0.000 000 000 002 12 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 000 002 12(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
0.000 000 000 002 12(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: 0.000 000 000 002 12.

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) 0.000 000 000 002 12 × 2 = 0 + 0.000 000 000 004 24;
  • 2) 0.000 000 000 004 24 × 2 = 0 + 0.000 000 000 008 48;
  • 3) 0.000 000 000 008 48 × 2 = 0 + 0.000 000 000 016 96;
  • 4) 0.000 000 000 016 96 × 2 = 0 + 0.000 000 000 033 92;
  • 5) 0.000 000 000 033 92 × 2 = 0 + 0.000 000 000 067 84;
  • 6) 0.000 000 000 067 84 × 2 = 0 + 0.000 000 000 135 68;
  • 7) 0.000 000 000 135 68 × 2 = 0 + 0.000 000 000 271 36;
  • 8) 0.000 000 000 271 36 × 2 = 0 + 0.000 000 000 542 72;
  • 9) 0.000 000 000 542 72 × 2 = 0 + 0.000 000 001 085 44;
  • 10) 0.000 000 001 085 44 × 2 = 0 + 0.000 000 002 170 88;
  • 11) 0.000 000 002 170 88 × 2 = 0 + 0.000 000 004 341 76;
  • 12) 0.000 000 004 341 76 × 2 = 0 + 0.000 000 008 683 52;
  • 13) 0.000 000 008 683 52 × 2 = 0 + 0.000 000 017 367 04;
  • 14) 0.000 000 017 367 04 × 2 = 0 + 0.000 000 034 734 08;
  • 15) 0.000 000 034 734 08 × 2 = 0 + 0.000 000 069 468 16;
  • 16) 0.000 000 069 468 16 × 2 = 0 + 0.000 000 138 936 32;
  • 17) 0.000 000 138 936 32 × 2 = 0 + 0.000 000 277 872 64;
  • 18) 0.000 000 277 872 64 × 2 = 0 + 0.000 000 555 745 28;
  • 19) 0.000 000 555 745 28 × 2 = 0 + 0.000 001 111 490 56;
  • 20) 0.000 001 111 490 56 × 2 = 0 + 0.000 002 222 981 12;
  • 21) 0.000 002 222 981 12 × 2 = 0 + 0.000 004 445 962 24;
  • 22) 0.000 004 445 962 24 × 2 = 0 + 0.000 008 891 924 48;
  • 23) 0.000 008 891 924 48 × 2 = 0 + 0.000 017 783 848 96;
  • 24) 0.000 017 783 848 96 × 2 = 0 + 0.000 035 567 697 92;
  • 25) 0.000 035 567 697 92 × 2 = 0 + 0.000 071 135 395 84;
  • 26) 0.000 071 135 395 84 × 2 = 0 + 0.000 142 270 791 68;
  • 27) 0.000 142 270 791 68 × 2 = 0 + 0.000 284 541 583 36;
  • 28) 0.000 284 541 583 36 × 2 = 0 + 0.000 569 083 166 72;
  • 29) 0.000 569 083 166 72 × 2 = 0 + 0.001 138 166 333 44;
  • 30) 0.001 138 166 333 44 × 2 = 0 + 0.002 276 332 666 88;
  • 31) 0.002 276 332 666 88 × 2 = 0 + 0.004 552 665 333 76;
  • 32) 0.004 552 665 333 76 × 2 = 0 + 0.009 105 330 667 52;
  • 33) 0.009 105 330 667 52 × 2 = 0 + 0.018 210 661 335 04;
  • 34) 0.018 210 661 335 04 × 2 = 0 + 0.036 421 322 670 08;
  • 35) 0.036 421 322 670 08 × 2 = 0 + 0.072 842 645 340 16;
  • 36) 0.072 842 645 340 16 × 2 = 0 + 0.145 685 290 680 32;
  • 37) 0.145 685 290 680 32 × 2 = 0 + 0.291 370 581 360 64;
  • 38) 0.291 370 581 360 64 × 2 = 0 + 0.582 741 162 721 28;
  • 39) 0.582 741 162 721 28 × 2 = 1 + 0.165 482 325 442 56;
  • 40) 0.165 482 325 442 56 × 2 = 0 + 0.330 964 650 885 12;
  • 41) 0.330 964 650 885 12 × 2 = 0 + 0.661 929 301 770 24;
  • 42) 0.661 929 301 770 24 × 2 = 1 + 0.323 858 603 540 48;
  • 43) 0.323 858 603 540 48 × 2 = 0 + 0.647 717 207 080 96;
  • 44) 0.647 717 207 080 96 × 2 = 1 + 0.295 434 414 161 92;
  • 45) 0.295 434 414 161 92 × 2 = 0 + 0.590 868 828 323 84;
  • 46) 0.590 868 828 323 84 × 2 = 1 + 0.181 737 656 647 68;
  • 47) 0.181 737 656 647 68 × 2 = 0 + 0.363 475 313 295 36;
  • 48) 0.363 475 313 295 36 × 2 = 0 + 0.726 950 626 590 72;
  • 49) 0.726 950 626 590 72 × 2 = 1 + 0.453 901 253 181 44;
  • 50) 0.453 901 253 181 44 × 2 = 0 + 0.907 802 506 362 88;
  • 51) 0.907 802 506 362 88 × 2 = 1 + 0.815 605 012 725 76;
  • 52) 0.815 605 012 725 76 × 2 = 1 + 0.631 210 025 451 52;
  • 53) 0.631 210 025 451 52 × 2 = 1 + 0.262 420 050 903 04;
  • 54) 0.262 420 050 903 04 × 2 = 0 + 0.524 840 101 806 08;
  • 55) 0.524 840 101 806 08 × 2 = 1 + 0.049 680 203 612 16;
  • 56) 0.049 680 203 612 16 × 2 = 0 + 0.099 360 407 224 32;
  • 57) 0.099 360 407 224 32 × 2 = 0 + 0.198 720 814 448 64;
  • 58) 0.198 720 814 448 64 × 2 = 0 + 0.397 441 628 897 28;
  • 59) 0.397 441 628 897 28 × 2 = 0 + 0.794 883 257 794 56;
  • 60) 0.794 883 257 794 56 × 2 = 1 + 0.589 766 515 589 12;
  • 61) 0.589 766 515 589 12 × 2 = 1 + 0.179 533 031 178 24;
  • 62) 0.179 533 031 178 24 × 2 = 0 + 0.359 066 062 356 48;
  • 63) 0.359 066 062 356 48 × 2 = 0 + 0.718 132 124 712 96;
  • 64) 0.718 132 124 712 96 × 2 = 1 + 0.436 264 249 425 92;
  • 65) 0.436 264 249 425 92 × 2 = 0 + 0.872 528 498 851 84;
  • 66) 0.872 528 498 851 84 × 2 = 1 + 0.745 056 997 703 68;
  • 67) 0.745 056 997 703 68 × 2 = 1 + 0.490 113 995 407 36;
  • 68) 0.490 113 995 407 36 × 2 = 0 + 0.980 227 990 814 72;
  • 69) 0.980 227 990 814 72 × 2 = 1 + 0.960 455 981 629 44;
  • 70) 0.960 455 981 629 44 × 2 = 1 + 0.920 911 963 258 88;
  • 71) 0.920 911 963 258 88 × 2 = 1 + 0.841 823 926 517 76;
  • 72) 0.841 823 926 517 76 × 2 = 1 + 0.683 647 853 035 52;
  • 73) 0.683 647 853 035 52 × 2 = 1 + 0.367 295 706 071 04;
  • 74) 0.367 295 706 071 04 × 2 = 0 + 0.734 591 412 142 08;
  • 75) 0.734 591 412 142 08 × 2 = 1 + 0.469 182 824 284 16;
  • 76) 0.469 182 824 284 16 × 2 = 0 + 0.938 365 648 568 32;
  • 77) 0.938 365 648 568 32 × 2 = 1 + 0.876 731 297 136 64;
  • 78) 0.876 731 297 136 64 × 2 = 1 + 0.753 462 594 273 28;
  • 79) 0.753 462 594 273 28 × 2 = 1 + 0.506 925 188 546 56;
  • 80) 0.506 925 188 546 56 × 2 = 1 + 0.013 850 377 093 12;
  • 81) 0.013 850 377 093 12 × 2 = 0 + 0.027 700 754 186 24;
  • 82) 0.027 700 754 186 24 × 2 = 0 + 0.055 401 508 372 48;
  • 83) 0.055 401 508 372 48 × 2 = 0 + 0.110 803 016 744 96;
  • 84) 0.110 803 016 744 96 × 2 = 0 + 0.221 606 033 489 92;
  • 85) 0.221 606 033 489 92 × 2 = 0 + 0.443 212 066 979 84;
  • 86) 0.443 212 066 979 84 × 2 = 0 + 0.886 424 133 959 68;
  • 87) 0.886 424 133 959 68 × 2 = 1 + 0.772 848 267 919 36;
  • 88) 0.772 848 267 919 36 × 2 = 1 + 0.545 696 535 838 72;
  • 89) 0.545 696 535 838 72 × 2 = 1 + 0.091 393 071 677 44;
  • 90) 0.091 393 071 677 44 × 2 = 0 + 0.182 786 143 354 88;
  • 91) 0.182 786 143 354 88 × 2 = 0 + 0.365 572 286 709 76;

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:


0.000 000 000 002 12(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 0101 0100 1011 1010 0001 1001 0110 1111 1010 1111 0000 0011 100(2)

5. Positive number before normalization:

0.000 000 000 002 12(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 0101 0100 1011 1010 0001 1001 0110 1111 1010 1111 0000 0011 100(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 39 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 002 12(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 0101 0100 1011 1010 0001 1001 0110 1111 1010 1111 0000 0011 100(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 0101 0100 1011 1010 0001 1001 0110 1111 1010 1111 0000 0011 100(2) × 20 =

1.0010 1010 0101 1101 0000 1100 1011 0111 1101 0111 1000 0001 1100(2) × 2-39


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): -39

Mantissa (not normalized):
1.0010 1010 0101 1101 0000 1100 1011 0111 1101 0111 1000 0001 1100


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-39 + 2(11-1) - 1 =

(-39 + 1 023)(10) =

984(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;
  • 984 ÷ 2 = 492 + 0;
  • 492 ÷ 2 = 246 + 0;
  • 246 ÷ 2 = 123 + 0;
  • 123 ÷ 2 = 61 + 1;
  • 61 ÷ 2 = 30 + 1;
  • 30 ÷ 2 = 15 + 0;
  • 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) =

984(10) =

011 1101 1000(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 1010 0101 1101 0000 1100 1011 0111 1101 0111 1000 0001 1100 =

0010 1010 0101 1101 0000 1100 1011 0111 1101 0111 1000 0001 1100


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 1101 1000

Mantissa (52 bits) =
0010 1010 0101 1101 0000 1100 1011 0111 1101 0111 1000 0001 1100


Decimal number 0.000 000 000 002 12 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1101 1000 - 0010 1010 0101 1101 0000 1100 1011 0111 1101 0111 1000 0001 1100

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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