2 000.142 857 142 857 142 857 142 857 21 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 2 000.142 857 142 857 142 857 142 857 21(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
2 000.142 857 142 857 142 857 142 857 21(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: 0.142 857 142 857 142 857 142 857 21.

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) 0.142 857 142 857 142 857 142 857 21 × 2 = 0 + 0.285 714 285 714 285 714 285 714 42;
  • 2) 0.285 714 285 714 285 714 285 714 42 × 2 = 0 + 0.571 428 571 428 571 428 571 428 84;
  • 3) 0.571 428 571 428 571 428 571 428 84 × 2 = 1 + 0.142 857 142 857 142 857 142 857 68;
  • 4) 0.142 857 142 857 142 857 142 857 68 × 2 = 0 + 0.285 714 285 714 285 714 285 715 36;
  • 5) 0.285 714 285 714 285 714 285 715 36 × 2 = 0 + 0.571 428 571 428 571 428 571 430 72;
  • 6) 0.571 428 571 428 571 428 571 430 72 × 2 = 1 + 0.142 857 142 857 142 857 142 861 44;
  • 7) 0.142 857 142 857 142 857 142 861 44 × 2 = 0 + 0.285 714 285 714 285 714 285 722 88;
  • 8) 0.285 714 285 714 285 714 285 722 88 × 2 = 0 + 0.571 428 571 428 571 428 571 445 76;
  • 9) 0.571 428 571 428 571 428 571 445 76 × 2 = 1 + 0.142 857 142 857 142 857 142 891 52;
  • 10) 0.142 857 142 857 142 857 142 891 52 × 2 = 0 + 0.285 714 285 714 285 714 285 783 04;
  • 11) 0.285 714 285 714 285 714 285 783 04 × 2 = 0 + 0.571 428 571 428 571 428 571 566 08;
  • 12) 0.571 428 571 428 571 428 571 566 08 × 2 = 1 + 0.142 857 142 857 142 857 143 132 16;
  • 13) 0.142 857 142 857 142 857 143 132 16 × 2 = 0 + 0.285 714 285 714 285 714 286 264 32;
  • 14) 0.285 714 285 714 285 714 286 264 32 × 2 = 0 + 0.571 428 571 428 571 428 572 528 64;
  • 15) 0.571 428 571 428 571 428 572 528 64 × 2 = 1 + 0.142 857 142 857 142 857 145 057 28;
  • 16) 0.142 857 142 857 142 857 145 057 28 × 2 = 0 + 0.285 714 285 714 285 714 290 114 56;
  • 17) 0.285 714 285 714 285 714 290 114 56 × 2 = 0 + 0.571 428 571 428 571 428 580 229 12;
  • 18) 0.571 428 571 428 571 428 580 229 12 × 2 = 1 + 0.142 857 142 857 142 857 160 458 24;
  • 19) 0.142 857 142 857 142 857 160 458 24 × 2 = 0 + 0.285 714 285 714 285 714 320 916 48;
  • 20) 0.285 714 285 714 285 714 320 916 48 × 2 = 0 + 0.571 428 571 428 571 428 641 832 96;
  • 21) 0.571 428 571 428 571 428 641 832 96 × 2 = 1 + 0.142 857 142 857 142 857 283 665 92;
  • 22) 0.142 857 142 857 142 857 283 665 92 × 2 = 0 + 0.285 714 285 714 285 714 567 331 84;
  • 23) 0.285 714 285 714 285 714 567 331 84 × 2 = 0 + 0.571 428 571 428 571 429 134 663 68;
  • 24) 0.571 428 571 428 571 429 134 663 68 × 2 = 1 + 0.142 857 142 857 142 858 269 327 36;
  • 25) 0.142 857 142 857 142 858 269 327 36 × 2 = 0 + 0.285 714 285 714 285 716 538 654 72;
  • 26) 0.285 714 285 714 285 716 538 654 72 × 2 = 0 + 0.571 428 571 428 571 433 077 309 44;
  • 27) 0.571 428 571 428 571 433 077 309 44 × 2 = 1 + 0.142 857 142 857 142 866 154 618 88;
  • 28) 0.142 857 142 857 142 866 154 618 88 × 2 = 0 + 0.285 714 285 714 285 732 309 237 76;
  • 29) 0.285 714 285 714 285 732 309 237 76 × 2 = 0 + 0.571 428 571 428 571 464 618 475 52;
  • 30) 0.571 428 571 428 571 464 618 475 52 × 2 = 1 + 0.142 857 142 857 142 929 236 951 04;
  • 31) 0.142 857 142 857 142 929 236 951 04 × 2 = 0 + 0.285 714 285 714 285 858 473 902 08;
  • 32) 0.285 714 285 714 285 858 473 902 08 × 2 = 0 + 0.571 428 571 428 571 716 947 804 16;
  • 33) 0.571 428 571 428 571 716 947 804 16 × 2 = 1 + 0.142 857 142 857 143 433 895 608 32;
  • 34) 0.142 857 142 857 143 433 895 608 32 × 2 = 0 + 0.285 714 285 714 286 867 791 216 64;
  • 35) 0.285 714 285 714 286 867 791 216 64 × 2 = 0 + 0.571 428 571 428 573 735 582 433 28;
  • 36) 0.571 428 571 428 573 735 582 433 28 × 2 = 1 + 0.142 857 142 857 147 471 164 866 56;
  • 37) 0.142 857 142 857 147 471 164 866 56 × 2 = 0 + 0.285 714 285 714 294 942 329 733 12;
  • 38) 0.285 714 285 714 294 942 329 733 12 × 2 = 0 + 0.571 428 571 428 589 884 659 466 24;
  • 39) 0.571 428 571 428 589 884 659 466 24 × 2 = 1 + 0.142 857 142 857 179 769 318 932 48;
  • 40) 0.142 857 142 857 179 769 318 932 48 × 2 = 0 + 0.285 714 285 714 359 538 637 864 96;
  • 41) 0.285 714 285 714 359 538 637 864 96 × 2 = 0 + 0.571 428 571 428 719 077 275 729 92;
  • 42) 0.571 428 571 428 719 077 275 729 92 × 2 = 1 + 0.142 857 142 857 438 154 551 459 84;
  • 43) 0.142 857 142 857 438 154 551 459 84 × 2 = 0 + 0.285 714 285 714 876 309 102 919 68;
  • 44) 0.285 714 285 714 876 309 102 919 68 × 2 = 0 + 0.571 428 571 429 752 618 205 839 36;
  • 45) 0.571 428 571 429 752 618 205 839 36 × 2 = 1 + 0.142 857 142 859 505 236 411 678 72;
  • 46) 0.142 857 142 859 505 236 411 678 72 × 2 = 0 + 0.285 714 285 719 010 472 823 357 44;
  • 47) 0.285 714 285 719 010 472 823 357 44 × 2 = 0 + 0.571 428 571 438 020 945 646 714 88;
  • 48) 0.571 428 571 438 020 945 646 714 88 × 2 = 1 + 0.142 857 142 876 041 891 293 429 76;
  • 49) 0.142 857 142 876 041 891 293 429 76 × 2 = 0 + 0.285 714 285 752 083 782 586 859 52;
  • 50) 0.285 714 285 752 083 782 586 859 52 × 2 = 0 + 0.571 428 571 504 167 565 173 719 04;
  • 51) 0.571 428 571 504 167 565 173 719 04 × 2 = 1 + 0.142 857 143 008 335 130 347 438 08;
  • 52) 0.142 857 143 008 335 130 347 438 08 × 2 = 0 + 0.285 714 286 016 670 260 694 876 16;
  • 53) 0.285 714 286 016 670 260 694 876 16 × 2 = 0 + 0.571 428 572 033 340 521 389 752 32;

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:


0.142 857 142 857 142 857 142 857 21(10) =

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

5. Positive number before normalization:

2 000.142 857 142 857 142 857 142 857 21(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:


2 000.142 857 142 857 142 857 142 857 21(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 2 000.142 857 142 857 142 857 142 857 21 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