## How to convert decimal numbers from the decimal system 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 number to be converted is negative, we start with the positive version of the number.
- 2. First convert the integer part. Divide repeatedly by 2 the positive representation of the integer number that is to be converted to binary, keeping track of each remainder. STOP when the last quotient of the operations is ZERO.
- 3. Construct the base 2 representation of the positive integer part of the number, by taking all the remainders 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 repeatedly by 2, 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 constructed list (in the order they have been calculated).
- 6. Normalize the binary representation of the number, shifting the decimal point n positions to the left (or, if the case, to the right) so that only one non zero decimal place is left in the number.
- 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) and adjust its length to 23 bits, by removing the excess bits, from the right (losing precision...) or adding '0' bits.
- 9. Sign (1 bit) is 1, if it's a negative number or 0, if it's a positive number.

### Example: convert 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 repeatedly by 2, keeping track of each remainder:
iteration division quotient remainder 1 25 : 2 = 12 **1**2 12 : 2 = 6 **0**3 6 : 2 = 3 **0**4 3 : 2 = 1 **1**5 1 : 2 = 0 **1**Last quotient is ZERO => FULL STOP - 3. Construct the base 2 representation of the integer part of the number, by taking all the remainders 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:
# multiplying integer fractional part 1 0.347 x 2 = **0**0.694 2 0.694 x 2 = **1**0.388 3 0.388 x 2 = **0**0.776 4 0.776 x 2 = **1**0.552 5 0.552 x 2 = **1**0.104 6 0.104 x 2 = **0**0.208 7 0.208 x 2 = **0**0.416 8 0.416 x 2 = **0**0.832 9 0.832 x 2 = **1**0.664 10 0.664 x 2 = **1**0.328 11 0.328 x 2 = **0**0.656 12 0.656 x 2 = **1**0.312 13 0.312 x 2 = **0**0.624 14 0.624 x 2 = **1**0.248 15 0.248 x 2 = **0**0.496 16 0.496 x 2 = **0**0.992 17 0.992 x 2 = **1**0.984 18 0.984 x 2 = **1**0.968 19 0.968 x 2 = **1**0.936 20 0.936 x 2 = **1**0.872 21 0.872 x 2 = **1**0.744 22 0.744 x 2 = **1**0.488 23 0.488 x 2 = **0**0.976 24 0.976 x 2 = **1**0.952 No fractionary part that was equal to zero has been obtained through calculations. But we have enough iterations (over Mantissa limit = 23) and at least one integer part different of zero was calculated => 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 multiplying operations, starting from the top of the constructed list:

0.347_{(10)}= 0.0101 1000 1101 0100 1111 1101_{(2)} - 6. Recap - 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 decimal place is in the number:

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

= 1 1001.0101 1000 1101 0100 1111 1101_{(2)}* 2^{0}=

= 1.1001 0101 1000 1101 0100 1111 1101_{(2)}* 2^{4} - 8. Up to this moment, there are the following elements that will feed into the 32 bit single precision IEEE 754 binary floating point:

Sign: 1 (a negative number)

Exponent (unadjusted): 4

Mantissa (non-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 by 2, as shown above:

Exponent (adjusted) = Exponent (unadjusted) + 2^{(8-1)}- 1 =

= (4 + 127)_{(10)}= 131_{(10)}= 1000 0011_{(2)} - 10. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign) and adjust its length to 23 bits, by removing the excess bits, from the right (losing precision...):

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