0.142 857 142 857 142 8 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.142 857 142 857 142 8(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 8(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 8.

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 8 × 2 = 0 + 0.285 714 285 714 285 6;
  • 2) 0.285 714 285 714 285 6 × 2 = 0 + 0.571 428 571 428 571 2;
  • 3) 0.571 428 571 428 571 2 × 2 = 1 + 0.142 857 142 857 142 4;
  • 4) 0.142 857 142 857 142 4 × 2 = 0 + 0.285 714 285 714 284 8;
  • 5) 0.285 714 285 714 284 8 × 2 = 0 + 0.571 428 571 428 569 6;
  • 6) 0.571 428 571 428 569 6 × 2 = 1 + 0.142 857 142 857 139 2;
  • 7) 0.142 857 142 857 139 2 × 2 = 0 + 0.285 714 285 714 278 4;
  • 8) 0.285 714 285 714 278 4 × 2 = 0 + 0.571 428 571 428 556 8;
  • 9) 0.571 428 571 428 556 8 × 2 = 1 + 0.142 857 142 857 113 6;
  • 10) 0.142 857 142 857 113 6 × 2 = 0 + 0.285 714 285 714 227 2;
  • 11) 0.285 714 285 714 227 2 × 2 = 0 + 0.571 428 571 428 454 4;
  • 12) 0.571 428 571 428 454 4 × 2 = 1 + 0.142 857 142 856 908 8;
  • 13) 0.142 857 142 856 908 8 × 2 = 0 + 0.285 714 285 713 817 6;
  • 14) 0.285 714 285 713 817 6 × 2 = 0 + 0.571 428 571 427 635 2;
  • 15) 0.571 428 571 427 635 2 × 2 = 1 + 0.142 857 142 855 270 4;
  • 16) 0.142 857 142 855 270 4 × 2 = 0 + 0.285 714 285 710 540 8;
  • 17) 0.285 714 285 710 540 8 × 2 = 0 + 0.571 428 571 421 081 6;
  • 18) 0.571 428 571 421 081 6 × 2 = 1 + 0.142 857 142 842 163 2;
  • 19) 0.142 857 142 842 163 2 × 2 = 0 + 0.285 714 285 684 326 4;
  • 20) 0.285 714 285 684 326 4 × 2 = 0 + 0.571 428 571 368 652 8;
  • 21) 0.571 428 571 368 652 8 × 2 = 1 + 0.142 857 142 737 305 6;
  • 22) 0.142 857 142 737 305 6 × 2 = 0 + 0.285 714 285 474 611 2;
  • 23) 0.285 714 285 474 611 2 × 2 = 0 + 0.571 428 570 949 222 4;
  • 24) 0.571 428 570 949 222 4 × 2 = 1 + 0.142 857 141 898 444 8;
  • 25) 0.142 857 141 898 444 8 × 2 = 0 + 0.285 714 283 796 889 6;
  • 26) 0.285 714 283 796 889 6 × 2 = 0 + 0.571 428 567 593 779 2;
  • 27) 0.571 428 567 593 779 2 × 2 = 1 + 0.142 857 135 187 558 4;
  • 28) 0.142 857 135 187 558 4 × 2 = 0 + 0.285 714 270 375 116 8;
  • 29) 0.285 714 270 375 116 8 × 2 = 0 + 0.571 428 540 750 233 6;
  • 30) 0.571 428 540 750 233 6 × 2 = 1 + 0.142 857 081 500 467 2;
  • 31) 0.142 857 081 500 467 2 × 2 = 0 + 0.285 714 163 000 934 4;
  • 32) 0.285 714 163 000 934 4 × 2 = 0 + 0.571 428 326 001 868 8;
  • 33) 0.571 428 326 001 868 8 × 2 = 1 + 0.142 856 652 003 737 6;
  • 34) 0.142 856 652 003 737 6 × 2 = 0 + 0.285 713 304 007 475 2;
  • 35) 0.285 713 304 007 475 2 × 2 = 0 + 0.571 426 608 014 950 4;
  • 36) 0.571 426 608 014 950 4 × 2 = 1 + 0.142 853 216 029 900 8;
  • 37) 0.142 853 216 029 900 8 × 2 = 0 + 0.285 706 432 059 801 6;
  • 38) 0.285 706 432 059 801 6 × 2 = 0 + 0.571 412 864 119 603 2;
  • 39) 0.571 412 864 119 603 2 × 2 = 1 + 0.142 825 728 239 206 4;
  • 40) 0.142 825 728 239 206 4 × 2 = 0 + 0.285 651 456 478 412 8;
  • 41) 0.285 651 456 478 412 8 × 2 = 0 + 0.571 302 912 956 825 6;
  • 42) 0.571 302 912 956 825 6 × 2 = 1 + 0.142 605 825 913 651 2;
  • 43) 0.142 605 825 913 651 2 × 2 = 0 + 0.285 211 651 827 302 4;
  • 44) 0.285 211 651 827 302 4 × 2 = 0 + 0.570 423 303 654 604 8;
  • 45) 0.570 423 303 654 604 8 × 2 = 1 + 0.140 846 607 309 209 6;
  • 46) 0.140 846 607 309 209 6 × 2 = 0 + 0.281 693 214 618 419 2;
  • 47) 0.281 693 214 618 419 2 × 2 = 0 + 0.563 386 429 236 838 4;
  • 48) 0.563 386 429 236 838 4 × 2 = 1 + 0.126 772 858 473 676 8;
  • 49) 0.126 772 858 473 676 8 × 2 = 0 + 0.253 545 716 947 353 6;
  • 50) 0.253 545 716 947 353 6 × 2 = 0 + 0.507 091 433 894 707 2;
  • 51) 0.507 091 433 894 707 2 × 2 = 1 + 0.014 182 867 789 414 4;
  • 52) 0.014 182 867 789 414 4 × 2 = 0 + 0.028 365 735 578 828 8;
  • 53) 0.028 365 735 578 828 8 × 2 = 0 + 0.056 731 471 157 657 6;
  • 54) 0.056 731 471 157 657 6 × 2 = 0 + 0.113 462 942 315 315 2;
  • 55) 0.113 462 942 315 315 2 × 2 = 0 + 0.226 925 884 630 630 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 8(10) =

0.0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 000(2)

5. Positive number before normalization:

0.142 857 142 857 142 8(10) =

0.0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 000(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 8(10) =

0.0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 000(2) =

0.0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 000(2) × 20 =

1.0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0000(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 0000


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 0000 =

0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0000


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 0000


Decimal number 0.142 857 142 857 142 8 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 0000

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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