Calculating Decimal points of modulo operations I am not too sure of the step that has occurred here (I found this form a cryptography book but I do not understand how he finds the second 9).
The function found in the  image is
$(2 \cdot 1)^{-1}(3 \cdot 5^2 + 2) = 2^{-1} \cdot 9 \equiv 9 \cdot 9 \equiv 13\,\mod\,17$
My calculations are that:
$(2 \cdot 1)^{-1}(3 \cdot 5^2 + 2) = 2^{-1} \cdot 9 = 4.5 \not\equiv 9 \cdot 9 \equiv 13\,\mod\,17$
Equation from Book

In modulo arithimetic then expression $(2^{-1})$ does not mean $0.5$ which is completely out of the purview of modular arithmetic (which is about equivalence classes of integers).
Instead $(2^{-1})$ refers to the the solution to  $2x \equiv 1\pmod n$
And $2x \equiv 1 \pmod {17}\iff x \equiv 9 \pmod {17}$[1]
So $2^{-1}\equiv 9 \pmod {17}$.
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[1]  $2x \equiv 1 \pmod {17} \implies \exists k\in \mathbb Z$ so that $2x = 1 + 17k$ so $2|1+17k$ but $2\not \mid 1$ so $2\not \mid 17k$ so $k$ is odd.  If $k = 2m + 1$ then $2x = 1 + (2m+1)17 = 18 + 34m$ which means $x = 9 + 17m$ or $x \equiv 9 \pmod {17}$ is the only solution ($\pmod {17}$).  
