0.142 857 142 857 135 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.142 857 142 857 135 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
0.142 857 142 857 135 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: 0.142 857 142 857 135 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) 0.142 857 142 857 135 3 × 2 = 0 + 0.285 714 285 714 270 6;
  • 2) 0.285 714 285 714 270 6 × 2 = 0 + 0.571 428 571 428 541 2;
  • 3) 0.571 428 571 428 541 2 × 2 = 1 + 0.142 857 142 857 082 4;
  • 4) 0.142 857 142 857 082 4 × 2 = 0 + 0.285 714 285 714 164 8;
  • 5) 0.285 714 285 714 164 8 × 2 = 0 + 0.571 428 571 428 329 6;
  • 6) 0.571 428 571 428 329 6 × 2 = 1 + 0.142 857 142 856 659 2;
  • 7) 0.142 857 142 856 659 2 × 2 = 0 + 0.285 714 285 713 318 4;
  • 8) 0.285 714 285 713 318 4 × 2 = 0 + 0.571 428 571 426 636 8;
  • 9) 0.571 428 571 426 636 8 × 2 = 1 + 0.142 857 142 853 273 6;
  • 10) 0.142 857 142 853 273 6 × 2 = 0 + 0.285 714 285 706 547 2;
  • 11) 0.285 714 285 706 547 2 × 2 = 0 + 0.571 428 571 413 094 4;
  • 12) 0.571 428 571 413 094 4 × 2 = 1 + 0.142 857 142 826 188 8;
  • 13) 0.142 857 142 826 188 8 × 2 = 0 + 0.285 714 285 652 377 6;
  • 14) 0.285 714 285 652 377 6 × 2 = 0 + 0.571 428 571 304 755 2;
  • 15) 0.571 428 571 304 755 2 × 2 = 1 + 0.142 857 142 609 510 4;
  • 16) 0.142 857 142 609 510 4 × 2 = 0 + 0.285 714 285 219 020 8;
  • 17) 0.285 714 285 219 020 8 × 2 = 0 + 0.571 428 570 438 041 6;
  • 18) 0.571 428 570 438 041 6 × 2 = 1 + 0.142 857 140 876 083 2;
  • 19) 0.142 857 140 876 083 2 × 2 = 0 + 0.285 714 281 752 166 4;
  • 20) 0.285 714 281 752 166 4 × 2 = 0 + 0.571 428 563 504 332 8;
  • 21) 0.571 428 563 504 332 8 × 2 = 1 + 0.142 857 127 008 665 6;
  • 22) 0.142 857 127 008 665 6 × 2 = 0 + 0.285 714 254 017 331 2;
  • 23) 0.285 714 254 017 331 2 × 2 = 0 + 0.571 428 508 034 662 4;
  • 24) 0.571 428 508 034 662 4 × 2 = 1 + 0.142 857 016 069 324 8;
  • 25) 0.142 857 016 069 324 8 × 2 = 0 + 0.285 714 032 138 649 6;
  • 26) 0.285 714 032 138 649 6 × 2 = 0 + 0.571 428 064 277 299 2;
  • 27) 0.571 428 064 277 299 2 × 2 = 1 + 0.142 856 128 554 598 4;
  • 28) 0.142 856 128 554 598 4 × 2 = 0 + 0.285 712 257 109 196 8;
  • 29) 0.285 712 257 109 196 8 × 2 = 0 + 0.571 424 514 218 393 6;
  • 30) 0.571 424 514 218 393 6 × 2 = 1 + 0.142 849 028 436 787 2;
  • 31) 0.142 849 028 436 787 2 × 2 = 0 + 0.285 698 056 873 574 4;
  • 32) 0.285 698 056 873 574 4 × 2 = 0 + 0.571 396 113 747 148 8;
  • 33) 0.571 396 113 747 148 8 × 2 = 1 + 0.142 792 227 494 297 6;
  • 34) 0.142 792 227 494 297 6 × 2 = 0 + 0.285 584 454 988 595 2;
  • 35) 0.285 584 454 988 595 2 × 2 = 0 + 0.571 168 909 977 190 4;
  • 36) 0.571 168 909 977 190 4 × 2 = 1 + 0.142 337 819 954 380 8;
  • 37) 0.142 337 819 954 380 8 × 2 = 0 + 0.284 675 639 908 761 6;
  • 38) 0.284 675 639 908 761 6 × 2 = 0 + 0.569 351 279 817 523 2;
  • 39) 0.569 351 279 817 523 2 × 2 = 1 + 0.138 702 559 635 046 4;
  • 40) 0.138 702 559 635 046 4 × 2 = 0 + 0.277 405 119 270 092 8;
  • 41) 0.277 405 119 270 092 8 × 2 = 0 + 0.554 810 238 540 185 6;
  • 42) 0.554 810 238 540 185 6 × 2 = 1 + 0.109 620 477 080 371 2;
  • 43) 0.109 620 477 080 371 2 × 2 = 0 + 0.219 240 954 160 742 4;
  • 44) 0.219 240 954 160 742 4 × 2 = 0 + 0.438 481 908 321 484 8;
  • 45) 0.438 481 908 321 484 8 × 2 = 0 + 0.876 963 816 642 969 6;
  • 46) 0.876 963 816 642 969 6 × 2 = 1 + 0.753 927 633 285 939 2;
  • 47) 0.753 927 633 285 939 2 × 2 = 1 + 0.507 855 266 571 878 4;
  • 48) 0.507 855 266 571 878 4 × 2 = 1 + 0.015 710 533 143 756 8;
  • 49) 0.015 710 533 143 756 8 × 2 = 0 + 0.031 421 066 287 513 6;
  • 50) 0.031 421 066 287 513 6 × 2 = 0 + 0.062 842 132 575 027 2;
  • 51) 0.062 842 132 575 027 2 × 2 = 0 + 0.125 684 265 150 054 4;
  • 52) 0.125 684 265 150 054 4 × 2 = 0 + 0.251 368 530 300 108 8;
  • 53) 0.251 368 530 300 108 8 × 2 = 0 + 0.502 737 060 600 217 6;
  • 54) 0.502 737 060 600 217 6 × 2 = 1 + 0.005 474 121 200 435 2;
  • 55) 0.005 474 121 200 435 2 × 2 = 0 + 0.010 948 242 400 870 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:


0.142 857 142 857 135 3(10) =

0.0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 0111 0000 010(2)

5. Positive number before normalization:

0.142 857 142 857 135 3(10) =

0.0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 0111 0000 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:


0.142 857 142 857 135 3(10) =

0.0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 0111 0000 010(2) =

0.0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 0111 0000 010(2) × 20 =

1.0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0011 1000 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 0011 1000 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 0011 1000 0010 =

0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0011 1000 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 0011 1000 0010


Decimal number 0.142 857 142 857 135 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 0011 1000 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