-0.000 002 06 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 002 06(10) to 32 bit single precision IEEE 754 binary floating point representation standard (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

What are the steps to convert decimal number
-0.000 002 06(10) to 32 bit single precision IEEE 754 binary floating point representation (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

1. Start with the positive version of the number:

|-0.000 002 06| = 0.000 002 06


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: 0.000 002 06.

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) 0.000 002 06 × 2 = 0 + 0.000 004 12;
  • 2) 0.000 004 12 × 2 = 0 + 0.000 008 24;
  • 3) 0.000 008 24 × 2 = 0 + 0.000 016 48;
  • 4) 0.000 016 48 × 2 = 0 + 0.000 032 96;
  • 5) 0.000 032 96 × 2 = 0 + 0.000 065 92;
  • 6) 0.000 065 92 × 2 = 0 + 0.000 131 84;
  • 7) 0.000 131 84 × 2 = 0 + 0.000 263 68;
  • 8) 0.000 263 68 × 2 = 0 + 0.000 527 36;
  • 9) 0.000 527 36 × 2 = 0 + 0.001 054 72;
  • 10) 0.001 054 72 × 2 = 0 + 0.002 109 44;
  • 11) 0.002 109 44 × 2 = 0 + 0.004 218 88;
  • 12) 0.004 218 88 × 2 = 0 + 0.008 437 76;
  • 13) 0.008 437 76 × 2 = 0 + 0.016 875 52;
  • 14) 0.016 875 52 × 2 = 0 + 0.033 751 04;
  • 15) 0.033 751 04 × 2 = 0 + 0.067 502 08;
  • 16) 0.067 502 08 × 2 = 0 + 0.135 004 16;
  • 17) 0.135 004 16 × 2 = 0 + 0.270 008 32;
  • 18) 0.270 008 32 × 2 = 0 + 0.540 016 64;
  • 19) 0.540 016 64 × 2 = 1 + 0.080 033 28;
  • 20) 0.080 033 28 × 2 = 0 + 0.160 066 56;
  • 21) 0.160 066 56 × 2 = 0 + 0.320 133 12;
  • 22) 0.320 133 12 × 2 = 0 + 0.640 266 24;
  • 23) 0.640 266 24 × 2 = 1 + 0.280 532 48;
  • 24) 0.280 532 48 × 2 = 0 + 0.561 064 96;
  • 25) 0.561 064 96 × 2 = 1 + 0.122 129 92;
  • 26) 0.122 129 92 × 2 = 0 + 0.244 259 84;
  • 27) 0.244 259 84 × 2 = 0 + 0.488 519 68;
  • 28) 0.488 519 68 × 2 = 0 + 0.977 039 36;
  • 29) 0.977 039 36 × 2 = 1 + 0.954 078 72;
  • 30) 0.954 078 72 × 2 = 1 + 0.908 157 44;
  • 31) 0.908 157 44 × 2 = 1 + 0.816 314 88;
  • 32) 0.816 314 88 × 2 = 1 + 0.632 629 76;
  • 33) 0.632 629 76 × 2 = 1 + 0.265 259 52;
  • 34) 0.265 259 52 × 2 = 0 + 0.530 519 04;
  • 35) 0.530 519 04 × 2 = 1 + 0.061 038 08;
  • 36) 0.061 038 08 × 2 = 0 + 0.122 076 16;
  • 37) 0.122 076 16 × 2 = 0 + 0.244 152 32;
  • 38) 0.244 152 32 × 2 = 0 + 0.488 304 64;
  • 39) 0.488 304 64 × 2 = 0 + 0.976 609 28;
  • 40) 0.976 609 28 × 2 = 1 + 0.953 218 56;
  • 41) 0.953 218 56 × 2 = 1 + 0.906 437 12;
  • 42) 0.906 437 12 × 2 = 1 + 0.812 874 24;

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:


0.000 002 06(10) =

0.0000 0000 0000 0000 0010 0010 1000 1111 1010 0001 11(2)

6. Positive number before normalization:

0.000 002 06(10) =

0.0000 0000 0000 0000 0010 0010 1000 1111 1010 0001 11(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 19 positions to the right, so that only one non zero digit remains to the left of it:


0.000 002 06(10) =

0.0000 0000 0000 0000 0010 0010 1000 1111 1010 0001 11(2) =

0.0000 0000 0000 0000 0010 0010 1000 1111 1010 0001 11(2) × 20 =

1.0001 0100 0111 1101 0000 111(2) × 2-19


8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point representation:

Sign 1 (a negative number)

Exponent (unadjusted): -19

Mantissa (not normalized):
1.0001 0100 0111 1101 0000 111


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(8-1) - 1 =

-19 + 2(8-1) - 1 =

(-19 + 127)(10) =

108(10)


10. Convert the adjusted exponent from the decimal (base 10) to 8 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 108 ÷ 2 = 54 + 0;
  • 54 ÷ 2 = 27 + 0;
  • 27 ÷ 2 = 13 + 1;
  • 13 ÷ 2 = 6 + 1;
  • 6 ÷ 2 = 3 + 0;
  • 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) =

108(10) =

0110 1100(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 23 bits, only if necessary (not the case here).


Mantissa (normalized) =

1. 000 1010 0011 1110 1000 0111 =

000 1010 0011 1110 1000 0111


13. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (8 bits) =
0110 1100

Mantissa (23 bits) =
000 1010 0011 1110 1000 0111


Decimal number -0.000 002 06 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 1100 - 000 1010 0011 1110 1000 0111

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How to convert decimal numbers from base ten to 32 bit single precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 32 bit single 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 base ten 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 of the previous dividing 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 previous multiplying operations, starting from the top of the constructed list 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, by shifting the decimal point (or if you prefer, the decimal mark) "n" positions either to the left or to the right, so that only one non zero digit remains to the left of the decimal point.
  • 7. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign if the case) and adjust its length to 23 bits, either by removing the excess bits from the right (losing precision...) or by adding extra '0' bits 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 -25.347 from decimal system (base ten) to 32 bit single precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-25.347| = 25.347

  • 2. First convert the integer part, 25. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 25 ÷ 2 = 12 + 1;
    • 12 ÷ 2 = 6 + 0;
    • 6 ÷ 2 = 3 + 0;
    • 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:

    25(10) = 1 1001(2)

  • 4. Then convert the fractional part, 0.347. 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) 0.347 × 2 = 0 + 0.694;
    • 2) 0.694 × 2 = 1 + 0.388;
    • 3) 0.388 × 2 = 0 + 0.776;
    • 4) 0.776 × 2 = 1 + 0.552;
    • 5) 0.552 × 2 = 1 + 0.104;
    • 6) 0.104 × 2 = 0 + 0.208;
    • 7) 0.208 × 2 = 0 + 0.416;
    • 8) 0.416 × 2 = 0 + 0.832;
    • 9) 0.832 × 2 = 1 + 0.664;
    • 10) 0.664 × 2 = 1 + 0.328;
    • 11) 0.328 × 2 = 0 + 0.656;
    • 12) 0.656 × 2 = 1 + 0.312;
    • 13) 0.312 × 2 = 0 + 0.624;
    • 14) 0.624 × 2 = 1 + 0.248;
    • 15) 0.248 × 2 = 0 + 0.496;
    • 16) 0.496 × 2 = 0 + 0.992;
    • 17) 0.992 × 2 = 1 + 0.984;
    • 18) 0.984 × 2 = 1 + 0.968;
    • 19) 0.968 × 2 = 1 + 0.936;
    • 20) 0.936 × 2 = 1 + 0.872;
    • 21) 0.872 × 2 = 1 + 0.744;
    • 22) 0.744 × 2 = 1 + 0.488;
    • 23) 0.488 × 2 = 0 + 0.976;
    • 24) 0.976 × 2 = 1 + 0.952;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 23) 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:

    0.347(10) = 0.0101 1000 1101 0100 1111 1101(2)

  • 6. Summarizing - the positive number before normalization:

    25.347(10) = 1 1001.0101 1000 1101 0100 1111 1101(2)

  • 7. Normalize the binary representation of the number, shifting the decimal point 4 positions to the left so that only one non-zero digit stays to the left of the decimal point:

    25.347(10) =
    1 1001.0101 1000 1101 0100 1111 1101(2) =
    1 1001.0101 1000 1101 0100 1111 1101(2) × 20 =
    1.1001 0101 1000 1101 0100 1111 1101(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

  • 9. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as already demonstrated above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1 = (4 + 127)(10) = 131(10) =
    1000 0011(2)

  • 10. Normalize the mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal point) and adjust its length to 23 bits, by removing the excess bits from the right (losing precision...):

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

    Mantissa (normalized): 100 1010 1100 0110 1010 0111

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 1000 0011

    Mantissa (23 bits) = 100 1010 1100 0110 1010 0111

  • Number -25.347, converted from the decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point =
    1 - 1000 0011 - 100 1010 1100 0110 1010 0111