 0 + 0.285 714 285 747 404 8;
  • 14) 0.285 714 285 747 404 8 × 2 = 0 + 0.571 428 571 494 809 6;
  • 15) 0.571 428 571 494 809 6 × 2 = 1 + 0.142 857 142 989 619 2;
  • 16) 0.142 857 142 989 619 2 × 2 = 0 + 0.285 714 285 979 238 4;
  • 17) 0.285 714 285 979 238 4 × 2 = 0 + 0.571 428 571 958 476 8;
  • 18) 0.571 428 571 958 476 8 × 2 = 1 + 0.142 857 143 916 953 6;
  • 19) 0.142 857 143 916 953 6 × 2 = 0 + 0.285 714 287 833 907 2;
  • 20) 0.285 714 287 833 907 2 × 2 = 0 + 0.571 428 575 667 814 4;
  • 21) 0.571 428 575 667 814 4 × 2 = 1 + 0.142 857 151 335 628 8;
  • 22) 0.142 857 151 335 628 8 × 2 = 0 + 0.285 714 302 671 257 6;
  • 23) 0.285 714 302 671 257 6 × 2 = 0 + 0.571 428 605 342 515 2;
  • 24) 0.571 428 605 342 515 2 × 2 = 1 + 0.142 857 210 685 030 4;
  • 25) 0.142 857 210 685 030 4 × 2 = 0 + 0.285 714 421 370 060 8;
  • 26) 0.285 714 421 370 060 8 × 2 = 0 + 0.571 428 842 740 121 6;
  • 27) 0.571 428 842 740 121 6 × 2 = 1 + 0.142 857 685 480 243 2;
  • 28) 0.142 857 685 480 243 2 × 2 = 0 + 0.285 715 370 960 486 4;
  • 29) 0.285 715 370 960 486 4 × 2 = 0 + 0.571 430 741 920 972 8;
  • 30) 0.571 430 741 920 972 8 × 2 = 1 + 0.142 861 483 841 945 6;
  • 31) 0.142 861 483 841 945 6 × 2 = 0 + 0.285 722 967 683 891 2;
  • 32) 0.285 722 967 683 891 2 × 2 = 0 + 0.571 445 935 367 782 4;
  • 33) 0.571 445 935 367 782 4 × 2 = 1 + 0.142 891 870 735 564 8;
  • 34) 0.142 891 870 735 564 8 × 2 = 0 + 0.285 783 741 471 129 6;
  • 35) 0.285 783 741 471 129 6 × 2 = 0 + 0.571 567 482 942 259 2;
  • 36) 0.571 567 482 942 259 2 × 2 = 1 + 0.143 134 965 884 518 4;
  • 37) 0.143 134 965 884 518 4 × 2 = 0 + 0.286 269 931 769 036 8;
  • 38) 0.286 269 931 769 036 8 × 2 = 0 + 0.572 539 863 538 073 6;
  • 39) 0.572 539 863 538 073 6 × 2 = 1 + 0.145 079 727 076 147 2;
  • 40) 0.145 079 727 076 147 2 × 2 = 0 + 0.290 159 454 152 294 4;
  • 41) 0.290 159 454 152 294 4 × 2 = 0 + 0.580 318 908 304 588 8;
  • 42) 0.580 318 908 304 588 8 × 2 = 1 + 0.160 637 816 609 177 6;
  • 43) 0.160 637 816 609 177 6 × 2 = 0 + 0.321 275 633 218 355 2;
  • 44) 0.321 275 633 218 355 2 × 2 = 0 + 0.642 551 266 436 710 4;
  • 45) 0.642 551 266 436 710 4 × 2 = 1 + 0.285 102 532 873 420 8;
  • 46) 0.285 102 532 873 420 8 × 2 = 0 + 0.570 205 065 746 841 6;
  • 47) 0.570 205 065 746 841 6 × 2 = 1 + 0.140 410 131 493 683 2;
  • 48) 0.140 410 131 493 683 2 × 2 = 0 + 0.280 820 262 987 366 4;
  • 49) 0.280 820 262 987 366 4 × 2 = 0 + 0.561 640 525 974 732 8;
  • 50) 0.561 640 525 974 732 8 × 2 = 1 + 0.123 281 051 949 465 6;
  • 51) 0.123 281 051 949 465 6 × 2 = 0 + 0.246 562 103 898 931 2;
  • 52) 0.246 562 103 898 931 2 × 2 = 0 + 0.493 124 207 797 862 4;
  • 53) 0.493 124 207 797 862 4 × 2 = 0 + 0.986 248 415 595 724 8;
  • 54) 0.986 248 415 595 724 8 × 2 = 1 + 0.972 496 831 191 449 6;
  • 55) 0.972 496 831 191 449 6 × 2 = 1 + 0.944 993 662 382 899 2;

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:


0.142 857 142 857 146 9(10) =

0.0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1010 0100 011(2)

5. Positive number before normalization:

0.142 857 142 857 146 9(10) =

0.0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1010 0100 011(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:


0.142 857 142 857 146 9(10) =

0.0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1010 0100 011(2) =

0.0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1010 0100 011(2) × 20 =

1.0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0101 0010 0011(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 0101 0010 0011


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 0101 0010 0011 =

0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0101 0010 0011


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 0101 0010 0011


Decimal number 0.142 857 142 857 146 9 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 0101 0010 0011

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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