-0.009 826 505 2 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.009 826 505 2(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
-0.009 826 505 2(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:

|-0.009 826 505 2| = 0.009 826 505 2


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: 0.009 826 505 2.

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) 0.009 826 505 2 × 2 = 0 + 0.019 653 010 4;
  • 2) 0.019 653 010 4 × 2 = 0 + 0.039 306 020 8;
  • 3) 0.039 306 020 8 × 2 = 0 + 0.078 612 041 6;
  • 4) 0.078 612 041 6 × 2 = 0 + 0.157 224 083 2;
  • 5) 0.157 224 083 2 × 2 = 0 + 0.314 448 166 4;
  • 6) 0.314 448 166 4 × 2 = 0 + 0.628 896 332 8;
  • 7) 0.628 896 332 8 × 2 = 1 + 0.257 792 665 6;
  • 8) 0.257 792 665 6 × 2 = 0 + 0.515 585 331 2;
  • 9) 0.515 585 331 2 × 2 = 1 + 0.031 170 662 4;
  • 10) 0.031 170 662 4 × 2 = 0 + 0.062 341 324 8;
  • 11) 0.062 341 324 8 × 2 = 0 + 0.124 682 649 6;
  • 12) 0.124 682 649 6 × 2 = 0 + 0.249 365 299 2;
  • 13) 0.249 365 299 2 × 2 = 0 + 0.498 730 598 4;
  • 14) 0.498 730 598 4 × 2 = 0 + 0.997 461 196 8;
  • 15) 0.997 461 196 8 × 2 = 1 + 0.994 922 393 6;
  • 16) 0.994 922 393 6 × 2 = 1 + 0.989 844 787 2;
  • 17) 0.989 844 787 2 × 2 = 1 + 0.979 689 574 4;
  • 18) 0.979 689 574 4 × 2 = 1 + 0.959 379 148 8;
  • 19) 0.959 379 148 8 × 2 = 1 + 0.918 758 297 6;
  • 20) 0.918 758 297 6 × 2 = 1 + 0.837 516 595 2;
  • 21) 0.837 516 595 2 × 2 = 1 + 0.675 033 190 4;
  • 22) 0.675 033 190 4 × 2 = 1 + 0.350 066 380 8;
  • 23) 0.350 066 380 8 × 2 = 0 + 0.700 132 761 6;
  • 24) 0.700 132 761 6 × 2 = 1 + 0.400 265 523 2;
  • 25) 0.400 265 523 2 × 2 = 0 + 0.800 531 046 4;
  • 26) 0.800 531 046 4 × 2 = 1 + 0.601 062 092 8;
  • 27) 0.601 062 092 8 × 2 = 1 + 0.202 124 185 6;
  • 28) 0.202 124 185 6 × 2 = 0 + 0.404 248 371 2;
  • 29) 0.404 248 371 2 × 2 = 0 + 0.808 496 742 4;
  • 30) 0.808 496 742 4 × 2 = 1 + 0.616 993 484 8;
  • 31) 0.616 993 484 8 × 2 = 1 + 0.233 986 969 6;
  • 32) 0.233 986 969 6 × 2 = 0 + 0.467 973 939 2;
  • 33) 0.467 973 939 2 × 2 = 0 + 0.935 947 878 4;
  • 34) 0.935 947 878 4 × 2 = 1 + 0.871 895 756 8;
  • 35) 0.871 895 756 8 × 2 = 1 + 0.743 791 513 6;
  • 36) 0.743 791 513 6 × 2 = 1 + 0.487 583 027 2;
  • 37) 0.487 583 027 2 × 2 = 0 + 0.975 166 054 4;
  • 38) 0.975 166 054 4 × 2 = 1 + 0.950 332 108 8;
  • 39) 0.950 332 108 8 × 2 = 1 + 0.900 664 217 6;
  • 40) 0.900 664 217 6 × 2 = 1 + 0.801 328 435 2;
  • 41) 0.801 328 435 2 × 2 = 1 + 0.602 656 870 4;
  • 42) 0.602 656 870 4 × 2 = 1 + 0.205 313 740 8;
  • 43) 0.205 313 740 8 × 2 = 0 + 0.410 627 481 6;
  • 44) 0.410 627 481 6 × 2 = 0 + 0.821 254 963 2;
  • 45) 0.821 254 963 2 × 2 = 1 + 0.642 509 926 4;
  • 46) 0.642 509 926 4 × 2 = 1 + 0.285 019 852 8;
  • 47) 0.285 019 852 8 × 2 = 0 + 0.570 039 705 6;
  • 48) 0.570 039 705 6 × 2 = 1 + 0.140 079 411 2;
  • 49) 0.140 079 411 2 × 2 = 0 + 0.280 158 822 4;
  • 50) 0.280 158 822 4 × 2 = 0 + 0.560 317 644 8;
  • 51) 0.560 317 644 8 × 2 = 1 + 0.120 635 289 6;
  • 52) 0.120 635 289 6 × 2 = 0 + 0.241 270 579 2;
  • 53) 0.241 270 579 2 × 2 = 0 + 0.482 541 158 4;
  • 54) 0.482 541 158 4 × 2 = 0 + 0.965 082 316 8;
  • 55) 0.965 082 316 8 × 2 = 1 + 0.930 164 633 6;
  • 56) 0.930 164 633 6 × 2 = 1 + 0.860 329 267 2;
  • 57) 0.860 329 267 2 × 2 = 1 + 0.720 658 534 4;
  • 58) 0.720 658 534 4 × 2 = 1 + 0.441 317 068 8;
  • 59) 0.441 317 068 8 × 2 = 0 + 0.882 634 137 6;

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:


0.009 826 505 2(10) =

0.0000 0010 1000 0011 1111 1101 0110 0110 0111 0111 1100 1101 0010 0011 110(2)

6. Positive number before normalization:

0.009 826 505 2(10) =

0.0000 0010 1000 0011 1111 1101 0110 0110 0111 0111 1100 1101 0010 0011 110(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 7 positions to the right, so that only one non zero digit remains to the left of it:


0.009 826 505 2(10) =

0.0000 0010 1000 0011 1111 1101 0110 0110 0111 0111 1100 1101 0010 0011 110(2) =

0.0000 0010 1000 0011 1111 1101 0110 0110 0111 0111 1100 1101 0010 0011 110(2) × 20 =

1.0100 0001 1111 1110 1011 0011 0011 1011 1110 0110 1001 0001 1110(2) × 2-7


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): -7

Mantissa (not normalized):
1.0100 0001 1111 1110 1011 0011 0011 1011 1110 0110 1001 0001 1110


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-7 + 2(11-1) - 1 =

(-7 + 1 023)(10) =

1 016(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;
  • 1 016 ÷ 2 = 508 + 0;
  • 508 ÷ 2 = 254 + 0;
  • 254 ÷ 2 = 127 + 0;
  • 127 ÷ 2 = 63 + 1;
  • 63 ÷ 2 = 31 + 1;
  • 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) =

1016(10) =

011 1111 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. 0100 0001 1111 1110 1011 0011 0011 1011 1110 0110 1001 0001 1110 =

0100 0001 1111 1110 1011 0011 0011 1011 1110 0110 1001 0001 1110


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 1111 1000

Mantissa (52 bits) =
0100 0001 1111 1110 1011 0011 0011 1011 1110 0110 1001 0001 1110


Decimal number -0.009 826 505 2 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1000 - 0100 0001 1111 1110 1011 0011 0011 1011 1110 0110 1001 0001 1110

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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