2 000.142 857 142 857 143 053 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 2 000.142 857 142 857 143 053(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
2 000.142 857 142 857 143 053(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: 0.142 857 142 857 143 053.

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) 0.142 857 142 857 143 053 × 2 = 0 + 0.285 714 285 714 286 106;
  • 2) 0.285 714 285 714 286 106 × 2 = 0 + 0.571 428 571 428 572 212;
  • 3) 0.571 428 571 428 572 212 × 2 = 1 + 0.142 857 142 857 144 424;
  • 4) 0.142 857 142 857 144 424 × 2 = 0 + 0.285 714 285 714 288 848;
  • 5) 0.285 714 285 714 288 848 × 2 = 0 + 0.571 428 571 428 577 696;
  • 6) 0.571 428 571 428 577 696 × 2 = 1 + 0.142 857 142 857 155 392;
  • 7) 0.142 857 142 857 155 392 × 2 = 0 + 0.285 714 285 714 310 784;
  • 8) 0.285 714 285 714 310 784 × 2 = 0 + 0.571 428 571 428 621 568;
  • 9) 0.571 428 571 428 621 568 × 2 = 1 + 0.142 857 142 857 243 136;
  • 10) 0.142 857 142 857 243 136 × 2 = 0 + 0.285 714 285 714 486 272;
  • 11) 0.285 714 285 714 486 272 × 2 = 0 + 0.571 428 571 428 972 544;
  • 12) 0.571 428 571 428 972 544 × 2 = 1 + 0.142 857 142 857 945 088;
  • 13) 0.142 857 142 857 945 088 × 2 = 0 + 0.285 714 285 715 890 176;
  • 14) 0.285 714 285 715 890 176 × 2 = 0 + 0.571 428 571 431 780 352;
  • 15) 0.571 428 571 431 780 352 × 2 = 1 + 0.142 857 142 863 560 704;
  • 16) 0.142 857 142 863 560 704 × 2 = 0 + 0.285 714 285 727 121 408;
  • 17) 0.285 714 285 727 121 408 × 2 = 0 + 0.571 428 571 454 242 816;
  • 18) 0.571 428 571 454 242 816 × 2 = 1 + 0.142 857 142 908 485 632;
  • 19) 0.142 857 142 908 485 632 × 2 = 0 + 0.285 714 285 816 971 264;
  • 20) 0.285 714 285 816 971 264 × 2 = 0 + 0.571 428 571 633 942 528;
  • 21) 0.571 428 571 633 942 528 × 2 = 1 + 0.142 857 143 267 885 056;
  • 22) 0.142 857 143 267 885 056 × 2 = 0 + 0.285 714 286 535 770 112;
  • 23) 0.285 714 286 535 770 112 × 2 = 0 + 0.571 428 573 071 540 224;
  • 24) 0.571 428 573 071 540 224 × 2 = 1 + 0.142 857 146 143 080 448;
  • 25) 0.142 857 146 143 080 448 × 2 = 0 + 0.285 714 292 286 160 896;
  • 26) 0.285 714 292 286 160 896 × 2 = 0 + 0.571 428 584 572 321 792;
  • 27) 0.571 428 584 572 321 792 × 2 = 1 + 0.142 857 169 144 643 584;
  • 28) 0.142 857 169 144 643 584 × 2 = 0 + 0.285 714 338 289 287 168;
  • 29) 0.285 714 338 289 287 168 × 2 = 0 + 0.571 428 676 578 574 336;
  • 30) 0.571 428 676 578 574 336 × 2 = 1 + 0.142 857 353 157 148 672;
  • 31) 0.142 857 353 157 148 672 × 2 = 0 + 0.285 714 706 314 297 344;
  • 32) 0.285 714 706 314 297 344 × 2 = 0 + 0.571 429 412 628 594 688;
  • 33) 0.571 429 412 628 594 688 × 2 = 1 + 0.142 858 825 257 189 376;
  • 34) 0.142 858 825 257 189 376 × 2 = 0 + 0.285 717 650 514 378 752;
  • 35) 0.285 717 650 514 378 752 × 2 = 0 + 0.571 435 301 028 757 504;
  • 36) 0.571 435 301 028 757 504 × 2 = 1 + 0.142 870 602 057 515 008;
  • 37) 0.142 870 602 057 515 008 × 2 = 0 + 0.285 741 204 115 030 016;
  • 38) 0.285 741 204 115 030 016 × 2 = 0 + 0.571 482 408 230 060 032;
  • 39) 0.571 482 408 230 060 032 × 2 = 1 + 0.142 964 816 460 120 064;
  • 40) 0.142 964 816 460 120 064 × 2 = 0 + 0.285 929 632 920 240 128;
  • 41) 0.285 929 632 920 240 128 × 2 = 0 + 0.571 859 265 840 480 256;
  • 42) 0.571 859 265 840 480 256 × 2 = 1 + 0.143 718 531 680 960 512;
  • 43) 0.143 718 531 680 960 512 × 2 = 0 + 0.287 437 063 361 921 024;
  • 44) 0.287 437 063 361 921 024 × 2 = 0 + 0.574 874 126 723 842 048;
  • 45) 0.574 874 126 723 842 048 × 2 = 1 + 0.149 748 253 447 684 096;
  • 46) 0.149 748 253 447 684 096 × 2 = 0 + 0.299 496 506 895 368 192;
  • 47) 0.299 496 506 895 368 192 × 2 = 0 + 0.598 993 013 790 736 384;
  • 48) 0.598 993 013 790 736 384 × 2 = 1 + 0.197 986 027 581 472 768;
  • 49) 0.197 986 027 581 472 768 × 2 = 0 + 0.395 972 055 162 945 536;
  • 50) 0.395 972 055 162 945 536 × 2 = 0 + 0.791 944 110 325 891 072;
  • 51) 0.791 944 110 325 891 072 × 2 = 1 + 0.583 888 220 651 782 144;
  • 52) 0.583 888 220 651 782 144 × 2 = 1 + 0.167 776 441 303 564 288;
  • 53) 0.167 776 441 303 564 288 × 2 = 0 + 0.335 552 882 607 128 576;

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:


0.142 857 142 857 143 053(10) =

0.0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0011 0(2)

5. Positive number before normalization:

2 000.142 857 142 857 143 053(10) =

111 1101 0000.0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0011 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:


2 000.142 857 142 857 143 053(10) =

111 1101 0000.0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0011 0(2) =

111 1101 0000.0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0011 0(2) × 20 =

1.1111 0100 0000 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 110(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 110


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 0110 =

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 2 000.142 857 142 857 143 053 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