2 000.142 857 142 857 142 857 142 856 47 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 2 000.142 857 142 857 142 857 142 856 47(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
2 000.142 857 142 857 142 857 142 856 47(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: 2 000.
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;
  • 2 000 ÷ 2 = 1 000 + 0;
  • 1 000 ÷ 2 = 500 + 0;
  • 500 ÷ 2 = 250 + 0;
  • 250 ÷ 2 = 125 + 0;
  • 125 ÷ 2 = 62 + 1;
  • 62 ÷ 2 = 31 + 0;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

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.

2 000(10) =

111 1101 0000(2)


3. Convert to binary (base 2) the fractional part: 0.142 857 142 857 142 857 142 856 47.

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) 0.142 857 142 857 142 857 142 856 47 × 2 = 0 + 0.285 714 285 714 285 714 285 712 94;
  • 2) 0.285 714 285 714 285 714 285 712 94 × 2 = 0 + 0.571 428 571 428 571 428 571 425 88;
  • 3) 0.571 428 571 428 571 428 571 425 88 × 2 = 1 + 0.142 857 142 857 142 857 142 851 76;
  • 4) 0.142 857 142 857 142 857 142 851 76 × 2 = 0 + 0.285 714 285 714 285 714 285 703 52;
  • 5) 0.285 714 285 714 285 714 285 703 52 × 2 = 0 + 0.571 428 571 428 571 428 571 407 04;
  • 6) 0.571 428 571 428 571 428 571 407 04 × 2 = 1 + 0.142 857 142 857 142 857 142 814 08;
  • 7) 0.142 857 142 857 142 857 142 814 08 × 2 = 0 + 0.285 714 285 714 285 714 285 628 16;
  • 8) 0.285 714 285 714 285 714 285 628 16 × 2 = 0 + 0.571 428 571 428 571 428 571 256 32;
  • 9) 0.571 428 571 428 571 428 571 256 32 × 2 = 1 + 0.142 857 142 857 142 857 142 512 64;
  • 10) 0.142 857 142 857 142 857 142 512 64 × 2 = 0 + 0.285 714 285 714 285 714 285 025 28;
  • 11) 0.285 714 285 714 285 714 285 025 28 × 2 = 0 + 0.571 428 571 428 571 428 570 050 56;
  • 12) 0.571 428 571 428 571 428 570 050 56 × 2 = 1 + 0.142 857 142 857 142 857 140 101 12;
  • 13) 0.142 857 142 857 142 857 140 101 12 × 2 = 0 + 0.285 714 285 714 285 714 280 202 24;
  • 14) 0.285 714 285 714 285 714 280 202 24 × 2 = 0 + 0.571 428 571 428 571 428 560 404 48;
  • 15) 0.571 428 571 428 571 428 560 404 48 × 2 = 1 + 0.142 857 142 857 142 857 120 808 96;
  • 16) 0.142 857 142 857 142 857 120 808 96 × 2 = 0 + 0.285 714 285 714 285 714 241 617 92;
  • 17) 0.285 714 285 714 285 714 241 617 92 × 2 = 0 + 0.571 428 571 428 571 428 483 235 84;
  • 18) 0.571 428 571 428 571 428 483 235 84 × 2 = 1 + 0.142 857 142 857 142 856 966 471 68;
  • 19) 0.142 857 142 857 142 856 966 471 68 × 2 = 0 + 0.285 714 285 714 285 713 932 943 36;
  • 20) 0.285 714 285 714 285 713 932 943 36 × 2 = 0 + 0.571 428 571 428 571 427 865 886 72;
  • 21) 0.571 428 571 428 571 427 865 886 72 × 2 = 1 + 0.142 857 142 857 142 855 731 773 44;
  • 22) 0.142 857 142 857 142 855 731 773 44 × 2 = 0 + 0.285 714 285 714 285 711 463 546 88;
  • 23) 0.285 714 285 714 285 711 463 546 88 × 2 = 0 + 0.571 428 571 428 571 422 927 093 76;
  • 24) 0.571 428 571 428 571 422 927 093 76 × 2 = 1 + 0.142 857 142 857 142 845 854 187 52;
  • 25) 0.142 857 142 857 142 845 854 187 52 × 2 = 0 + 0.285 714 285 714 285 691 708 375 04;
  • 26) 0.285 714 285 714 285 691 708 375 04 × 2 = 0 + 0.571 428 571 428 571 383 416 750 08;
  • 27) 0.571 428 571 428 571 383 416 750 08 × 2 = 1 + 0.142 857 142 857 142 766 833 500 16;
  • 28) 0.142 857 142 857 142 766 833 500 16 × 2 = 0 + 0.285 714 285 714 285 533 667 000 32;
  • 29) 0.285 714 285 714 285 533 667 000 32 × 2 = 0 + 0.571 428 571 428 571 067 334 000 64;
  • 30) 0.571 428 571 428 571 067 334 000 64 × 2 = 1 + 0.142 857 142 857 142 134 668 001 28;
  • 31) 0.142 857 142 857 142 134 668 001 28 × 2 = 0 + 0.285 714 285 714 284 269 336 002 56;
  • 32) 0.285 714 285 714 284 269 336 002 56 × 2 = 0 + 0.571 428 571 428 568 538 672 005 12;
  • 33) 0.571 428 571 428 568 538 672 005 12 × 2 = 1 + 0.142 857 142 857 137 077 344 010 24;
  • 34) 0.142 857 142 857 137 077 344 010 24 × 2 = 0 + 0.285 714 285 714 274 154 688 020 48;
  • 35) 0.285 714 285 714 274 154 688 020 48 × 2 = 0 + 0.571 428 571 428 548 309 376 040 96;
  • 36) 0.571 428 571 428 548 309 376 040 96 × 2 = 1 + 0.142 857 142 857 096 618 752 081 92;
  • 37) 0.142 857 142 857 096 618 752 081 92 × 2 = 0 + 0.285 714 285 714 193 237 504 163 84;
  • 38) 0.285 714 285 714 193 237 504 163 84 × 2 = 0 + 0.571 428 571 428 386 475 008 327 68;
  • 39) 0.571 428 571 428 386 475 008 327 68 × 2 = 1 + 0.142 857 142 856 772 950 016 655 36;
  • 40) 0.142 857 142 856 772 950 016 655 36 × 2 = 0 + 0.285 714 285 713 545 900 033 310 72;
  • 41) 0.285 714 285 713 545 900 033 310 72 × 2 = 0 + 0.571 428 571 427 091 800 066 621 44;
  • 42) 0.571 428 571 427 091 800 066 621 44 × 2 = 1 + 0.142 857 142 854 183 600 133 242 88;
  • 43) 0.142 857 142 854 183 600 133 242 88 × 2 = 0 + 0.285 714 285 708 367 200 266 485 76;
  • 44) 0.285 714 285 708 367 200 266 485 76 × 2 = 0 + 0.571 428 571 416 734 400 532 971 52;
  • 45) 0.571 428 571 416 734 400 532 971 52 × 2 = 1 + 0.142 857 142 833 468 801 065 943 04;
  • 46) 0.142 857 142 833 468 801 065 943 04 × 2 = 0 + 0.285 714 285 666 937 602 131 886 08;
  • 47) 0.285 714 285 666 937 602 131 886 08 × 2 = 0 + 0.571 428 571 333 875 204 263 772 16;
  • 48) 0.571 428 571 333 875 204 263 772 16 × 2 = 1 + 0.142 857 142 667 750 408 527 544 32;
  • 49) 0.142 857 142 667 750 408 527 544 32 × 2 = 0 + 0.285 714 285 335 500 817 055 088 64;
  • 50) 0.285 714 285 335 500 817 055 088 64 × 2 = 0 + 0.571 428 570 671 001 634 110 177 28;
  • 51) 0.571 428 570 671 001 634 110 177 28 × 2 = 1 + 0.142 857 141 342 003 268 220 354 56;
  • 52) 0.142 857 141 342 003 268 220 354 56 × 2 = 0 + 0.285 714 282 684 006 536 440 709 12;
  • 53) 0.285 714 282 684 006 536 440 709 12 × 2 = 0 + 0.571 428 565 368 013 072 881 418 24;

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:


0.142 857 142 857 142 857 142 856 47(10) =

0.0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0(2)

5. Positive number before normalization:

2 000.142 857 142 857 142 857 142 856 47(10) =

111 1101 0000.0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 10 positions to the left, so that only one non zero digit remains to the left of it:


2 000.142 857 142 857 142 857 142 856 47(10) =

111 1101 0000.0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0(2) =

111 1101 0000.0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0(2) × 20 =

1.1111 0100 0000 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 100(2) × 210


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): 10

Mantissa (not normalized):
1.1111 0100 0000 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 100


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

10 + 2(11-1) - 1 =

(10 + 1 023)(10) =

1 033(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;
  • 1 033 ÷ 2 = 516 + 1;
  • 516 ÷ 2 = 258 + 0;
  • 258 ÷ 2 = 129 + 0;
  • 129 ÷ 2 = 64 + 1;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 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) =

1033(10) =

100 0000 1001(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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =

1. 1111 0100 0000 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 001 0010 0100 =

1111 0100 0000 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001


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) =
100 0000 1001

Mantissa (52 bits) =
1111 0100 0000 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001


Decimal number 2 000.142 857 142 857 142 857 142 856 47 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 1001 - 1111 0100 0000 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001

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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