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?
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3Apply the definitions.... $R$ is reflexive iff : for every $x \in A : xRx$. – Mauro ALLEGRANZA Jun 12 '20 at 08:18
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1What have you tried? – JMP Jun 12 '20 at 08:19
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1"I was thinking something like..." It is enough to check if the conditions of the corresponding definitions are satisfied or not. – Mauro ALLEGRANZA Jun 12 '20 at 08:32
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@MauroALLEGRANZA i suggest you post an answer because i think you have already answered it here! :) – Omar Shaaban Jun 12 '20 at 08:59
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Please provide additional context, which ideally explains why the question is relevant to you and the community. Some forms of context include background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc. – Sahiba Arora Jun 12 '20 at 09:31
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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.
NimaJan
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