Intro to Proofs: Properties of Relations

[Dembadon]

Hello, I would like to check my arguments for this problem.

Homework Statement

Consider the relation [itex] R = \{(x,y) \in \mathbb{R} \times \mathbb{R}: x-y \in \mathbb{Z}\}[/itex] on [itex]\mathbb{R} [/itex]. Prove that this relation is symmetric, reflexive, and transitive.

Homework Equations

Supposing a relation [itex]R[/itex] on a set [itex]A[/itex].

Reflexivity: Relation [itex] R [/itex] is reflexive if [itex] \forall x \in A, xRx [/itex].

Symmetry: Relation [itex] R [/itex] is symmetric if [itex] \forall x,y \in A, xRy \Rightarrow yRx [/itex].

Transitivity: Relation [itex] R [/itex] is transitive if [itex] \forall x,y,z \in A, \left((xRy) \land (yRz)\right) \Rightarrow xRz [/itex].

The Attempt at a Solution

Reflexivity:

We choose any [itex] x \in \mathbb{R} [/itex] and discover that [itex] x - x = 0 [/itex], which is in [itex] \mathbb{Z} [/itex]. Therefore, we have [itex] xRx [/itex], showing that [itex] R [/itex] is reflexive on [itex] \mathbb{R} [/itex].

Symmetry:

We can argue directly by assuming [itex] xRy [/itex]. This relation means we have [itex] (x - y) \in \mathbb{Z} [/itex]. It follows that [itex] -(x - y) [/itex], which is [itex] yRx [/itex], is also in [itex] \mathbb{Z} [/itex].

Transitivity:

If [itex] xRy [/itex] and [itex] yRz [/itex] are both integers, then adding them yields another integer.

 

[vela]

Hello, I would like to check my arguments for this problem.

Homework Statement

Consider the relation [itex] R = \{(x,y) \in \mathbb{R} \times \mathbb{R}: x-y \in \mathbb{Z}\}[/itex] on [itex]\mathbb{R} [/itex]. Prove that this relation is symmetric, reflexive, and transitive.

Homework Equations

Supposing a relation [itex]R[/itex] on a set [itex]A[/itex].

Reflexivity: Relation [itex] R [/itex] is reflexive if [itex] \forall x \in A, xRx [/itex].

Symmetry: Relation [itex] R [/itex] is symmetric if [itex] \forall x,y \in A, xRy \Rightarrow yRx [/itex].

Transitivity: Relation [itex] R [/itex] is transitive if [itex] \forall x,y,z \in A, \left((xRy) \land (yRz)\right) \Rightarrow xRz [/itex].

The Attempt at a Solution

Reflexivity:

We choose any [itex] x \in \mathbb{R} [/itex] and discover that [itex] x - x = 0 [/itex], which is in [itex] \mathbb{Z} [/itex]. Therefore, we have [itex] xRx [/itex], showing that [itex] R [/itex] is reflexive on [itex] \mathbb{R} [/itex].

Symmetry:

We can argue directly by assuming [itex] xRy [/itex]. This relation means we have [itex] (x - y) \in \mathbb{Z} [/itex]. It follows that [itex] -(x - y) [/itex], which is [itex] yRx [/itex], is also in [itex] \mathbb{Z} [/itex].

It depends on how nit-picky you want to get, but it's probably better to say ##y-x \in \mathbb{Z}##, rather than ##-(x-y) \in \mathbb{Z}##, implies yRx.

Transitivity:

If [itex] xRy [/itex] and [itex] yRz [/itex] are both integers, then adding them yields another integer.

Right idea, but you should write out explicitly how xRy and yRz imply x-z is in Z.

 

[LCKurtz]

Hello, I would like to check my arguments for this problem.

Homework Statement

Consider the relation [itex] R = \{(x,y) \in \mathbb{R} \times \mathbb{R}: x-y \in \mathbb{Z}\}[/itex] on [itex]\mathbb{R} [/itex]. Prove that this relation is symmetric, reflexive, and transitive.

Homework Equations

Supposing a relation [itex]R[/itex] on a set [itex]A[/itex].

Reflexivity: Relation [itex] R [/itex] is reflexive if [itex] \forall x \in A, xRx [/itex].

Symmetry: Relation [itex] R [/itex] is symmetric if [itex] \forall x,y \in A, xRy \Rightarrow yRx [/itex].

Transitivity: Relation [itex] R [/itex] is transitive if [itex] \forall x,y,z \in A, \left((xRy) \land (yRz)\right) \Rightarrow xRz [/itex].


Transitivity:

If [itex] xRy [/itex] and [itex] yRz [/itex] are both integers, then adding them yields another integer.

You have the idea but you need to rewrite the transitive one. xRy is not an integer and neither is yRz, and you can't add them. Write more carefully what you mean.

 

[Dembadon]

It depends on how nit-picky you want to get, but it's probably better to say ##y-x \in \mathbb{Z}##, rather than ##-(x-y) \in \mathbb{Z}##, implies yRx.

Gotcha.

Right idea, but you should write out explicitly how xRy and yRz implies x-z is in Z.

I understand. This is what I had in my head but omitted from my argument:

(x - y) + (y - z) = x - z, which is xRz.

I appreciate your input.