-0.009 826 499 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.009 826 499 3(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
-0.009 826 499 3(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:

|-0.009 826 499 3| = 0.009 826 499 3


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: 0.009 826 499 3.

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) 0.009 826 499 3 × 2 = 0 + 0.019 652 998 6;
  • 2) 0.019 652 998 6 × 2 = 0 + 0.039 305 997 2;
  • 3) 0.039 305 997 2 × 2 = 0 + 0.078 611 994 4;
  • 4) 0.078 611 994 4 × 2 = 0 + 0.157 223 988 8;
  • 5) 0.157 223 988 8 × 2 = 0 + 0.314 447 977 6;
  • 6) 0.314 447 977 6 × 2 = 0 + 0.628 895 955 2;
  • 7) 0.628 895 955 2 × 2 = 1 + 0.257 791 910 4;
  • 8) 0.257 791 910 4 × 2 = 0 + 0.515 583 820 8;
  • 9) 0.515 583 820 8 × 2 = 1 + 0.031 167 641 6;
  • 10) 0.031 167 641 6 × 2 = 0 + 0.062 335 283 2;
  • 11) 0.062 335 283 2 × 2 = 0 + 0.124 670 566 4;
  • 12) 0.124 670 566 4 × 2 = 0 + 0.249 341 132 8;
  • 13) 0.249 341 132 8 × 2 = 0 + 0.498 682 265 6;
  • 14) 0.498 682 265 6 × 2 = 0 + 0.997 364 531 2;
  • 15) 0.997 364 531 2 × 2 = 1 + 0.994 729 062 4;
  • 16) 0.994 729 062 4 × 2 = 1 + 0.989 458 124 8;
  • 17) 0.989 458 124 8 × 2 = 1 + 0.978 916 249 6;
  • 18) 0.978 916 249 6 × 2 = 1 + 0.957 832 499 2;
  • 19) 0.957 832 499 2 × 2 = 1 + 0.915 664 998 4;
  • 20) 0.915 664 998 4 × 2 = 1 + 0.831 329 996 8;
  • 21) 0.831 329 996 8 × 2 = 1 + 0.662 659 993 6;
  • 22) 0.662 659 993 6 × 2 = 1 + 0.325 319 987 2;
  • 23) 0.325 319 987 2 × 2 = 0 + 0.650 639 974 4;
  • 24) 0.650 639 974 4 × 2 = 1 + 0.301 279 948 8;
  • 25) 0.301 279 948 8 × 2 = 0 + 0.602 559 897 6;
  • 26) 0.602 559 897 6 × 2 = 1 + 0.205 119 795 2;
  • 27) 0.205 119 795 2 × 2 = 0 + 0.410 239 590 4;
  • 28) 0.410 239 590 4 × 2 = 0 + 0.820 479 180 8;
  • 29) 0.820 479 180 8 × 2 = 1 + 0.640 958 361 6;
  • 30) 0.640 958 361 6 × 2 = 1 + 0.281 916 723 2;
  • 31) 0.281 916 723 2 × 2 = 0 + 0.563 833 446 4;
  • 32) 0.563 833 446 4 × 2 = 1 + 0.127 666 892 8;
  • 33) 0.127 666 892 8 × 2 = 0 + 0.255 333 785 6;
  • 34) 0.255 333 785 6 × 2 = 0 + 0.510 667 571 2;
  • 35) 0.510 667 571 2 × 2 = 1 + 0.021 335 142 4;
  • 36) 0.021 335 142 4 × 2 = 0 + 0.042 670 284 8;
  • 37) 0.042 670 284 8 × 2 = 0 + 0.085 340 569 6;
  • 38) 0.085 340 569 6 × 2 = 0 + 0.170 681 139 2;
  • 39) 0.170 681 139 2 × 2 = 0 + 0.341 362 278 4;
  • 40) 0.341 362 278 4 × 2 = 0 + 0.682 724 556 8;
  • 41) 0.682 724 556 8 × 2 = 1 + 0.365 449 113 6;
  • 42) 0.365 449 113 6 × 2 = 0 + 0.730 898 227 2;
  • 43) 0.730 898 227 2 × 2 = 1 + 0.461 796 454 4;
  • 44) 0.461 796 454 4 × 2 = 0 + 0.923 592 908 8;
  • 45) 0.923 592 908 8 × 2 = 1 + 0.847 185 817 6;
  • 46) 0.847 185 817 6 × 2 = 1 + 0.694 371 635 2;
  • 47) 0.694 371 635 2 × 2 = 1 + 0.388 743 270 4;
  • 48) 0.388 743 270 4 × 2 = 0 + 0.777 486 540 8;
  • 49) 0.777 486 540 8 × 2 = 1 + 0.554 973 081 6;
  • 50) 0.554 973 081 6 × 2 = 1 + 0.109 946 163 2;
  • 51) 0.109 946 163 2 × 2 = 0 + 0.219 892 326 4;
  • 52) 0.219 892 326 4 × 2 = 0 + 0.439 784 652 8;
  • 53) 0.439 784 652 8 × 2 = 0 + 0.879 569 305 6;
  • 54) 0.879 569 305 6 × 2 = 1 + 0.759 138 611 2;
  • 55) 0.759 138 611 2 × 2 = 1 + 0.518 277 222 4;
  • 56) 0.518 277 222 4 × 2 = 1 + 0.036 554 444 8;
  • 57) 0.036 554 444 8 × 2 = 0 + 0.073 108 889 6;
  • 58) 0.073 108 889 6 × 2 = 0 + 0.146 217 779 2;
  • 59) 0.146 217 779 2 × 2 = 0 + 0.292 435 558 4;

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:


0.009 826 499 3(10) =

0.0000 0010 1000 0011 1111 1101 0100 1101 0010 0000 1010 1110 1100 0111 000(2)

6. Positive number before normalization:

0.009 826 499 3(10) =

0.0000 0010 1000 0011 1111 1101 0100 1101 0010 0000 1010 1110 1100 0111 000(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:


0.009 826 499 3(10) =

0.0000 0010 1000 0011 1111 1101 0100 1101 0010 0000 1010 1110 1100 0111 000(2) =

0.0000 0010 1000 0011 1111 1101 0100 1101 0010 0000 1010 1110 1100 0111 000(2) × 20 =

1.0100 0001 1111 1110 1010 0110 1001 0000 0101 0111 0110 0011 1000(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 1010 0110 1001 0000 0101 0111 0110 0011 1000


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 1010 0110 1001 0000 0101 0111 0110 0011 1000 =

0100 0001 1111 1110 1010 0110 1001 0000 0101 0111 0110 0011 1000


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 1010 0110 1001 0000 0101 0111 0110 0011 1000


Decimal number -0.009 826 499 3 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1000 - 0100 0001 1111 1110 1010 0110 1001 0000 0101 0111 0110 0011 1000

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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