0.142 857 142 857 142 7 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.142 857 142 857 142 7(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 7(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 7.

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 7 × 2 = 0 + 0.285 714 285 714 285 4;
  • 2) 0.285 714 285 714 285 4 × 2 = 0 + 0.571 428 571 428 570 8;
  • 3) 0.571 428 571 428 570 8 × 2 = 1 + 0.142 857 142 857 141 6;
  • 4) 0.142 857 142 857 141 6 × 2 = 0 + 0.285 714 285 714 283 2;
  • 5) 0.285 714 285 714 283 2 × 2 = 0 + 0.571 428 571 428 566 4;
  • 6) 0.571 428 571 428 566 4 × 2 = 1 + 0.142 857 142 857 132 8;
  • 7) 0.142 857 142 857 132 8 × 2 = 0 + 0.285 714 285 714 265 6;
  • 8) 0.285 714 285 714 265 6 × 2 = 0 + 0.571 428 571 428 531 2;
  • 9) 0.571 428 571 428 531 2 × 2 = 1 + 0.142 857 142 857 062 4;
  • 10) 0.142 857 142 857 062 4 × 2 = 0 + 0.285 714 285 714 124 8;
  • 11) 0.285 714 285 714 124 8 × 2 = 0 + 0.571 428 571 428 249 6;
  • 12) 0.571 428 571 428 249 6 × 2 = 1 + 0.142 857 142 856 499 2;
  • 13) 0.142 857 142 856 499 2 × 2 = 0 + 0.285 714 285 712 998 4;
  • 14) 0.285 714 285 712 998 4 × 2 = 0 + 0.571 428 571 425 996 8;
  • 15) 0.571 428 571 425 996 8 × 2 = 1 + 0.142 857 142 851 993 6;
  • 16) 0.142 857 142 851 993 6 × 2 = 0 + 0.285 714 285 703 987 2;
  • 17) 0.285 714 285 703 987 2 × 2 = 0 + 0.571 428 571 407 974 4;
  • 18) 0.571 428 571 407 974 4 × 2 = 1 + 0.142 857 142 815 948 8;
  • 19) 0.142 857 142 815 948 8 × 2 = 0 + 0.285 714 285 631 897 6;
  • 20) 0.285 714 285 631 897 6 × 2 = 0 + 0.571 428 571 263 795 2;
  • 21) 0.571 428 571 263 795 2 × 2 = 1 + 0.142 857 142 527 590 4;
  • 22) 0.142 857 142 527 590 4 × 2 = 0 + 0.285 714 285 055 180 8;
  • 23) 0.285 714 285 055 180 8 × 2 = 0 + 0.571 428 570 110 361 6;
  • 24) 0.571 428 570 110 361 6 × 2 = 1 + 0.142 857 140 220 723 2;
  • 25) 0.142 857 140 220 723 2 × 2 = 0 + 0.285 714 280 441 446 4;
  • 26) 0.285 714 280 441 446 4 × 2 = 0 + 0.571 428 560 882 892 8;
  • 27) 0.571 428 560 882 892 8 × 2 = 1 + 0.142 857 121 765 785 6;
  • 28) 0.142 857 121 765 785 6 × 2 = 0 + 0.285 714 243 531 571 2;
  • 29) 0.285 714 243 531 571 2 × 2 = 0 + 0.571 428 487 063 142 4;
  • 30) 0.571 428 487 063 142 4 × 2 = 1 + 0.142 856 974 126 284 8;
  • 31) 0.142 856 974 126 284 8 × 2 = 0 + 0.285 713 948 252 569 6;
  • 32) 0.285 713 948 252 569 6 × 2 = 0 + 0.571 427 896 505 139 2;
  • 33) 0.571 427 896 505 139 2 × 2 = 1 + 0.142 855 793 010 278 4;
  • 34) 0.142 855 793 010 278 4 × 2 = 0 + 0.285 711 586 020 556 8;
  • 35) 0.285 711 586 020 556 8 × 2 = 0 + 0.571 423 172 041 113 6;
  • 36) 0.571 423 172 041 113 6 × 2 = 1 + 0.142 846 344 082 227 2;
  • 37) 0.142 846 344 082 227 2 × 2 = 0 + 0.285 692 688 164 454 4;
  • 38) 0.285 692 688 164 454 4 × 2 = 0 + 0.571 385 376 328 908 8;
  • 39) 0.571 385 376 328 908 8 × 2 = 1 + 0.142 770 752 657 817 6;
  • 40) 0.142 770 752 657 817 6 × 2 = 0 + 0.285 541 505 315 635 2;
  • 41) 0.285 541 505 315 635 2 × 2 = 0 + 0.571 083 010 631 270 4;
  • 42) 0.571 083 010 631 270 4 × 2 = 1 + 0.142 166 021 262 540 8;
  • 43) 0.142 166 021 262 540 8 × 2 = 0 + 0.284 332 042 525 081 6;
  • 44) 0.284 332 042 525 081 6 × 2 = 0 + 0.568 664 085 050 163 2;
  • 45) 0.568 664 085 050 163 2 × 2 = 1 + 0.137 328 170 100 326 4;
  • 46) 0.137 328 170 100 326 4 × 2 = 0 + 0.274 656 340 200 652 8;
  • 47) 0.274 656 340 200 652 8 × 2 = 0 + 0.549 312 680 401 305 6;
  • 48) 0.549 312 680 401 305 6 × 2 = 1 + 0.098 625 360 802 611 2;
  • 49) 0.098 625 360 802 611 2 × 2 = 0 + 0.197 250 721 605 222 4;
  • 50) 0.197 250 721 605 222 4 × 2 = 0 + 0.394 501 443 210 444 8;
  • 51) 0.394 501 443 210 444 8 × 2 = 0 + 0.789 002 886 420 889 6;
  • 52) 0.789 002 886 420 889 6 × 2 = 1 + 0.578 005 772 841 779 2;
  • 53) 0.578 005 772 841 779 2 × 2 = 1 + 0.156 011 545 683 558 4;
  • 54) 0.156 011 545 683 558 4 × 2 = 0 + 0.312 023 091 367 116 8;
  • 55) 0.312 023 091 367 116 8 × 2 = 0 + 0.624 046 182 734 233 6;

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 7(10) =

0.0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0001 100(2)

5. Positive number before normalization:

0.142 857 142 857 142 7(10) =

0.0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0001 100(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 7(10) =

0.0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0001 100(2) =

0.0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0001 100(2) × 20 =

1.0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1000 1100(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 1000 1100


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 1000 1100 =

0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1000 1100


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 1000 1100


Decimal number 0.142 857 142 857 142 7 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 1000 1100

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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