-0.000 266 5 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 266 5(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
-0.000 266 5(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:

|-0.000 266 5| = 0.000 266 5


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: 0.000 266 5.

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) 0.000 266 5 × 2 = 0 + 0.000 533;
  • 2) 0.000 533 × 2 = 0 + 0.001 066;
  • 3) 0.001 066 × 2 = 0 + 0.002 132;
  • 4) 0.002 132 × 2 = 0 + 0.004 264;
  • 5) 0.004 264 × 2 = 0 + 0.008 528;
  • 6) 0.008 528 × 2 = 0 + 0.017 056;
  • 7) 0.017 056 × 2 = 0 + 0.034 112;
  • 8) 0.034 112 × 2 = 0 + 0.068 224;
  • 9) 0.068 224 × 2 = 0 + 0.136 448;
  • 10) 0.136 448 × 2 = 0 + 0.272 896;
  • 11) 0.272 896 × 2 = 0 + 0.545 792;
  • 12) 0.545 792 × 2 = 1 + 0.091 584;
  • 13) 0.091 584 × 2 = 0 + 0.183 168;
  • 14) 0.183 168 × 2 = 0 + 0.366 336;
  • 15) 0.366 336 × 2 = 0 + 0.732 672;
  • 16) 0.732 672 × 2 = 1 + 0.465 344;
  • 17) 0.465 344 × 2 = 0 + 0.930 688;
  • 18) 0.930 688 × 2 = 1 + 0.861 376;
  • 19) 0.861 376 × 2 = 1 + 0.722 752;
  • 20) 0.722 752 × 2 = 1 + 0.445 504;
  • 21) 0.445 504 × 2 = 0 + 0.891 008;
  • 22) 0.891 008 × 2 = 1 + 0.782 016;
  • 23) 0.782 016 × 2 = 1 + 0.564 032;
  • 24) 0.564 032 × 2 = 1 + 0.128 064;
  • 25) 0.128 064 × 2 = 0 + 0.256 128;
  • 26) 0.256 128 × 2 = 0 + 0.512 256;
  • 27) 0.512 256 × 2 = 1 + 0.024 512;
  • 28) 0.024 512 × 2 = 0 + 0.049 024;
  • 29) 0.049 024 × 2 = 0 + 0.098 048;
  • 30) 0.098 048 × 2 = 0 + 0.196 096;
  • 31) 0.196 096 × 2 = 0 + 0.392 192;
  • 32) 0.392 192 × 2 = 0 + 0.784 384;
  • 33) 0.784 384 × 2 = 1 + 0.568 768;
  • 34) 0.568 768 × 2 = 1 + 0.137 536;
  • 35) 0.137 536 × 2 = 0 + 0.275 072;
  • 36) 0.275 072 × 2 = 0 + 0.550 144;
  • 37) 0.550 144 × 2 = 1 + 0.100 288;
  • 38) 0.100 288 × 2 = 0 + 0.200 576;
  • 39) 0.200 576 × 2 = 0 + 0.401 152;
  • 40) 0.401 152 × 2 = 0 + 0.802 304;
  • 41) 0.802 304 × 2 = 1 + 0.604 608;
  • 42) 0.604 608 × 2 = 1 + 0.209 216;
  • 43) 0.209 216 × 2 = 0 + 0.418 432;
  • 44) 0.418 432 × 2 = 0 + 0.836 864;
  • 45) 0.836 864 × 2 = 1 + 0.673 728;
  • 46) 0.673 728 × 2 = 1 + 0.347 456;
  • 47) 0.347 456 × 2 = 0 + 0.694 912;
  • 48) 0.694 912 × 2 = 1 + 0.389 824;
  • 49) 0.389 824 × 2 = 0 + 0.779 648;
  • 50) 0.779 648 × 2 = 1 + 0.559 296;
  • 51) 0.559 296 × 2 = 1 + 0.118 592;
  • 52) 0.118 592 × 2 = 0 + 0.237 184;
  • 53) 0.237 184 × 2 = 0 + 0.474 368;
  • 54) 0.474 368 × 2 = 0 + 0.948 736;
  • 55) 0.948 736 × 2 = 1 + 0.897 472;
  • 56) 0.897 472 × 2 = 1 + 0.794 944;
  • 57) 0.794 944 × 2 = 1 + 0.589 888;
  • 58) 0.589 888 × 2 = 1 + 0.179 776;
  • 59) 0.179 776 × 2 = 0 + 0.359 552;
  • 60) 0.359 552 × 2 = 0 + 0.719 104;
  • 61) 0.719 104 × 2 = 1 + 0.438 208;
  • 62) 0.438 208 × 2 = 0 + 0.876 416;
  • 63) 0.876 416 × 2 = 1 + 0.752 832;
  • 64) 0.752 832 × 2 = 1 + 0.505 664;

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:


0.000 266 5(10) =

0.0000 0000 0001 0001 0111 0111 0010 0000 1100 1000 1100 1101 0110 0011 1100 1011(2)

6. Positive number before normalization:

0.000 266 5(10) =

0.0000 0000 0001 0001 0111 0111 0010 0000 1100 1000 1100 1101 0110 0011 1100 1011(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 12 positions to the right, so that only one non zero digit remains to the left of it:


0.000 266 5(10) =

0.0000 0000 0001 0001 0111 0111 0010 0000 1100 1000 1100 1101 0110 0011 1100 1011(2) =

0.0000 0000 0001 0001 0111 0111 0010 0000 1100 1000 1100 1101 0110 0011 1100 1011(2) × 20 =

1.0001 0111 0111 0010 0000 1100 1000 1100 1101 0110 0011 1100 1011(2) × 2-12


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): -12

Mantissa (not normalized):
1.0001 0111 0111 0010 0000 1100 1000 1100 1101 0110 0011 1100 1011


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-12 + 2(11-1) - 1 =

(-12 + 1 023)(10) =

1 011(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;
  • 1 011 ÷ 2 = 505 + 1;
  • 505 ÷ 2 = 252 + 1;
  • 252 ÷ 2 = 126 + 0;
  • 126 ÷ 2 = 63 + 0;
  • 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) =

1011(10) =

011 1111 0011(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. 0001 0111 0111 0010 0000 1100 1000 1100 1101 0110 0011 1100 1011 =

0001 0111 0111 0010 0000 1100 1000 1100 1101 0110 0011 1100 1011


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 0011

Mantissa (52 bits) =
0001 0111 0111 0010 0000 1100 1000 1100 1101 0110 0011 1100 1011


Decimal number -0.000 266 5 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0001 0111 0111 0010 0000 1100 1000 1100 1101 0110 0011 1100 1011

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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