is f(x)=1/x reflexive/symmetric/transitive? Determine whether this relationship is reflexive/symmetric/transitive.
Im trying to think of it interms of the graph but i seem to be getting no where. Could someone explain how they would test these properties on the function. 
 A: Firstly you probably mean $x\in\Bbb R\setminus\{0\}$.
It's clearly not symmetric, $(2,2)$ is not in there, for example.
Symmetric is easy ${1\over x^{-1}}=x$ proves that.
Transitive is also easy since the only two ordered pairs with a given $x_0$ in it are $(x_0, x_0^{-1})$ and $(x_0^{-1}, x_0)$, so the only case of $(a,b)$ and $(b,c)$ is when $a=c$, but we already verified symmetry, so transitivity follows.
A: Let the symbol $\sim$ denote a relationship between pairs of
numbers $x$ and $y$, writing $x \sim y$ if $x$ has that relationship to $y$.
In graphical terms, if you plot a point on a Cartesian plane at coordinates
$(x,y)$ for each pair of numbers $x$ and $y$ such that $x \sim y$,
you will have the graph of a function if the relationship $\sim$
has the properties of a function, that is, for each $x$ there is
exactly one $y$ such that $x \sim y$. This is sometimes called the
vertical line test.
Graphically, the relationship $\sim$ is reflexive if the plot of
$x \sim y$ includes every point on the line through the origin at
$45$ degrees sloping upward to the right, namely $y=x$.
If the relationship is only over a subset of the real numbers,
then we only need to see that the plot contains all the points $y=x$
such that $x$ is in the domain of the relationship.
A quick sketch of $y = \frac1x$ immediately shows that it is not reflexive.
Graphically, the relationship $\sim$ is symmetric if the plot of
$x \sim y$ is its own mirror image around the line $y=x$.
That is, we are looking to see if we can flip the graph over that diagonal
line so that the $x$ and $y$ axes trade places, and still have
the exact same graph we started with.
In the case of $y=\frac1x$, a careful sketch of the graph supports the
conclusion that this relationship is symmetric.
Transitivity is harder to explain graphically. It's so much harder to
visualize transitivity this way that I would recommend not to try,
but to use formulas and logic instead.
