32bit IEEE 754: Decimal ↗ Single Precision Floating Point Binary: -46 955 Convert the Number to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number

Number -46 955(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:

|-46 955| = 46 955

2. 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;
  • 46 955 ÷ 2 = 23 477 + 1;
  • 23 477 ÷ 2 = 11 738 + 1;
  • 11 738 ÷ 2 = 5 869 + 0;
  • 5 869 ÷ 2 = 2 934 + 1;
  • 2 934 ÷ 2 = 1 467 + 0;
  • 1 467 ÷ 2 = 733 + 1;
  • 733 ÷ 2 = 366 + 1;
  • 366 ÷ 2 = 183 + 0;
  • 183 ÷ 2 = 91 + 1;
  • 91 ÷ 2 = 45 + 1;
  • 45 ÷ 2 = 22 + 1;
  • 22 ÷ 2 = 11 + 0;
  • 11 ÷ 2 = 5 + 1;
  • 5 ÷ 2 = 2 + 1;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

3. Construct the base 2 representation of the positive number.

Take all the remainders starting from the bottom of the list constructed above.


46 955(10) =


1011 0111 0110 1011(2)


4. Normalize the binary representation of the number.

Shift the decimal mark 15 positions to the left, so that only one non zero digit remains to the left of it:


46 955(10) =


1011 0111 0110 1011(2) =


1011 0111 0110 1011(2) × 20 =


1.0110 1110 1101 011(2) × 215


5. 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): 15


Mantissa (not normalized):
1.0110 1110 1101 011


6. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =


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


15 + 2(8-1) - 1 =


(15 + 127)(10) =


142(10)


7. 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;
  • 142 ÷ 2 = 71 + 0;
  • 71 ÷ 2 = 35 + 1;
  • 35 ÷ 2 = 17 + 1;
  • 17 ÷ 2 = 8 + 1;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

8. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =


142(10) =


1000 1110(2)


9. 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 adding the necessary number of zeros to the right.


Mantissa (normalized) =


1. 011 0111 0110 1011 0000 0000 =


011 0111 0110 1011 0000 0000


10. 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 1110


Mantissa (23 bits) =
011 0111 0110 1011 0000 0000


The base ten decimal number -46 955 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
1 - 1000 1110 - 011 0111 0110 1011 0000 0000

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