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

Convert decimal 0.142 857 142 857 142 857 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
0.142 857 142 857 142 857 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: 0.142 857 142 857 142 857 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) 0.142 857 142 857 142 857 7 × 2 = 0 + 0.285 714 285 714 285 715 4;
  • 2) 0.285 714 285 714 285 715 4 × 2 = 0 + 0.571 428 571 428 571 430 8;
  • 3) 0.571 428 571 428 571 430 8 × 2 = 1 + 0.142 857 142 857 142 861 6;
  • 4) 0.142 857 142 857 142 861 6 × 2 = 0 + 0.285 714 285 714 285 723 2;
  • 5) 0.285 714 285 714 285 723 2 × 2 = 0 + 0.571 428 571 428 571 446 4;
  • 6) 0.571 428 571 428 571 446 4 × 2 = 1 + 0.142 857 142 857 142 892 8;
  • 7) 0.142 857 142 857 142 892 8 × 2 = 0 + 0.285 714 285 714 285 785 6;
  • 8) 0.285 714 285 714 285 785 6 × 2 = 0 + 0.571 428 571 428 571 571 2;
  • 9) 0.571 428 571 428 571 571 2 × 2 = 1 + 0.142 857 142 857 143 142 4;
  • 10) 0.142 857 142 857 143 142 4 × 2 = 0 + 0.285 714 285 714 286 284 8;
  • 11) 0.285 714 285 714 286 284 8 × 2 = 0 + 0.571 428 571 428 572 569 6;
  • 12) 0.571 428 571 428 572 569 6 × 2 = 1 + 0.142 857 142 857 145 139 2;
  • 13) 0.142 857 142 857 145 139 2 × 2 = 0 + 0.285 714 285 714 290 278 4;
  • 14) 0.285 714 285 714 290 278 4 × 2 = 0 + 0.571 428 571 428 580 556 8;
  • 15) 0.571 428 571 428 580 556 8 × 2 = 1 + 0.142 857 142 857 161 113 6;
  • 16) 0.142 857 142 857 161 113 6 × 2 = 0 + 0.285 714 285 714 322 227 2;
  • 17) 0.285 714 285 714 322 227 2 × 2 = 0 + 0.571 428 571 428 644 454 4;
  • 18) 0.571 428 571 428 644 454 4 × 2 = 1 + 0.142 857 142 857 288 908 8;
  • 19) 0.142 857 142 857 288 908 8 × 2 = 0 + 0.285 714 285 714 577 817 6;
  • 20) 0.285 714 285 714 577 817 6 × 2 = 0 + 0.571 428 571 429 155 635 2;
  • 21) 0.571 428 571 429 155 635 2 × 2 = 1 + 0.142 857 142 858 311 270 4;
  • 22) 0.142 857 142 858 311 270 4 × 2 = 0 + 0.285 714 285 716 622 540 8;
  • 23) 0.285 714 285 716 622 540 8 × 2 = 0 + 0.571 428 571 433 245 081 6;
  • 24) 0.571 428 571 433 245 081 6 × 2 = 1 + 0.142 857 142 866 490 163 2;
  • 25) 0.142 857 142 866 490 163 2 × 2 = 0 + 0.285 714 285 732 980 326 4;
  • 26) 0.285 714 285 732 980 326 4 × 2 = 0 + 0.571 428 571 465 960 652 8;
  • 27) 0.571 428 571 465 960 652 8 × 2 = 1 + 0.142 857 142 931 921 305 6;
  • 28) 0.142 857 142 931 921 305 6 × 2 = 0 + 0.285 714 285 863 842 611 2;
  • 29) 0.285 714 285 863 842 611 2 × 2 = 0 + 0.571 428 571 727 685 222 4;
  • 30) 0.571 428 571 727 685 222 4 × 2 = 1 + 0.142 857 143 455 370 444 8;
  • 31) 0.142 857 143 455 370 444 8 × 2 = 0 + 0.285 714 286 910 740 889 6;
  • 32) 0.285 714 286 910 740 889 6 × 2 = 0 + 0.571 428 573 821 481 779 2;
  • 33) 0.571 428 573 821 481 779 2 × 2 = 1 + 0.142 857 147 642 963 558 4;
  • 34) 0.142 857 147 642 963 558 4 × 2 = 0 + 0.285 714 295 285 927 116 8;
  • 35) 0.285 714 295 285 927 116 8 × 2 = 0 + 0.571 428 590 571 854 233 6;
  • 36) 0.571 428 590 571 854 233 6 × 2 = 1 + 0.142 857 181 143 708 467 2;
  • 37) 0.142 857 181 143 708 467 2 × 2 = 0 + 0.285 714 362 287 416 934 4;
  • 38) 0.285 714 362 287 416 934 4 × 2 = 0 + 0.571 428 724 574 833 868 8;
  • 39) 0.571 428 724 574 833 868 8 × 2 = 1 + 0.142 857 449 149 667 737 6;
  • 40) 0.142 857 449 149 667 737 6 × 2 = 0 + 0.285 714 898 299 335 475 2;
  • 41) 0.285 714 898 299 335 475 2 × 2 = 0 + 0.571 429 796 598 670 950 4;
  • 42) 0.571 429 796 598 670 950 4 × 2 = 1 + 0.142 859 593 197 341 900 8;
  • 43) 0.142 859 593 197 341 900 8 × 2 = 0 + 0.285 719 186 394 683 801 6;
  • 44) 0.285 719 186 394 683 801 6 × 2 = 0 + 0.571 438 372 789 367 603 2;
  • 45) 0.571 438 372 789 367 603 2 × 2 = 1 + 0.142 876 745 578 735 206 4;
  • 46) 0.142 876 745 578 735 206 4 × 2 = 0 + 0.285 753 491 157 470 412 8;
  • 47) 0.285 753 491 157 470 412 8 × 2 = 0 + 0.571 506 982 314 940 825 6;
  • 48) 0.571 506 982 314 940 825 6 × 2 = 1 + 0.143 013 964 629 881 651 2;
  • 49) 0.143 013 964 629 881 651 2 × 2 = 0 + 0.286 027 929 259 763 302 4;
  • 50) 0.286 027 929 259 763 302 4 × 2 = 0 + 0.572 055 858 519 526 604 8;
  • 51) 0.572 055 858 519 526 604 8 × 2 = 1 + 0.144 111 717 039 053 209 6;
  • 52) 0.144 111 717 039 053 209 6 × 2 = 0 + 0.288 223 434 078 106 419 2;
  • 53) 0.288 223 434 078 106 419 2 × 2 = 0 + 0.576 446 868 156 212 838 4;
  • 54) 0.576 446 868 156 212 838 4 × 2 = 1 + 0.152 893 736 312 425 676 8;
  • 55) 0.152 893 736 312 425 676 8 × 2 = 0 + 0.305 787 472 624 851 353 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:


0.142 857 142 857 142 857 7(10) =

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

5. Positive number before normalization:

0.142 857 142 857 142 857 7(10) =

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


0.142 857 142 857 142 857 7(10) =

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

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

1.0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 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 0100 1001 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 0100 1001 0010 =

0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 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 0100 1001 0010


Decimal number 0.142 857 142 857 142 857 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 1001 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