-0.000 018 1 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 018 1(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
-0.000 018 1(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:

|-0.000 018 1| = 0.000 018 1


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: 0.000 018 1.

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) 0.000 018 1 × 2 = 0 + 0.000 036 2;
  • 2) 0.000 036 2 × 2 = 0 + 0.000 072 4;
  • 3) 0.000 072 4 × 2 = 0 + 0.000 144 8;
  • 4) 0.000 144 8 × 2 = 0 + 0.000 289 6;
  • 5) 0.000 289 6 × 2 = 0 + 0.000 579 2;
  • 6) 0.000 579 2 × 2 = 0 + 0.001 158 4;
  • 7) 0.001 158 4 × 2 = 0 + 0.002 316 8;
  • 8) 0.002 316 8 × 2 = 0 + 0.004 633 6;
  • 9) 0.004 633 6 × 2 = 0 + 0.009 267 2;
  • 10) 0.009 267 2 × 2 = 0 + 0.018 534 4;
  • 11) 0.018 534 4 × 2 = 0 + 0.037 068 8;
  • 12) 0.037 068 8 × 2 = 0 + 0.074 137 6;
  • 13) 0.074 137 6 × 2 = 0 + 0.148 275 2;
  • 14) 0.148 275 2 × 2 = 0 + 0.296 550 4;
  • 15) 0.296 550 4 × 2 = 0 + 0.593 100 8;
  • 16) 0.593 100 8 × 2 = 1 + 0.186 201 6;
  • 17) 0.186 201 6 × 2 = 0 + 0.372 403 2;
  • 18) 0.372 403 2 × 2 = 0 + 0.744 806 4;
  • 19) 0.744 806 4 × 2 = 1 + 0.489 612 8;
  • 20) 0.489 612 8 × 2 = 0 + 0.979 225 6;
  • 21) 0.979 225 6 × 2 = 1 + 0.958 451 2;
  • 22) 0.958 451 2 × 2 = 1 + 0.916 902 4;
  • 23) 0.916 902 4 × 2 = 1 + 0.833 804 8;
  • 24) 0.833 804 8 × 2 = 1 + 0.667 609 6;
  • 25) 0.667 609 6 × 2 = 1 + 0.335 219 2;
  • 26) 0.335 219 2 × 2 = 0 + 0.670 438 4;
  • 27) 0.670 438 4 × 2 = 1 + 0.340 876 8;
  • 28) 0.340 876 8 × 2 = 0 + 0.681 753 6;
  • 29) 0.681 753 6 × 2 = 1 + 0.363 507 2;
  • 30) 0.363 507 2 × 2 = 0 + 0.727 014 4;
  • 31) 0.727 014 4 × 2 = 1 + 0.454 028 8;
  • 32) 0.454 028 8 × 2 = 0 + 0.908 057 6;
  • 33) 0.908 057 6 × 2 = 1 + 0.816 115 2;
  • 34) 0.816 115 2 × 2 = 1 + 0.632 230 4;
  • 35) 0.632 230 4 × 2 = 1 + 0.264 460 8;
  • 36) 0.264 460 8 × 2 = 0 + 0.528 921 6;
  • 37) 0.528 921 6 × 2 = 1 + 0.057 843 2;
  • 38) 0.057 843 2 × 2 = 0 + 0.115 686 4;
  • 39) 0.115 686 4 × 2 = 0 + 0.231 372 8;

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:


0.000 018 1(10) =

0.0000 0000 0000 0001 0010 1111 1010 1010 1110 100(2)

6. Positive number before normalization:

0.000 018 1(10) =

0.0000 0000 0000 0001 0010 1111 1010 1010 1110 100(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 16 positions to the right, so that only one non zero digit remains to the left of it:


0.000 018 1(10) =

0.0000 0000 0000 0001 0010 1111 1010 1010 1110 100(2) =

0.0000 0000 0000 0001 0010 1111 1010 1010 1110 100(2) × 20 =

1.0010 1111 1010 1010 1110 100(2) × 2-16


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): -16

Mantissa (not normalized):
1.0010 1111 1010 1010 1110 100


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-16 + 2(8-1) - 1 =

(-16 + 127)(10) =

111(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;
  • 111 ÷ 2 = 55 + 1;
  • 55 ÷ 2 = 27 + 1;
  • 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) =

111(10) =

0110 1111(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. 001 0111 1101 0101 0111 0100 =

001 0111 1101 0101 0111 0100


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 1111

Mantissa (23 bits) =
001 0111 1101 0101 0111 0100


Decimal number -0.000 018 1 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 1111 - 001 0111 1101 0101 0111 0100

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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