-0.000 000 165 732 1 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 165 732 1(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
-0.000 000 165 732 1(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. Start with the positive version of the number:

|-0.000 000 165 732 1| = 0.000 000 165 732 1


2. 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;

3. 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)


4. Convert to binary (base 2) the fractional part: 0.000 000 165 732 1.

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) 0.000 000 165 732 1 × 2 = 0 + 0.000 000 331 464 2;
  • 2) 0.000 000 331 464 2 × 2 = 0 + 0.000 000 662 928 4;
  • 3) 0.000 000 662 928 4 × 2 = 0 + 0.000 001 325 856 8;
  • 4) 0.000 001 325 856 8 × 2 = 0 + 0.000 002 651 713 6;
  • 5) 0.000 002 651 713 6 × 2 = 0 + 0.000 005 303 427 2;
  • 6) 0.000 005 303 427 2 × 2 = 0 + 0.000 010 606 854 4;
  • 7) 0.000 010 606 854 4 × 2 = 0 + 0.000 021 213 708 8;
  • 8) 0.000 021 213 708 8 × 2 = 0 + 0.000 042 427 417 6;
  • 9) 0.000 042 427 417 6 × 2 = 0 + 0.000 084 854 835 2;
  • 10) 0.000 084 854 835 2 × 2 = 0 + 0.000 169 709 670 4;
  • 11) 0.000 169 709 670 4 × 2 = 0 + 0.000 339 419 340 8;
  • 12) 0.000 339 419 340 8 × 2 = 0 + 0.000 678 838 681 6;
  • 13) 0.000 678 838 681 6 × 2 = 0 + 0.001 357 677 363 2;
  • 14) 0.001 357 677 363 2 × 2 = 0 + 0.002 715 354 726 4;
  • 15) 0.002 715 354 726 4 × 2 = 0 + 0.005 430 709 452 8;
  • 16) 0.005 430 709 452 8 × 2 = 0 + 0.010 861 418 905 6;
  • 17) 0.010 861 418 905 6 × 2 = 0 + 0.021 722 837 811 2;
  • 18) 0.021 722 837 811 2 × 2 = 0 + 0.043 445 675 622 4;
  • 19) 0.043 445 675 622 4 × 2 = 0 + 0.086 891 351 244 8;
  • 20) 0.086 891 351 244 8 × 2 = 0 + 0.173 782 702 489 6;
  • 21) 0.173 782 702 489 6 × 2 = 0 + 0.347 565 404 979 2;
  • 22) 0.347 565 404 979 2 × 2 = 0 + 0.695 130 809 958 4;
  • 23) 0.695 130 809 958 4 × 2 = 1 + 0.390 261 619 916 8;
  • 24) 0.390 261 619 916 8 × 2 = 0 + 0.780 523 239 833 6;
  • 25) 0.780 523 239 833 6 × 2 = 1 + 0.561 046 479 667 2;
  • 26) 0.561 046 479 667 2 × 2 = 1 + 0.122 092 959 334 4;
  • 27) 0.122 092 959 334 4 × 2 = 0 + 0.244 185 918 668 8;
  • 28) 0.244 185 918 668 8 × 2 = 0 + 0.488 371 837 337 6;
  • 29) 0.488 371 837 337 6 × 2 = 0 + 0.976 743 674 675 2;
  • 30) 0.976 743 674 675 2 × 2 = 1 + 0.953 487 349 350 4;
  • 31) 0.953 487 349 350 4 × 2 = 1 + 0.906 974 698 700 8;
  • 32) 0.906 974 698 700 8 × 2 = 1 + 0.813 949 397 401 6;
  • 33) 0.813 949 397 401 6 × 2 = 1 + 0.627 898 794 803 2;
  • 34) 0.627 898 794 803 2 × 2 = 1 + 0.255 797 589 606 4;
  • 35) 0.255 797 589 606 4 × 2 = 0 + 0.511 595 179 212 8;
  • 36) 0.511 595 179 212 8 × 2 = 1 + 0.023 190 358 425 6;
  • 37) 0.023 190 358 425 6 × 2 = 0 + 0.046 380 716 851 2;
  • 38) 0.046 380 716 851 2 × 2 = 0 + 0.092 761 433 702 4;
  • 39) 0.092 761 433 702 4 × 2 = 0 + 0.185 522 867 404 8;
  • 40) 0.185 522 867 404 8 × 2 = 0 + 0.371 045 734 809 6;
  • 41) 0.371 045 734 809 6 × 2 = 0 + 0.742 091 469 619 2;
  • 42) 0.742 091 469 619 2 × 2 = 1 + 0.484 182 939 238 4;
  • 43) 0.484 182 939 238 4 × 2 = 0 + 0.968 365 878 476 8;
  • 44) 0.968 365 878 476 8 × 2 = 1 + 0.936 731 756 953 6;
  • 45) 0.936 731 756 953 6 × 2 = 1 + 0.873 463 513 907 2;
  • 46) 0.873 463 513 907 2 × 2 = 1 + 0.746 927 027 814 4;
  • 47) 0.746 927 027 814 4 × 2 = 1 + 0.493 854 055 628 8;
  • 48) 0.493 854 055 628 8 × 2 = 0 + 0.987 708 111 257 6;
  • 49) 0.987 708 111 257 6 × 2 = 1 + 0.975 416 222 515 2;
  • 50) 0.975 416 222 515 2 × 2 = 1 + 0.950 832 445 030 4;
  • 51) 0.950 832 445 030 4 × 2 = 1 + 0.901 664 890 060 8;
  • 52) 0.901 664 890 060 8 × 2 = 1 + 0.803 329 780 121 6;
  • 53) 0.803 329 780 121 6 × 2 = 1 + 0.606 659 560 243 2;
  • 54) 0.606 659 560 243 2 × 2 = 1 + 0.213 319 120 486 4;
  • 55) 0.213 319 120 486 4 × 2 = 0 + 0.426 638 240 972 8;
  • 56) 0.426 638 240 972 8 × 2 = 0 + 0.853 276 481 945 6;
  • 57) 0.853 276 481 945 6 × 2 = 1 + 0.706 552 963 891 2;
  • 58) 0.706 552 963 891 2 × 2 = 1 + 0.413 105 927 782 4;
  • 59) 0.413 105 927 782 4 × 2 = 0 + 0.826 211 855 564 8;
  • 60) 0.826 211 855 564 8 × 2 = 1 + 0.652 423 711 129 6;
  • 61) 0.652 423 711 129 6 × 2 = 1 + 0.304 847 422 259 2;
  • 62) 0.304 847 422 259 2 × 2 = 0 + 0.609 694 844 518 4;
  • 63) 0.609 694 844 518 4 × 2 = 1 + 0.219 389 689 036 8;
  • 64) 0.219 389 689 036 8 × 2 = 0 + 0.438 779 378 073 6;
  • 65) 0.438 779 378 073 6 × 2 = 0 + 0.877 558 756 147 2;
  • 66) 0.877 558 756 147 2 × 2 = 1 + 0.755 117 512 294 4;
  • 67) 0.755 117 512 294 4 × 2 = 1 + 0.510 235 024 588 8;
  • 68) 0.510 235 024 588 8 × 2 = 1 + 0.020 470 049 177 6;
  • 69) 0.020 470 049 177 6 × 2 = 0 + 0.040 940 098 355 2;
  • 70) 0.040 940 098 355 2 × 2 = 0 + 0.081 880 196 710 4;
  • 71) 0.081 880 196 710 4 × 2 = 0 + 0.163 760 393 420 8;
  • 72) 0.163 760 393 420 8 × 2 = 0 + 0.327 520 786 841 6;
  • 73) 0.327 520 786 841 6 × 2 = 0 + 0.655 041 573 683 2;
  • 74) 0.655 041 573 683 2 × 2 = 1 + 0.310 083 147 366 4;
  • 75) 0.310 083 147 366 4 × 2 = 0 + 0.620 166 294 732 8;

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).


5. 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:


0.000 000 165 732 1(10) =

0.0000 0000 0000 0000 0000 0010 1100 0111 1101 0000 0101 1110 1111 1100 1101 1010 0111 0000 010(2)

6. Positive number before normalization:

0.000 000 165 732 1(10) =

0.0000 0000 0000 0000 0000 0010 1100 0111 1101 0000 0101 1110 1111 1100 1101 1010 0111 0000 010(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 23 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 165 732 1(10) =

0.0000 0000 0000 0000 0000 0010 1100 0111 1101 0000 0101 1110 1111 1100 1101 1010 0111 0000 010(2) =

0.0000 0000 0000 0000 0000 0010 1100 0111 1101 0000 0101 1110 1111 1100 1101 1010 0111 0000 010(2) × 20 =

1.0110 0011 1110 1000 0010 1111 0111 1110 0110 1101 0011 1000 0010(2) × 2-23


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): -23

Mantissa (not normalized):
1.0110 0011 1110 1000 0010 1111 0111 1110 0110 1101 0011 1000 0010


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-23 + 2(11-1) - 1 =

(-23 + 1 023)(10) =

1 000(10)


10. 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;
  • 1 000 ÷ 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;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

1000(10) =

011 1110 1000(2)


12. 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. 0110 0011 1110 1000 0010 1111 0111 1110 0110 1101 0011 1000 0010 =

0110 0011 1110 1000 0010 1111 0111 1110 0110 1101 0011 1000 0010


13. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (11 bits) =
011 1110 1000

Mantissa (52 bits) =
0110 0011 1110 1000 0010 1111 0111 1110 0110 1101 0011 1000 0010


Decimal number -0.000 000 165 732 1 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1110 1000 - 0110 0011 1110 1000 0010 1111 0111 1110 0110 1101 0011 1000 0010

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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