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Addition of two numbers in IEEE 754 format

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A) a=3f200000, b=be600000
B) a=3f200000, b=ff800000
C) a=01100000, b=80e00000

I have no idea were to start really. Any help will be great

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A): a=3f200000, b=be600000
representing these in binary form we have,
a= 0 01111110 01000000000000000000000
b= 1 01111100 11000000000000000000000

The first bit is representing the sign of the number, the next 8 bits are representing the exponent and the next 23 bits are showing the mantissa.
Now for addition the first step is aligning the binary point.

Step 1:
Binary points are aligned as,
compare the exponent part of both numbers and right shift the mantissa part of the number whose exponent part is small. This shift will be equal to the difference between the two exponents.

In our case,
exponent of a > exponent of b
01111110 > 01111100
difference is: 01111110 - 01111100= 010 (or 2 in decimal number)
right shift mantissa of b by two position
11000000000000000000000 will become 01110000000000000000000
note the first one in the new mantissa of b. This is the hidden bit of mantissa which is always one.

Step 2:
Note the sign ...

Solution Summary

This solution provides the addition of two numbers in IEEE 754 format.

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