How to prove reflexivity, symmetry or antisymmetry and transitivity

Question

Let $A = \lbrace 1, 2, 3, 4, 5, 6, 7 \rbrace$ Define the relation $R$ on $A$ by $xRy ⇔ xy ≥ 10$. Is the relation reflexive, symmetric, anti-symmetric or transitive?

Answer 1

We know that

• $R$ is reflexive if $ \forall x \in A: (x,x) \in R$.

  In your case $xx \geq 10$ has to be satisfied, but this is not reflexive because if we take $x=1$ then $1<10$.

• R is transitive if $\forall x,y,z \in A:((x,y) \in R \text{ and } (y,z) \in R \Rightarrow (x,z) \in R$.

  In your case we need to prove that if $xy \geq 10 \text{ and } yz \geq 10$, then $xz \geq 10$. We can give a counterexample again as if you take $(2,7) \in R \text{ and } (7,3) \in R$, then $(2,3) \notin R$ as $2 \times3=6 < 10$.

• R is symmetric if $\forall x,y \in A : (x,y) \in R \Rightarrow (y,x) \in R$.

  This is trivial in our case as multiplication is commutative.

Next time try to give your shots in the question as well, so we can see where you are struggling. The best tip I can give you is when you are trying to prove something is true and it takes too long for you to prove it, it means that it is probably false so you should start looking for a counterexample. Let me know if you have any questions in the comments.