32bit IEEE 754: Decimal ↗ Single Precision Floating Point Binary: -128.111 111 111 111 111 111 111 111 111 111 111 111 118 Convert the Number to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number

Number -128.111 111 111 111 111 111 111 111 111 111 111 111 118(10) converted and written in 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:

|-128.111 111 111 111 111 111 111 111 111 111 111 111 118| = 128.111 111 111 111 111 111 111 111 111 111 111 111 118

2. First, convert to binary (in base 2) the integer part: 128.
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;
  • 128 ÷ 2 = 64 + 0;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

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.


128(10) =


1000 0000(2)


4. Convert to binary (base 2) the fractional part: 0.111 111 111 111 111 111 111 111 111 111 111 111 118.

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) 0.111 111 111 111 111 111 111 111 111 111 111 111 118 × 2 = 0 + 0.222 222 222 222 222 222 222 222 222 222 222 222 236;
  • 2) 0.222 222 222 222 222 222 222 222 222 222 222 222 236 × 2 = 0 + 0.444 444 444 444 444 444 444 444 444 444 444 444 472;
  • 3) 0.444 444 444 444 444 444 444 444 444 444 444 444 472 × 2 = 0 + 0.888 888 888 888 888 888 888 888 888 888 888 888 944;
  • 4) 0.888 888 888 888 888 888 888 888 888 888 888 888 944 × 2 = 1 + 0.777 777 777 777 777 777 777 777 777 777 777 777 888;
  • 5) 0.777 777 777 777 777 777 777 777 777 777 777 777 888 × 2 = 1 + 0.555 555 555 555 555 555 555 555 555 555 555 555 776;
  • 6) 0.555 555 555 555 555 555 555 555 555 555 555 555 776 × 2 = 1 + 0.111 111 111 111 111 111 111 111 111 111 111 111 552;
  • 7) 0.111 111 111 111 111 111 111 111 111 111 111 111 552 × 2 = 0 + 0.222 222 222 222 222 222 222 222 222 222 222 223 104;
  • 8) 0.222 222 222 222 222 222 222 222 222 222 222 223 104 × 2 = 0 + 0.444 444 444 444 444 444 444 444 444 444 444 446 208;
  • 9) 0.444 444 444 444 444 444 444 444 444 444 444 446 208 × 2 = 0 + 0.888 888 888 888 888 888 888 888 888 888 888 892 416;
  • 10) 0.888 888 888 888 888 888 888 888 888 888 888 892 416 × 2 = 1 + 0.777 777 777 777 777 777 777 777 777 777 777 784 832;
  • 11) 0.777 777 777 777 777 777 777 777 777 777 777 784 832 × 2 = 1 + 0.555 555 555 555 555 555 555 555 555 555 555 569 664;
  • 12) 0.555 555 555 555 555 555 555 555 555 555 555 569 664 × 2 = 1 + 0.111 111 111 111 111 111 111 111 111 111 111 139 328;
  • 13) 0.111 111 111 111 111 111 111 111 111 111 111 139 328 × 2 = 0 + 0.222 222 222 222 222 222 222 222 222 222 222 278 656;
  • 14) 0.222 222 222 222 222 222 222 222 222 222 222 278 656 × 2 = 0 + 0.444 444 444 444 444 444 444 444 444 444 444 557 312;
  • 15) 0.444 444 444 444 444 444 444 444 444 444 444 557 312 × 2 = 0 + 0.888 888 888 888 888 888 888 888 888 888 889 114 624;
  • 16) 0.888 888 888 888 888 888 888 888 888 888 889 114 624 × 2 = 1 + 0.777 777 777 777 777 777 777 777 777 777 778 229 248;
  • 17) 0.777 777 777 777 777 777 777 777 777 777 778 229 248 × 2 = 1 + 0.555 555 555 555 555 555 555 555 555 555 556 458 496;
  • 18) 0.555 555 555 555 555 555 555 555 555 555 556 458 496 × 2 = 1 + 0.111 111 111 111 111 111 111 111 111 111 112 916 992;
  • 19) 0.111 111 111 111 111 111 111 111 111 111 112 916 992 × 2 = 0 + 0.222 222 222 222 222 222 222 222 222 222 225 833 984;
  • 20) 0.222 222 222 222 222 222 222 222 222 222 225 833 984 × 2 = 0 + 0.444 444 444 444 444 444 444 444 444 444 451 667 968;
  • 21) 0.444 444 444 444 444 444 444 444 444 444 451 667 968 × 2 = 0 + 0.888 888 888 888 888 888 888 888 888 888 903 335 936;
  • 22) 0.888 888 888 888 888 888 888 888 888 888 903 335 936 × 2 = 1 + 0.777 777 777 777 777 777 777 777 777 777 806 671 872;
  • 23) 0.777 777 777 777 777 777 777 777 777 777 806 671 872 × 2 = 1 + 0.555 555 555 555 555 555 555 555 555 555 613 343 744;
  • 24) 0.555 555 555 555 555 555 555 555 555 555 613 343 744 × 2 = 1 + 0.111 111 111 111 111 111 111 111 111 111 226 687 488;

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...)


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:


0.111 111 111 111 111 111 111 111 111 111 111 111 118(10) =


0.0001 1100 0111 0001 1100 0111(2)


6. Positive number before normalization:

128.111 111 111 111 111 111 111 111 111 111 111 111 118(10) =


1000 0000.0001 1100 0111 0001 1100 0111(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 7 positions to the left, so that only one non zero digit remains to the left of it:


128.111 111 111 111 111 111 111 111 111 111 111 111 118(10) =


1000 0000.0001 1100 0111 0001 1100 0111(2) =


1000 0000.0001 1100 0111 0001 1100 0111(2) × 20 =


1.0000 0000 0011 1000 1110 0011 1000 111(2) × 27


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): 7


Mantissa (not normalized):
1.0000 0000 0011 1000 1110 0011 1000 111


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =


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


7 + 2(8-1) - 1 =


(7 + 127)(10) =


134(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;
  • 134 ÷ 2 = 67 + 0;
  • 67 ÷ 2 = 33 + 1;
  • 33 ÷ 2 = 16 + 1;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 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) =


134(10) =


1000 0110(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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =


1. 000 0000 0001 1100 0111 0001 1100 0111 =


000 0000 0001 1100 0111 0001


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) =
1000 0110


Mantissa (23 bits) =
000 0000 0001 1100 0111 0001


The base ten decimal number -128.111 111 111 111 111 111 111 111 111 111 111 111 118 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
1 - 1000 0110 - 000 0000 0001 1100 0111 0001

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