0.142 857 142 857 155 7 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.142 857 142 857 155 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
0.142 857 142 857 155 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: 0.142 857 142 857 155 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) 0.142 857 142 857 155 7 × 2 = 0 + 0.285 714 285 714 311 4;
  • 2) 0.285 714 285 714 311 4 × 2 = 0 + 0.571 428 571 428 622 8;
  • 3) 0.571 428 571 428 622 8 × 2 = 1 + 0.142 857 142 857 245 6;
  • 4) 0.142 857 142 857 245 6 × 2 = 0 + 0.285 714 285 714 491 2;
  • 5) 0.285 714 285 714 491 2 × 2 = 0 + 0.571 428 571 428 982 4;
  • 6) 0.571 428 571 428 982 4 × 2 = 1 + 0.142 857 142 857 964 8;
  • 7) 0.142 857 142 857 964 8 × 2 = 0 + 0.285 714 285 715 929 6;
  • 8) 0.285 714 285 715 929 6 × 2 = 0 + 0.571 428 571 431 859 2;
  • 9) 0.571 428 571 431 859 2 × 2 = 1 + 0.142 857 142 863 718 4;
  • 10) 0.142 857 142 863 718 4 × 2 = 0 + 0.285 714 285 727 436 8;
  • 11) 0.285 714 285 727 436 8 × 2 = 0 + 0.571 428 571 454 873 6;
  • 12) 0.571 428 571 454 873 6 × 2 = 1 + 0.142 857 142 909 747 2;
  • 13) 0.142 857 142 909 747 2 × 2 = 0 + 0.285 714 285 819 494 4;
  • 14) 0.285 714 285 819 494 4 × 2 = 0 + 0.571 428 571 638 988 8;
  • 15) 0.571 428 571 638 988 8 × 2 = 1 + 0.142 857 143 277 977 6;
  • 16) 0.142 857 143 277 977 6 × 2 = 0 + 0.285 714 286 555 955 2;
  • 17) 0.285 714 286 555 955 2 × 2 = 0 + 0.571 428 573 111 910 4;
  • 18) 0.571 428 573 111 910 4 × 2 = 1 + 0.142 857 146 223 820 8;
  • 19) 0.142 857 146 223 820 8 × 2 = 0 + 0.285 714 292 447 641 6;
  • 20) 0.285 714 292 447 641 6 × 2 = 0 + 0.571 428 584 895 283 2;
  • 21) 0.571 428 584 895 283 2 × 2 = 1 + 0.142 857 169 790 566 4;
  • 22) 0.142 857 169 790 566 4 × 2 = 0 + 0.285 714 339 581 132 8;
  • 23) 0.285 714 339 581 132 8 × 2 = 0 + 0.571 428 679 162 265 6;
  • 24) 0.571 428 679 162 265 6 × 2 = 1 + 0.142 857 358 324 531 2;
  • 25) 0.142 857 358 324 531 2 × 2 = 0 + 0.285 714 716 649 062 4;
  • 26) 0.285 714 716 649 062 4 × 2 = 0 + 0.571 429 433 298 124 8;
  • 27) 0.571 429 433 298 124 8 × 2 = 1 + 0.142 858 866 596 249 6;
  • 28) 0.142 858 866 596 249 6 × 2 = 0 + 0.285 717 733 192 499 2;
  • 29) 0.285 717 733 192 499 2 × 2 = 0 + 0.571 435 466 384 998 4;
  • 30) 0.571 435 466 384 998 4 × 2 = 1 + 0.142 870 932 769 996 8;
  • 31) 0.142 870 932 769 996 8 × 2 = 0 + 0.285 741 865 539 993 6;
  • 32) 0.285 741 865 539 993 6 × 2 = 0 + 0.571 483 731 079 987 2;
  • 33) 0.571 483 731 079 987 2 × 2 = 1 + 0.142 967 462 159 974 4;
  • 34) 0.142 967 462 159 974 4 × 2 = 0 + 0.285 934 924 319 948 8;
  • 35) 0.285 934 924 319 948 8 × 2 = 0 + 0.571 869 848 639 897 6;
  • 36) 0.571 869 848 639 897 6 × 2 = 1 + 0.143 739 697 279 795 2;
  • 37) 0.143 739 697 279 795 2 × 2 = 0 + 0.287 479 394 559 590 4;
  • 38) 0.287 479 394 559 590 4 × 2 = 0 + 0.574 958 789 119 180 8;
  • 39) 0.574 958 789 119 180 8 × 2 = 1 + 0.149 917 578 238 361 6;
  • 40) 0.149 917 578 238 361 6 × 2 = 0 + 0.299 835 156 476 723 2;
  • 41) 0.299 835 156 476 723 2 × 2 = 0 + 0.599 670 312 953 446 4;
  • 42) 0.599 670 312 953 446 4 × 2 = 1 + 0.199 340 625 906 892 8;
  • 43) 0.199 340 625 906 892 8 × 2 = 0 + 0.398 681 251 813 785 6;
  • 44) 0.398 681 251 813 785 6 × 2 = 0 + 0.797 362 503 627 571 2;
  • 45) 0.797 362 503 627 571 2 × 2 = 1 + 0.594 725 007 255 142 4;
  • 46) 0.594 725 007 255 142 4 × 2 = 1 + 0.189 450 014 510 284 8;
  • 47) 0.189 450 014 510 284 8 × 2 = 0 + 0.378 900 029 020 569 6;
  • 48) 0.378 900 029 020 569 6 × 2 = 0 + 0.757 800 058 041 139 2;
  • 49) 0.757 800 058 041 139 2 × 2 = 1 + 0.515 600 116 082 278 4;
  • 50) 0.515 600 116 082 278 4 × 2 = 1 + 0.031 200 232 164 556 8;
  • 51) 0.031 200 232 164 556 8 × 2 = 0 + 0.062 400 464 329 113 6;
  • 52) 0.062 400 464 329 113 6 × 2 = 0 + 0.124 800 928 658 227 2;
  • 53) 0.124 800 928 658 227 2 × 2 = 0 + 0.249 601 857 316 454 4;
  • 54) 0.249 601 857 316 454 4 × 2 = 0 + 0.499 203 714 632 908 8;
  • 55) 0.499 203 714 632 908 8 × 2 = 0 + 0.998 407 429 265 817 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:


0.142 857 142 857 155 7(10) =

0.0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1100 1100 000(2)

5. Positive number before normalization:

0.142 857 142 857 155 7(10) =

0.0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1100 1100 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:


0.142 857 142 857 155 7(10) =

0.0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1100 1100 000(2) =

0.0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1100 1100 000(2) × 20 =

1.0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0110 0110 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 0110 0110 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 0110 0110 0000 =

0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0110 0110 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 0110 0110 0000


Decimal number 0.142 857 142 857 155 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 0110 0110 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