-0.000 007 820 86 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 007 820 86(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
-0.000 007 820 86(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:

|-0.000 007 820 86| = 0.000 007 820 86


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: 0.000 007 820 86.

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) 0.000 007 820 86 × 2 = 0 + 0.000 015 641 72;
  • 2) 0.000 015 641 72 × 2 = 0 + 0.000 031 283 44;
  • 3) 0.000 031 283 44 × 2 = 0 + 0.000 062 566 88;
  • 4) 0.000 062 566 88 × 2 = 0 + 0.000 125 133 76;
  • 5) 0.000 125 133 76 × 2 = 0 + 0.000 250 267 52;
  • 6) 0.000 250 267 52 × 2 = 0 + 0.000 500 535 04;
  • 7) 0.000 500 535 04 × 2 = 0 + 0.001 001 070 08;
  • 8) 0.001 001 070 08 × 2 = 0 + 0.002 002 140 16;
  • 9) 0.002 002 140 16 × 2 = 0 + 0.004 004 280 32;
  • 10) 0.004 004 280 32 × 2 = 0 + 0.008 008 560 64;
  • 11) 0.008 008 560 64 × 2 = 0 + 0.016 017 121 28;
  • 12) 0.016 017 121 28 × 2 = 0 + 0.032 034 242 56;
  • 13) 0.032 034 242 56 × 2 = 0 + 0.064 068 485 12;
  • 14) 0.064 068 485 12 × 2 = 0 + 0.128 136 970 24;
  • 15) 0.128 136 970 24 × 2 = 0 + 0.256 273 940 48;
  • 16) 0.256 273 940 48 × 2 = 0 + 0.512 547 880 96;
  • 17) 0.512 547 880 96 × 2 = 1 + 0.025 095 761 92;
  • 18) 0.025 095 761 92 × 2 = 0 + 0.050 191 523 84;
  • 19) 0.050 191 523 84 × 2 = 0 + 0.100 383 047 68;
  • 20) 0.100 383 047 68 × 2 = 0 + 0.200 766 095 36;
  • 21) 0.200 766 095 36 × 2 = 0 + 0.401 532 190 72;
  • 22) 0.401 532 190 72 × 2 = 0 + 0.803 064 381 44;
  • 23) 0.803 064 381 44 × 2 = 1 + 0.606 128 762 88;
  • 24) 0.606 128 762 88 × 2 = 1 + 0.212 257 525 76;
  • 25) 0.212 257 525 76 × 2 = 0 + 0.424 515 051 52;
  • 26) 0.424 515 051 52 × 2 = 0 + 0.849 030 103 04;
  • 27) 0.849 030 103 04 × 2 = 1 + 0.698 060 206 08;
  • 28) 0.698 060 206 08 × 2 = 1 + 0.396 120 412 16;
  • 29) 0.396 120 412 16 × 2 = 0 + 0.792 240 824 32;
  • 30) 0.792 240 824 32 × 2 = 1 + 0.584 481 648 64;
  • 31) 0.584 481 648 64 × 2 = 1 + 0.168 963 297 28;
  • 32) 0.168 963 297 28 × 2 = 0 + 0.337 926 594 56;
  • 33) 0.337 926 594 56 × 2 = 0 + 0.675 853 189 12;
  • 34) 0.675 853 189 12 × 2 = 1 + 0.351 706 378 24;
  • 35) 0.351 706 378 24 × 2 = 0 + 0.703 412 756 48;
  • 36) 0.703 412 756 48 × 2 = 1 + 0.406 825 512 96;
  • 37) 0.406 825 512 96 × 2 = 0 + 0.813 651 025 92;
  • 38) 0.813 651 025 92 × 2 = 1 + 0.627 302 051 84;
  • 39) 0.627 302 051 84 × 2 = 1 + 0.254 604 103 68;
  • 40) 0.254 604 103 68 × 2 = 0 + 0.509 208 207 36;
  • 41) 0.509 208 207 36 × 2 = 1 + 0.018 416 414 72;
  • 42) 0.018 416 414 72 × 2 = 0 + 0.036 832 829 44;
  • 43) 0.036 832 829 44 × 2 = 0 + 0.073 665 658 88;
  • 44) 0.073 665 658 88 × 2 = 0 + 0.147 331 317 76;
  • 45) 0.147 331 317 76 × 2 = 0 + 0.294 662 635 52;
  • 46) 0.294 662 635 52 × 2 = 0 + 0.589 325 271 04;
  • 47) 0.589 325 271 04 × 2 = 1 + 0.178 650 542 08;
  • 48) 0.178 650 542 08 × 2 = 0 + 0.357 301 084 16;
  • 49) 0.357 301 084 16 × 2 = 0 + 0.714 602 168 32;
  • 50) 0.714 602 168 32 × 2 = 1 + 0.429 204 336 64;
  • 51) 0.429 204 336 64 × 2 = 0 + 0.858 408 673 28;
  • 52) 0.858 408 673 28 × 2 = 1 + 0.716 817 346 56;
  • 53) 0.716 817 346 56 × 2 = 1 + 0.433 634 693 12;
  • 54) 0.433 634 693 12 × 2 = 0 + 0.867 269 386 24;
  • 55) 0.867 269 386 24 × 2 = 1 + 0.734 538 772 48;
  • 56) 0.734 538 772 48 × 2 = 1 + 0.469 077 544 96;
  • 57) 0.469 077 544 96 × 2 = 0 + 0.938 155 089 92;
  • 58) 0.938 155 089 92 × 2 = 1 + 0.876 310 179 84;
  • 59) 0.876 310 179 84 × 2 = 1 + 0.752 620 359 68;
  • 60) 0.752 620 359 68 × 2 = 1 + 0.505 240 719 36;
  • 61) 0.505 240 719 36 × 2 = 1 + 0.010 481 438 72;
  • 62) 0.010 481 438 72 × 2 = 0 + 0.020 962 877 44;
  • 63) 0.020 962 877 44 × 2 = 0 + 0.041 925 754 88;
  • 64) 0.041 925 754 88 × 2 = 0 + 0.083 851 509 76;
  • 65) 0.083 851 509 76 × 2 = 0 + 0.167 703 019 52;
  • 66) 0.167 703 019 52 × 2 = 0 + 0.335 406 039 04;
  • 67) 0.335 406 039 04 × 2 = 0 + 0.670 812 078 08;
  • 68) 0.670 812 078 08 × 2 = 1 + 0.341 624 156 16;
  • 69) 0.341 624 156 16 × 2 = 0 + 0.683 248 312 32;

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:


0.000 007 820 86(10) =

0.0000 0000 0000 0000 1000 0011 0011 0110 0101 0110 1000 0010 0101 1011 0111 1000 0001 0(2)

6. Positive number before normalization:

0.000 007 820 86(10) =

0.0000 0000 0000 0000 1000 0011 0011 0110 0101 0110 1000 0010 0101 1011 0111 1000 0001 0(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 17 positions to the right, so that only one non zero digit remains to the left of it:


0.000 007 820 86(10) =

0.0000 0000 0000 0000 1000 0011 0011 0110 0101 0110 1000 0010 0101 1011 0111 1000 0001 0(2) =

0.0000 0000 0000 0000 1000 0011 0011 0110 0101 0110 1000 0010 0101 1011 0111 1000 0001 0(2) × 20 =

1.0000 0110 0110 1100 1010 1101 0000 0100 1011 0110 1111 0000 0010(2) × 2-17


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): -17

Mantissa (not normalized):
1.0000 0110 0110 1100 1010 1101 0000 0100 1011 0110 1111 0000 0010


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-17 + 2(11-1) - 1 =

(-17 + 1 023)(10) =

1 006(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;
  • 1 006 ÷ 2 = 503 + 0;
  • 503 ÷ 2 = 251 + 1;
  • 251 ÷ 2 = 125 + 1;
  • 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) =

1006(10) =

011 1110 1110(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. 0000 0110 0110 1100 1010 1101 0000 0100 1011 0110 1111 0000 0010 =

0000 0110 0110 1100 1010 1101 0000 0100 1011 0110 1111 0000 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 1110

Mantissa (52 bits) =
0000 0110 0110 1100 1010 1101 0000 0100 1011 0110 1111 0000 0010


Decimal number -0.000 007 820 86 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1110 1110 - 0000 0110 0110 1100 1010 1101 0000 0100 1011 0110 1111 0000 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