0.142 857 142 857 165 6 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.142 857 142 857 165 6(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
0.142 857 142 857 165 6(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: 0.142 857 142 857 165 6.

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) 0.142 857 142 857 165 6 × 2 = 0 + 0.285 714 285 714 331 2;
  • 2) 0.285 714 285 714 331 2 × 2 = 0 + 0.571 428 571 428 662 4;
  • 3) 0.571 428 571 428 662 4 × 2 = 1 + 0.142 857 142 857 324 8;
  • 4) 0.142 857 142 857 324 8 × 2 = 0 + 0.285 714 285 714 649 6;
  • 5) 0.285 714 285 714 649 6 × 2 = 0 + 0.571 428 571 429 299 2;
  • 6) 0.571 428 571 429 299 2 × 2 = 1 + 0.142 857 142 858 598 4;
  • 7) 0.142 857 142 858 598 4 × 2 = 0 + 0.285 714 285 717 196 8;
  • 8) 0.285 714 285 717 196 8 × 2 = 0 + 0.571 428 571 434 393 6;
  • 9) 0.571 428 571 434 393 6 × 2 = 1 + 0.142 857 142 868 787 2;
  • 10) 0.142 857 142 868 787 2 × 2 = 0 + 0.285 714 285 737 574 4;
  • 11) 0.285 714 285 737 574 4 × 2 = 0 + 0.571 428 571 475 148 8;
  • 12) 0.571 428 571 475 148 8 × 2 = 1 + 0.142 857 142 950 297 6;
  • 13) 0.142 857 142 950 297 6 × 2 = 0 + 0.285 714 285 900 595 2;
  • 14) 0.285 714 285 900 595 2 × 2 = 0 + 0.571 428 571 801 190 4;
  • 15) 0.571 428 571 801 190 4 × 2 = 1 + 0.142 857 143 602 380 8;
  • 16) 0.142 857 143 602 380 8 × 2 = 0 + 0.285 714 287 204 761 6;
  • 17) 0.285 714 287 204 761 6 × 2 = 0 + 0.571 428 574 409 523 2;
  • 18) 0.571 428 574 409 523 2 × 2 = 1 + 0.142 857 148 819 046 4;
  • 19) 0.142 857 148 819 046 4 × 2 = 0 + 0.285 714 297 638 092 8;
  • 20) 0.285 714 297 638 092 8 × 2 = 0 + 0.571 428 595 276 185 6;
  • 21) 0.571 428 595 276 185 6 × 2 = 1 + 0.142 857 190 552 371 2;
  • 22) 0.142 857 190 552 371 2 × 2 = 0 + 0.285 714 381 104 742 4;
  • 23) 0.285 714 381 104 742 4 × 2 = 0 + 0.571 428 762 209 484 8;
  • 24) 0.571 428 762 209 484 8 × 2 = 1 + 0.142 857 524 418 969 6;
  • 25) 0.142 857 524 418 969 6 × 2 = 0 + 0.285 715 048 837 939 2;
  • 26) 0.285 715 048 837 939 2 × 2 = 0 + 0.571 430 097 675 878 4;
  • 27) 0.571 430 097 675 878 4 × 2 = 1 + 0.142 860 195 351 756 8;
  • 28) 0.142 860 195 351 756 8 × 2 = 0 + 0.285 720 390 703 513 6;
  • 29) 0.285 720 390 703 513 6 × 2 = 0 + 0.571 440 781 407 027 2;
  • 30) 0.571 440 781 407 027 2 × 2 = 1 + 0.142 881 562 814 054 4;
  • 31) 0.142 881 562 814 054 4 × 2 = 0 + 0.285 763 125 628 108 8;
  • 32) 0.285 763 125 628 108 8 × 2 = 0 + 0.571 526 251 256 217 6;
  • 33) 0.571 526 251 256 217 6 × 2 = 1 + 0.143 052 502 512 435 2;
  • 34) 0.143 052 502 512 435 2 × 2 = 0 + 0.286 105 005 024 870 4;
  • 35) 0.286 105 005 024 870 4 × 2 = 0 + 0.572 210 010 049 740 8;
  • 36) 0.572 210 010 049 740 8 × 2 = 1 + 0.144 420 020 099 481 6;
  • 37) 0.144 420 020 099 481 6 × 2 = 0 + 0.288 840 040 198 963 2;
  • 38) 0.288 840 040 198 963 2 × 2 = 0 + 0.577 680 080 397 926 4;
  • 39) 0.577 680 080 397 926 4 × 2 = 1 + 0.155 360 160 795 852 8;
  • 40) 0.155 360 160 795 852 8 × 2 = 0 + 0.310 720 321 591 705 6;
  • 41) 0.310 720 321 591 705 6 × 2 = 0 + 0.621 440 643 183 411 2;
  • 42) 0.621 440 643 183 411 2 × 2 = 1 + 0.242 881 286 366 822 4;
  • 43) 0.242 881 286 366 822 4 × 2 = 0 + 0.485 762 572 733 644 8;
  • 44) 0.485 762 572 733 644 8 × 2 = 0 + 0.971 525 145 467 289 6;
  • 45) 0.971 525 145 467 289 6 × 2 = 1 + 0.943 050 290 934 579 2;
  • 46) 0.943 050 290 934 579 2 × 2 = 1 + 0.886 100 581 869 158 4;
  • 47) 0.886 100 581 869 158 4 × 2 = 1 + 0.772 201 163 738 316 8;
  • 48) 0.772 201 163 738 316 8 × 2 = 1 + 0.544 402 327 476 633 6;
  • 49) 0.544 402 327 476 633 6 × 2 = 1 + 0.088 804 654 953 267 2;
  • 50) 0.088 804 654 953 267 2 × 2 = 0 + 0.177 609 309 906 534 4;
  • 51) 0.177 609 309 906 534 4 × 2 = 0 + 0.355 218 619 813 068 8;
  • 52) 0.355 218 619 813 068 8 × 2 = 0 + 0.710 437 239 626 137 6;
  • 53) 0.710 437 239 626 137 6 × 2 = 1 + 0.420 874 479 252 275 2;
  • 54) 0.420 874 479 252 275 2 × 2 = 0 + 0.841 748 958 504 550 4;
  • 55) 0.841 748 958 504 550 4 × 2 = 1 + 0.683 497 917 009 100 8;

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:


0.142 857 142 857 165 6(10) =

0.0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1111 1000 101(2)

5. Positive number before normalization:

0.142 857 142 857 165 6(10) =

0.0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1111 1000 101(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:


0.142 857 142 857 165 6(10) =

0.0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1111 1000 101(2) =

0.0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1111 1000 101(2) × 20 =

1.0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0111 1100 0101(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 0111 1100 0101


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 0111 1100 0101 =

0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0111 1100 0101


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 0111 1100 0101


Decimal number 0.142 857 142 857 165 6 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 0111 1100 0101

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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