is f(x)=1/x reflexive/symmetric/transitive?

Question

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.

Answer 1

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.

Answer 2

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.