What's the difference of $xRy=1$ on $\mathbb {Z}$ or $\mathbb {R}$?

Question

Suppose we define the relation xRy iff xy=1, is this relation:

Reflexive?

Irreflexive?

Symmetric?

Antisymmtric?

Transitive?

From my understanding on Z (integers), the relation is symmetric and transitive.

Symmetric: if xy=1 then yx=1 is true

Transitive: if xy=1 then yz=1, so xz=1

Neither Reflexive or irreflexive: xx=1 iff x=1, so there is at least one instance, not all instances, where xRx, so it's neither (if I am applying this correctly - it's reflexive if xx=1 for all x, and irreflexive if xx!=1 for all x)

However, what's the difference between asking if this is on the Real (R) set or the integer set (Z)? Are they the same?

Answer 1

Reflexive: Fails in both Integers and Reals.

Symmetic: Passes in both Integers and Reals.

Transitive: Passes in Integers but fails in Reals.

We can write the relation as $R_A=\{(x,x^{-1}): x\in A\}$ where $A$ is either $\Bbb Z$ or $\Bbb R$.   As you noticed, when under the Integers, the relation is $R_{\Bbb Z}=\{(1,1),(-1,-1)\}$ as only $1$ and $-1$ have multiplicative inverse (which are themselves) and this trivially transitive.   However, in the Reals there are many more entries in $R_\Bbb R$, which preserves symmetry but breaks transitivity [because if $xy=1\wedge yz=1$ then $x=z$, but $xx=1$ only if $x=\pm 1$].

Answer 2

There’s a big difference between doing this for real numbers vs integers. For one thing, in the integers $xRy\Rightarrow x=y$, which isn’t true for te real numbers. We have $2R0.5$, $0.5R2$, but not $2R2$ so for the reals it is not transitive.

You have the right answers but the wrong analysis for the integers. In addition to $1R1$ there is also $-1R-1$

Answer 3

Of course not. When you extend the set, you could never be guaranteed to keep the property of a relation. It's possible to keep the property only when the extension is somewhat natural and compatible with the property by which you defined the relation.

In this case, when you consider in $R$, the transitivity is broken, (you can easily notice that.)