-0.000 299 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 299(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
-0.000 299(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:

|-0.000 299| = 0.000 299


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: 0.000 299.

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) 0.000 299 × 2 = 0 + 0.000 598;
  • 2) 0.000 598 × 2 = 0 + 0.001 196;
  • 3) 0.001 196 × 2 = 0 + 0.002 392;
  • 4) 0.002 392 × 2 = 0 + 0.004 784;
  • 5) 0.004 784 × 2 = 0 + 0.009 568;
  • 6) 0.009 568 × 2 = 0 + 0.019 136;
  • 7) 0.019 136 × 2 = 0 + 0.038 272;
  • 8) 0.038 272 × 2 = 0 + 0.076 544;
  • 9) 0.076 544 × 2 = 0 + 0.153 088;
  • 10) 0.153 088 × 2 = 0 + 0.306 176;
  • 11) 0.306 176 × 2 = 0 + 0.612 352;
  • 12) 0.612 352 × 2 = 1 + 0.224 704;
  • 13) 0.224 704 × 2 = 0 + 0.449 408;
  • 14) 0.449 408 × 2 = 0 + 0.898 816;
  • 15) 0.898 816 × 2 = 1 + 0.797 632;
  • 16) 0.797 632 × 2 = 1 + 0.595 264;
  • 17) 0.595 264 × 2 = 1 + 0.190 528;
  • 18) 0.190 528 × 2 = 0 + 0.381 056;
  • 19) 0.381 056 × 2 = 0 + 0.762 112;
  • 20) 0.762 112 × 2 = 1 + 0.524 224;
  • 21) 0.524 224 × 2 = 1 + 0.048 448;
  • 22) 0.048 448 × 2 = 0 + 0.096 896;
  • 23) 0.096 896 × 2 = 0 + 0.193 792;
  • 24) 0.193 792 × 2 = 0 + 0.387 584;
  • 25) 0.387 584 × 2 = 0 + 0.775 168;
  • 26) 0.775 168 × 2 = 1 + 0.550 336;
  • 27) 0.550 336 × 2 = 1 + 0.100 672;
  • 28) 0.100 672 × 2 = 0 + 0.201 344;
  • 29) 0.201 344 × 2 = 0 + 0.402 688;
  • 30) 0.402 688 × 2 = 0 + 0.805 376;
  • 31) 0.805 376 × 2 = 1 + 0.610 752;
  • 32) 0.610 752 × 2 = 1 + 0.221 504;
  • 33) 0.221 504 × 2 = 0 + 0.443 008;
  • 34) 0.443 008 × 2 = 0 + 0.886 016;
  • 35) 0.886 016 × 2 = 1 + 0.772 032;
  • 36) 0.772 032 × 2 = 1 + 0.544 064;
  • 37) 0.544 064 × 2 = 1 + 0.088 128;
  • 38) 0.088 128 × 2 = 0 + 0.176 256;
  • 39) 0.176 256 × 2 = 0 + 0.352 512;
  • 40) 0.352 512 × 2 = 0 + 0.705 024;
  • 41) 0.705 024 × 2 = 1 + 0.410 048;
  • 42) 0.410 048 × 2 = 0 + 0.820 096;
  • 43) 0.820 096 × 2 = 1 + 0.640 192;
  • 44) 0.640 192 × 2 = 1 + 0.280 384;
  • 45) 0.280 384 × 2 = 0 + 0.560 768;
  • 46) 0.560 768 × 2 = 1 + 0.121 536;
  • 47) 0.121 536 × 2 = 0 + 0.243 072;
  • 48) 0.243 072 × 2 = 0 + 0.486 144;
  • 49) 0.486 144 × 2 = 0 + 0.972 288;
  • 50) 0.972 288 × 2 = 1 + 0.944 576;
  • 51) 0.944 576 × 2 = 1 + 0.889 152;
  • 52) 0.889 152 × 2 = 1 + 0.778 304;
  • 53) 0.778 304 × 2 = 1 + 0.556 608;
  • 54) 0.556 608 × 2 = 1 + 0.113 216;
  • 55) 0.113 216 × 2 = 0 + 0.226 432;
  • 56) 0.226 432 × 2 = 0 + 0.452 864;
  • 57) 0.452 864 × 2 = 0 + 0.905 728;
  • 58) 0.905 728 × 2 = 1 + 0.811 456;
  • 59) 0.811 456 × 2 = 1 + 0.622 912;
  • 60) 0.622 912 × 2 = 1 + 0.245 824;
  • 61) 0.245 824 × 2 = 0 + 0.491 648;
  • 62) 0.491 648 × 2 = 0 + 0.983 296;
  • 63) 0.983 296 × 2 = 1 + 0.966 592;
  • 64) 0.966 592 × 2 = 1 + 0.933 184;

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:


0.000 299(10) =

0.0000 0000 0001 0011 1001 1000 0110 0011 0011 1000 1011 0100 0111 1100 0111 0011(2)

6. Positive number before normalization:

0.000 299(10) =

0.0000 0000 0001 0011 1001 1000 0110 0011 0011 1000 1011 0100 0111 1100 0111 0011(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:


0.000 299(10) =

0.0000 0000 0001 0011 1001 1000 0110 0011 0011 1000 1011 0100 0111 1100 0111 0011(2) =

0.0000 0000 0001 0011 1001 1000 0110 0011 0011 1000 1011 0100 0111 1100 0111 0011(2) × 20 =

1.0011 1001 1000 0110 0011 0011 1000 1011 0100 0111 1100 0111 0011(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):
1.0011 1001 1000 0110 0011 0011 1000 1011 0100 0111 1100 0111 0011


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. 0011 1001 1000 0110 0011 0011 1000 1011 0100 0111 1100 0111 0011 =

0011 1001 1000 0110 0011 0011 1000 1011 0100 0111 1100 0111 0011


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) =
0011 1001 1000 0110 0011 0011 1000 1011 0100 0111 1100 0111 0011


Decimal number -0.000 299 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0011 1001 1000 0110 0011 0011 1000 1011 0100 0111 1100 0111 0011

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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