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I have to compare $ (2)_{16} $ and $ (1.\overline{F})_{16} $.

$ (2)_{16} $ is very simple to compute, e.g. $ (2)_{16} = 2 \cdot 16^0 = 2 $. On the other hand, I dont know how to find out what $ (1.\overline{F})_{16} $ is because I don't understand what the letter $ F $ means.

Can anyone help me?

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the numbers are equal –  OBDA Feb 24 at 21:41
    
Thanks. But how can I compute $ (1.\overline{F})_{16} $ to show that the two expressions are equal? I don't see how to do that –  n plus one Feb 24 at 21:44
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in base 16 you have $\{0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F\}$ you can use math.uzh.ch/index.php?file&key1=7451 –  OBDA Feb 24 at 21:46
    
Didn't you think to look up "base 16" first? Surely any discussion of it will include something about the notation. –  Tim Seguine Feb 24 at 22:05
    
@O.B.D.A. I am not so sure if the average reader here is going to find lecture notes in German to be very helpful –  Tim Seguine Feb 24 at 22:15
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3 Answers

In number system of base $16$ we use letters $A,B,C,D,E,F$ for the digits $10,\,11\,\dots15$.

So, the other number is $$ 1 + \frac{15}{16}+\frac{15}{16^2}+\frac{15}{16^3}\dots$$ (Just as $1.\overset\bullet9\ =\ 1+\displaystyle\frac9{10}+\frac9{100}+\dots$ in decimal.)

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This is the hexadecimal analgy to showing that $1.\bar{9} = 2$ in decimal.

The hexadecimal digits are $0-9, A, B, C, D, E, F$, where $A=10_{10}$, $B=11_{10}$, ... , $F=15_{10}$.

For the decimal case, you can express the repeating part as a geometric series, and you'd want to show that

$$\sum_{n=1}^{\infty} \frac{9 }{10^n} = 1.$$

For the hexadecimal case, replace the $9$ with a $15$, and the $10$ with a $16$.

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This is also true for any number base $r$: $$\sum_{n=1}^\infty\frac{r-1}{r^n}=1$$ –  Brian J. Fink Feb 24 at 22:05
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In base 16, F corresponds to the number 15 (the highest one-digit number).

Hence you can see this as comparing $2$ and $1.999...$ in base 10. Can you resolve this now?

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