-0.000 007 835 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 007 835(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
-0.000 007 835(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:

|-0.000 007 835| = 0.000 007 835


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: 0.000 007 835.

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) 0.000 007 835 × 2 = 0 + 0.000 015 67;
  • 2) 0.000 015 67 × 2 = 0 + 0.000 031 34;
  • 3) 0.000 031 34 × 2 = 0 + 0.000 062 68;
  • 4) 0.000 062 68 × 2 = 0 + 0.000 125 36;
  • 5) 0.000 125 36 × 2 = 0 + 0.000 250 72;
  • 6) 0.000 250 72 × 2 = 0 + 0.000 501 44;
  • 7) 0.000 501 44 × 2 = 0 + 0.001 002 88;
  • 8) 0.001 002 88 × 2 = 0 + 0.002 005 76;
  • 9) 0.002 005 76 × 2 = 0 + 0.004 011 52;
  • 10) 0.004 011 52 × 2 = 0 + 0.008 023 04;
  • 11) 0.008 023 04 × 2 = 0 + 0.016 046 08;
  • 12) 0.016 046 08 × 2 = 0 + 0.032 092 16;
  • 13) 0.032 092 16 × 2 = 0 + 0.064 184 32;
  • 14) 0.064 184 32 × 2 = 0 + 0.128 368 64;
  • 15) 0.128 368 64 × 2 = 0 + 0.256 737 28;
  • 16) 0.256 737 28 × 2 = 0 + 0.513 474 56;
  • 17) 0.513 474 56 × 2 = 1 + 0.026 949 12;
  • 18) 0.026 949 12 × 2 = 0 + 0.053 898 24;
  • 19) 0.053 898 24 × 2 = 0 + 0.107 796 48;
  • 20) 0.107 796 48 × 2 = 0 + 0.215 592 96;
  • 21) 0.215 592 96 × 2 = 0 + 0.431 185 92;
  • 22) 0.431 185 92 × 2 = 0 + 0.862 371 84;
  • 23) 0.862 371 84 × 2 = 1 + 0.724 743 68;
  • 24) 0.724 743 68 × 2 = 1 + 0.449 487 36;
  • 25) 0.449 487 36 × 2 = 0 + 0.898 974 72;
  • 26) 0.898 974 72 × 2 = 1 + 0.797 949 44;
  • 27) 0.797 949 44 × 2 = 1 + 0.595 898 88;
  • 28) 0.595 898 88 × 2 = 1 + 0.191 797 76;
  • 29) 0.191 797 76 × 2 = 0 + 0.383 595 52;
  • 30) 0.383 595 52 × 2 = 0 + 0.767 191 04;
  • 31) 0.767 191 04 × 2 = 1 + 0.534 382 08;
  • 32) 0.534 382 08 × 2 = 1 + 0.068 764 16;
  • 33) 0.068 764 16 × 2 = 0 + 0.137 528 32;
  • 34) 0.137 528 32 × 2 = 0 + 0.275 056 64;
  • 35) 0.275 056 64 × 2 = 0 + 0.550 113 28;
  • 36) 0.550 113 28 × 2 = 1 + 0.100 226 56;
  • 37) 0.100 226 56 × 2 = 0 + 0.200 453 12;
  • 38) 0.200 453 12 × 2 = 0 + 0.400 906 24;
  • 39) 0.400 906 24 × 2 = 0 + 0.801 812 48;
  • 40) 0.801 812 48 × 2 = 1 + 0.603 624 96;
  • 41) 0.603 624 96 × 2 = 1 + 0.207 249 92;
  • 42) 0.207 249 92 × 2 = 0 + 0.414 499 84;
  • 43) 0.414 499 84 × 2 = 0 + 0.828 999 68;
  • 44) 0.828 999 68 × 2 = 1 + 0.657 999 36;
  • 45) 0.657 999 36 × 2 = 1 + 0.315 998 72;
  • 46) 0.315 998 72 × 2 = 0 + 0.631 997 44;
  • 47) 0.631 997 44 × 2 = 1 + 0.263 994 88;
  • 48) 0.263 994 88 × 2 = 0 + 0.527 989 76;
  • 49) 0.527 989 76 × 2 = 1 + 0.055 979 52;
  • 50) 0.055 979 52 × 2 = 0 + 0.111 959 04;
  • 51) 0.111 959 04 × 2 = 0 + 0.223 918 08;
  • 52) 0.223 918 08 × 2 = 0 + 0.447 836 16;
  • 53) 0.447 836 16 × 2 = 0 + 0.895 672 32;
  • 54) 0.895 672 32 × 2 = 1 + 0.791 344 64;
  • 55) 0.791 344 64 × 2 = 1 + 0.582 689 28;
  • 56) 0.582 689 28 × 2 = 1 + 0.165 378 56;
  • 57) 0.165 378 56 × 2 = 0 + 0.330 757 12;
  • 58) 0.330 757 12 × 2 = 0 + 0.661 514 24;
  • 59) 0.661 514 24 × 2 = 1 + 0.323 028 48;
  • 60) 0.323 028 48 × 2 = 0 + 0.646 056 96;
  • 61) 0.646 056 96 × 2 = 1 + 0.292 113 92;
  • 62) 0.292 113 92 × 2 = 0 + 0.584 227 84;
  • 63) 0.584 227 84 × 2 = 1 + 0.168 455 68;
  • 64) 0.168 455 68 × 2 = 0 + 0.336 911 36;
  • 65) 0.336 911 36 × 2 = 0 + 0.673 822 72;
  • 66) 0.673 822 72 × 2 = 1 + 0.347 645 44;
  • 67) 0.347 645 44 × 2 = 0 + 0.695 290 88;
  • 68) 0.695 290 88 × 2 = 1 + 0.390 581 76;
  • 69) 0.390 581 76 × 2 = 0 + 0.781 163 52;

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:


0.000 007 835(10) =

0.0000 0000 0000 0000 1000 0011 0111 0011 0001 0001 1001 1010 1000 0111 0010 1010 0101 0(2)

6. Positive number before normalization:

0.000 007 835(10) =

0.0000 0000 0000 0000 1000 0011 0111 0011 0001 0001 1001 1010 1000 0111 0010 1010 0101 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:


0.000 007 835(10) =

0.0000 0000 0000 0000 1000 0011 0111 0011 0001 0001 1001 1010 1000 0111 0010 1010 0101 0(2) =

0.0000 0000 0000 0000 1000 0011 0111 0011 0001 0001 1001 1010 1000 0111 0010 1010 0101 0(2) × 20 =

1.0000 0110 1110 0110 0010 0011 0011 0101 0000 1110 0101 0100 1010(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 1110 0110 0010 0011 0011 0101 0000 1110 0101 0100 1010


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 1110 0110 0010 0011 0011 0101 0000 1110 0101 0100 1010 =

0000 0110 1110 0110 0010 0011 0011 0101 0000 1110 0101 0100 1010


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 1110 0110 0010 0011 0011 0101 0000 1110 0101 0100 1010


Decimal number -0.000 007 835 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1110 1110 - 0000 0110 1110 0110 0010 0011 0011 0101 0000 1110 0101 0100 1010

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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