64bit IEEE 754: Decimal ↗ Double Precision Floating Point Binary: 0.000 524 281 565 170 515 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number 0.000 524 281 565 170 515(10) converted and written in 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: 0.000 524 281 565 170 515.

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) 0.000 524 281 565 170 515 × 2 = 0 + 0.001 048 563 130 341 03;
  • 2) 0.001 048 563 130 341 03 × 2 = 0 + 0.002 097 126 260 682 06;
  • 3) 0.002 097 126 260 682 06 × 2 = 0 + 0.004 194 252 521 364 12;
  • 4) 0.004 194 252 521 364 12 × 2 = 0 + 0.008 388 505 042 728 24;
  • 5) 0.008 388 505 042 728 24 × 2 = 0 + 0.016 777 010 085 456 48;
  • 6) 0.016 777 010 085 456 48 × 2 = 0 + 0.033 554 020 170 912 96;
  • 7) 0.033 554 020 170 912 96 × 2 = 0 + 0.067 108 040 341 825 92;
  • 8) 0.067 108 040 341 825 92 × 2 = 0 + 0.134 216 080 683 651 84;
  • 9) 0.134 216 080 683 651 84 × 2 = 0 + 0.268 432 161 367 303 68;
  • 10) 0.268 432 161 367 303 68 × 2 = 0 + 0.536 864 322 734 607 36;
  • 11) 0.536 864 322 734 607 36 × 2 = 1 + 0.073 728 645 469 214 72;
  • 12) 0.073 728 645 469 214 72 × 2 = 0 + 0.147 457 290 938 429 44;
  • 13) 0.147 457 290 938 429 44 × 2 = 0 + 0.294 914 581 876 858 88;
  • 14) 0.294 914 581 876 858 88 × 2 = 0 + 0.589 829 163 753 717 76;
  • 15) 0.589 829 163 753 717 76 × 2 = 1 + 0.179 658 327 507 435 52;
  • 16) 0.179 658 327 507 435 52 × 2 = 0 + 0.359 316 655 014 871 04;
  • 17) 0.359 316 655 014 871 04 × 2 = 0 + 0.718 633 310 029 742 08;
  • 18) 0.718 633 310 029 742 08 × 2 = 1 + 0.437 266 620 059 484 16;
  • 19) 0.437 266 620 059 484 16 × 2 = 0 + 0.874 533 240 118 968 32;
  • 20) 0.874 533 240 118 968 32 × 2 = 1 + 0.749 066 480 237 936 64;
  • 21) 0.749 066 480 237 936 64 × 2 = 1 + 0.498 132 960 475 873 28;
  • 22) 0.498 132 960 475 873 28 × 2 = 0 + 0.996 265 920 951 746 56;
  • 23) 0.996 265 920 951 746 56 × 2 = 1 + 0.992 531 841 903 493 12;
  • 24) 0.992 531 841 903 493 12 × 2 = 1 + 0.985 063 683 806 986 24;
  • 25) 0.985 063 683 806 986 24 × 2 = 1 + 0.970 127 367 613 972 48;
  • 26) 0.970 127 367 613 972 48 × 2 = 1 + 0.940 254 735 227 944 96;
  • 27) 0.940 254 735 227 944 96 × 2 = 1 + 0.880 509 470 455 889 92;
  • 28) 0.880 509 470 455 889 92 × 2 = 1 + 0.761 018 940 911 779 84;
  • 29) 0.761 018 940 911 779 84 × 2 = 1 + 0.522 037 881 823 559 68;
  • 30) 0.522 037 881 823 559 68 × 2 = 1 + 0.044 075 763 647 119 36;
  • 31) 0.044 075 763 647 119 36 × 2 = 0 + 0.088 151 527 294 238 72;
  • 32) 0.088 151 527 294 238 72 × 2 = 0 + 0.176 303 054 588 477 44;
  • 33) 0.176 303 054 588 477 44 × 2 = 0 + 0.352 606 109 176 954 88;
  • 34) 0.352 606 109 176 954 88 × 2 = 0 + 0.705 212 218 353 909 76;
  • 35) 0.705 212 218 353 909 76 × 2 = 1 + 0.410 424 436 707 819 52;
  • 36) 0.410 424 436 707 819 52 × 2 = 0 + 0.820 848 873 415 639 04;
  • 37) 0.820 848 873 415 639 04 × 2 = 1 + 0.641 697 746 831 278 08;
  • 38) 0.641 697 746 831 278 08 × 2 = 1 + 0.283 395 493 662 556 16;
  • 39) 0.283 395 493 662 556 16 × 2 = 0 + 0.566 790 987 325 112 32;
  • 40) 0.566 790 987 325 112 32 × 2 = 1 + 0.133 581 974 650 224 64;
  • 41) 0.133 581 974 650 224 64 × 2 = 0 + 0.267 163 949 300 449 28;
  • 42) 0.267 163 949 300 449 28 × 2 = 0 + 0.534 327 898 600 898 56;
  • 43) 0.534 327 898 600 898 56 × 2 = 1 + 0.068 655 797 201 797 12;
  • 44) 0.068 655 797 201 797 12 × 2 = 0 + 0.137 311 594 403 594 24;
  • 45) 0.137 311 594 403 594 24 × 2 = 0 + 0.274 623 188 807 188 48;
  • 46) 0.274 623 188 807 188 48 × 2 = 0 + 0.549 246 377 614 376 96;
  • 47) 0.549 246 377 614 376 96 × 2 = 1 + 0.098 492 755 228 753 92;
  • 48) 0.098 492 755 228 753 92 × 2 = 0 + 0.196 985 510 457 507 84;
  • 49) 0.196 985 510 457 507 84 × 2 = 0 + 0.393 971 020 915 015 68;
  • 50) 0.393 971 020 915 015 68 × 2 = 0 + 0.787 942 041 830 031 36;
  • 51) 0.787 942 041 830 031 36 × 2 = 1 + 0.575 884 083 660 062 72;
  • 52) 0.575 884 083 660 062 72 × 2 = 1 + 0.151 768 167 320 125 44;
  • 53) 0.151 768 167 320 125 44 × 2 = 0 + 0.303 536 334 640 250 88;
  • 54) 0.303 536 334 640 250 88 × 2 = 0 + 0.607 072 669 280 501 76;
  • 55) 0.607 072 669 280 501 76 × 2 = 1 + 0.214 145 338 561 003 52;
  • 56) 0.214 145 338 561 003 52 × 2 = 0 + 0.428 290 677 122 007 04;
  • 57) 0.428 290 677 122 007 04 × 2 = 0 + 0.856 581 354 244 014 08;
  • 58) 0.856 581 354 244 014 08 × 2 = 1 + 0.713 162 708 488 028 16;
  • 59) 0.713 162 708 488 028 16 × 2 = 1 + 0.426 325 416 976 056 32;
  • 60) 0.426 325 416 976 056 32 × 2 = 0 + 0.852 650 833 952 112 64;
  • 61) 0.852 650 833 952 112 64 × 2 = 1 + 0.705 301 667 904 225 28;
  • 62) 0.705 301 667 904 225 28 × 2 = 1 + 0.410 603 335 808 450 56;
  • 63) 0.410 603 335 808 450 56 × 2 = 0 + 0.821 206 671 616 901 12;

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...)


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:


0.000 524 281 565 170 515(10) =

0.0000 0000 0010 0010 0101 1011 1111 1100 0010 1101 0010 0010 0011 0010 0110 110(2)


5. Positive number before normalization:

0.000 524 281 565 170 515(10) =

0.0000 0000 0010 0010 0101 1011 1111 1100 0010 1101 0010 0010 0011 0010 0110 110(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 11 positions to the right, so that only one non zero digit remains to the left of it:


0.000 524 281 565 170 515(10) =

0.0000 0000 0010 0010 0101 1011 1111 1100 0010 1101 0010 0010 0011 0010 0110 110(2) =

0.0000 0000 0010 0010 0101 1011 1111 1100 0010 1101 0010 0010 0011 0010 0110 110(2) × 20 =

1.0001 0010 1101 1111 1110 0001 0110 1001 0001 0001 1001 0011 0110(2) × 2-11


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): -11

Mantissa (not normalized):
1.0001 0010 1101 1111 1110 0001 0110 1001 0001 0001 1001 0011 0110


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

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

(-11 + 1 023)(10) =

1 012(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;
  • 1 012 ÷ 2 = 506 + 0;
  • 506 ÷ 2 = 253 + 0;
  • 253 ÷ 2 = 126 + 1;
  • 126 ÷ 2 = 63 + 0;
  • 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) =

1012(10) =

011 1111 0100(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. 0001 0010 1101 1111 1110 0001 0110 1001 0001 0001 1001 0011 0110 =

0001 0010 1101 1111 1110 0001 0110 1001 0001 0001 1001 0011 0110


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 0100

Mantissa (52 bits) =
0001 0010 1101 1111 1110 0001 0110 1001 0001 0001 1001 0011 0110


The base ten decimal number 0.000 524 281 565 170 515 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
0 - 011 1111 0100 - 0001 0010 1101 1111 1110 0001 0110 1001 0001 0001 1001 0011 0110

More operations with decimal numbers converted to 64 bit double precision IEEE 754 binary floating point representation:

» Number 0.000 524 281 565 170 514 converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point representation = ?

» Number 0.000 524 281 565 170 516 converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point representation = ?

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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