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Based on your answers to 3.32 and 3.33, does

3.984375×10-1+3.4375×10-1+1.771×103=3.984375×10-1+3.4375×10-1+1.771×103?

Short Answer

Expert verified

No, there is a difference between the two results.

Step by step solution

01

Define the concept

The decimal number system uses base 10 and the binary number system uses base 2.

For example, “12” is also equivalent to the value of the binary number system “1100”.

The decimal number 3.984375×10-1is equal to 1.1001100000×2-2in the binary number system.

The decimal number 3.4375×10-1is equal to 1.0110000000×2-2in the binary number system.

The decimal number 1.771×103is equal to -1.1011101011×210in the binary number system.

Given numbers are 3.984375×10-1, 3.4375×10-1and 1.771×103

Let, x=3.984375×10-1, y=3.4375×10-1and

x+y+z=3.984375×10-1+3.4375×10-1+1.771×103=1772x+y+z=3.984375×10-1+3.4375×10-1+1.771×103=1771

02

Determine the calculation

Given numbers are 3.984375×10-1, 3.4375×10-1and 1.771×103.

Let, x=3.984375×10-1, y=3.4375×10-1and z=1.771×103

x=1.1001100000×2-2y=+1.0110000000×2-2

----------------------------------------------

x+y=10.1111100000[Normalized]10.1111100000=1.01111100000×2-1x+y=1.0111110000z=+1.1011101011[.000000000010111110000]

----------------------------------------------------

localid="1654950526652" x+y+z=1.1011101011101[Roundedup]1.1011101100×210=0110101011101100=1772[decimal]x+y+z=3.984375×10-1+3.4375×10-1+1.771×103=1772y=0.0000000000010110000000z=+1.1011101011

----------------------------------------------------------

y+z=+1.1011101011x=0.0000000000011001100000

----------------------------------------------------------

x+y+z=+1.1011101011+1.1011101011=+1.1011101011×210=0110101011101011=1771[Indecimal]

x+y+z=3.984375×10-1+3.4375×10-1+1.771×103=1771

Hence, the two results are not same.

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Most popular questions from this chapter

Write down the binary representation of the decimal number 63.25 assuming it was stored using the single precision IBM format (base 16, instead of base 2, with 7 bits of exponent).

Question: Assume 151 and 214 are signed 8-bit decimal integers stored in two’s complement format. Calculate 151 - 214 using saturating arithmetic. The result should be written in decimal. Show your work.

Calculate 1.666015625×100×1.9760×104+1.666015625×100×-1.9744×104 by hand, assuming each of the values is stored in the 16-bit half-precision format described in Exercise 3.27 (and also described in the text). Assume 1 guard, 1 round bit, and 1 sticky bit, and round to the nearest even. Show all the steps, and write your answer in both the 16-bit floating-point format and in decimal.

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