0.142 857 142 857 148 9 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.142 857 142 857 148 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
0.142 857 142 857 148 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: 0.142 857 142 857 148 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) 0.142 857 142 857 148 9 × 2 = 0 + 0.285 714 285 714 297 8;
  • 2) 0.285 714 285 714 297 8 × 2 = 0 + 0.571 428 571 428 595 6;
  • 3) 0.571 428 571 428 595 6 × 2 = 1 + 0.142 857 142 857 191 2;
  • 4) 0.142 857 142 857 191 2 × 2 = 0 + 0.285 714 285 714 382 4;
  • 5) 0.285 714 285 714 382 4 × 2 = 0 + 0.571 428 571 428 764 8;
  • 6) 0.571 428 571 428 764 8 × 2 = 1 + 0.142 857 142 857 529 6;
  • 7) 0.142 857 142 857 529 6 × 2 = 0 + 0.285 714 285 715 059 2;
  • 8) 0.285 714 285 715 059 2 × 2 = 0 + 0.571 428 571 430 118 4;
  • 9) 0.571 428 571 430 118 4 × 2 = 1 + 0.142 857 142 860 236 8;
  • 10) 0.142 857 142 860 236 8 × 2 = 0 + 0.285 714 285 720 473 6;
  • 11) 0.285 714 285 720 473 6 × 2 = 0 + 0.571 428 571 440 947 2;
  • 12) 0.571 428 571 440 947 2 × 2 = 1 + 0.142 857 142 881 894 4;
  • 13) 0.142 857 142 881 894 4 × 2 = 0 + 0.285 714 285 763 788 8;
  • 14) 0.285 714 285 763 788 8 × 2 = 0 + 0.571 428 571 527 577 6;
  • 15) 0.571 428 571 527 577 6 × 2 = 1 + 0.142 857 143 055 155 2;
  • 16) 0.142 857 143 055 155 2 × 2 = 0 + 0.285 714 286 110 310 4;
  • 17) 0.285 714 286 110 310 4 × 2 = 0 + 0.571 428 572 220 620 8;
  • 18) 0.571 428 572 220 620 8 × 2 = 1 + 0.142 857 144 441 241 6;
  • 19) 0.142 857 144 441 241 6 × 2 = 0 + 0.285 714 288 882 483 2;
  • 20) 0.285 714 288 882 483 2 × 2 = 0 + 0.571 428 577 764 966 4;
  • 21) 0.571 428 577 764 966 4 × 2 = 1 + 0.142 857 155 529 932 8;
  • 22) 0.142 857 155 529 932 8 × 2 = 0 + 0.285 714 311 059 865 6;
  • 23) 0.285 714 311 059 865 6 × 2 = 0 + 0.571 428 622 119 731 2;
  • 24) 0.571 428 622 119 731 2 × 2 = 1 + 0.142 857 244 239 462 4;
  • 25) 0.142 857 244 239 462 4 × 2 = 0 + 0.285 714 488 478 924 8;
  • 26) 0.285 714 488 478 924 8 × 2 = 0 + 0.571 428 976 957 849 6;
  • 27) 0.571 428 976 957 849 6 × 2 = 1 + 0.142 857 953 915 699 2;
  • 28) 0.142 857 953 915 699 2 × 2 = 0 + 0.285 715 907 831 398 4;
  • 29) 0.285 715 907 831 398 4 × 2 = 0 + 0.571 431 815 662 796 8;
  • 30) 0.571 431 815 662 796 8 × 2 = 1 + 0.142 863 631 325 593 6;
  • 31) 0.142 863 631 325 593 6 × 2 = 0 + 0.285 727 262 651 187 2;
  • 32) 0.285 727 262 651 187 2 × 2 = 0 + 0.571 454 525 302 374 4;
  • 33) 0.571 454 525 302 374 4 × 2 = 1 + 0.142 909 050 604 748 8;
  • 34) 0.142 909 050 604 748 8 × 2 = 0 + 0.285 818 101 209 497 6;
  • 35) 0.285 818 101 209 497 6 × 2 = 0 + 0.571 636 202 418 995 2;
  • 36) 0.571 636 202 418 995 2 × 2 = 1 + 0.143 272 404 837 990 4;
  • 37) 0.143 272 404 837 990 4 × 2 = 0 + 0.286 544 809 675 980 8;
  • 38) 0.286 544 809 675 980 8 × 2 = 0 + 0.573 089 619 351 961 6;
  • 39) 0.573 089 619 351 961 6 × 2 = 1 + 0.146 179 238 703 923 2;
  • 40) 0.146 179 238 703 923 2 × 2 = 0 + 0.292 358 477 407 846 4;
  • 41) 0.292 358 477 407 846 4 × 2 = 0 + 0.584 716 954 815 692 8;
  • 42) 0.584 716 954 815 692 8 × 2 = 1 + 0.169 433 909 631 385 6;
  • 43) 0.169 433 909 631 385 6 × 2 = 0 + 0.338 867 819 262 771 2;
  • 44) 0.338 867 819 262 771 2 × 2 = 0 + 0.677 735 638 525 542 4;
  • 45) 0.677 735 638 525 542 4 × 2 = 1 + 0.355 471 277 051 084 8;
  • 46) 0.355 471 277 051 084 8 × 2 = 0 + 0.710 942 554 102 169 6;
  • 47) 0.710 942 554 102 169 6 × 2 = 1 + 0.421 885 108 204 339 2;
  • 48) 0.421 885 108 204 339 2 × 2 = 0 + 0.843 770 216 408 678 4;
  • 49) 0.843 770 216 408 678 4 × 2 = 1 + 0.687 540 432 817 356 8;
  • 50) 0.687 540 432 817 356 8 × 2 = 1 + 0.375 080 865 634 713 6;
  • 51) 0.375 080 865 634 713 6 × 2 = 0 + 0.750 161 731 269 427 2;
  • 52) 0.750 161 731 269 427 2 × 2 = 1 + 0.500 323 462 538 854 4;
  • 53) 0.500 323 462 538 854 4 × 2 = 1 + 0.000 646 925 077 708 8;
  • 54) 0.000 646 925 077 708 8 × 2 = 0 + 0.001 293 850 155 417 6;
  • 55) 0.001 293 850 155 417 6 × 2 = 0 + 0.002 587 700 310 835 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:


0.142 857 142 857 148 9(10) =

0.0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1010 1101 100(2)

5. Positive number before normalization:

0.142 857 142 857 148 9(10) =

0.0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1010 1101 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:


0.142 857 142 857 148 9(10) =

0.0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1010 1101 100(2) =

0.0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1010 1101 100(2) × 20 =

1.0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0101 0110 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 0101 0110 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 0101 0110 1100 =

0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0101 0110 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 0101 0110 1100


Decimal number 0.142 857 142 857 148 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 0110 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