Why does the professor stop converting at the fourth digit?

[beanryu]

I am sorry to bother you people again and I don't really know if this is the right place to put it.

my great professor shows us in pdf format that

0.83(decimal)=0.1101(binary) because

0.83 x 2 = 1.66 carry 1
0.66 x 2 = 1.32 carry 1
0.32 x 2 = 0.64 carry 0
0.64 x 2 = 1.28 carry 1

My question is why he stops doing it at the fourth digit?
it seems you can still do the algarithm from 1.28 onward. why he stops?

THanx a lot... I am drowning.

 

[Dmstifik8ion]

Sometimes you just have to stop sooner or later. Each iteration contributes half as much as the previous to the final precision of the answer. Since .28 will only add a zero value to the sum either the necessary precision has been obtained or the professor is satisfied that the point has been made well enough to satisfy your need for understanding the process? Just a guess.

 

[siddharth]

0.1101(binary) in decimal is,

$$1*2^{-1} + 1*2^{-2} + 0*2^{-3} + 1*2^{-4} = (0.8125)_{10}$$

which is not 0.83(decimal)

 

[beanryu]

So binary # can't exactly represent a decimal number?
THANKS a lot guys!

 

[NateTG]

So binary # can't exactly represent a decimal number?
THANKS a lot guys!

All numbers that can be represented in decimal notation can also be represented in binary notation. If a number has a terminating binary representation, then it also has a terminating decimal representation. However, some numbers, such as $0.2$(Decimal) have a terminating decimal representation, but have a repeating expression in binary.

 

[Integral]

Perhaps your teacher thought you, as the student, might want to continue the process? Have you?