# What is meant by “show the following relations over $A$” in this problem?

## Will someone explain what is being asked for by "show the following relations over $A$"?

Let $A = \{1, 2, 3, 4, 5, 6\}$

Show the following relations over $A$.

1. $R1 = \{(x, y) \mid \log2 x < \log2 y \}$

2. $R2 = \{(x, y) \mid \log2 x = 2 + \log2 y \}$

3. $R3 = \{(x, y) \mid \log2 x = \log2 y \}$

4. $R4 = \{(x, y) \mid 3 = \log2 x + \log2 y \}$

Part II

Will someone help me understand what is meant by the following:

For each of the above relations, indicate whether it is reflexive, symmetric, antisymmetric and/or transitive.

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I'm guessing that they want you define the sets $R_i$ such that the condition on the right holds (so for which $x$ and $y$ the condition holds). – Cameron Williams Jul 26 '13 at 3:19
So I plug in the values from set $A$ into the inequality conditions on the right for each set $R$, and list, for each set $R$, which values from set $A$ satisfied the inequality? – user87509 Jul 26 '13 at 3:21
That is my guess, yes. – Cameron Williams Jul 26 '13 at 3:22
What about part II of the question? We haven't covered any of this in class yet. I'm just trying to get a general understanding of next week's homework. – user87509 Jul 26 '13 at 3:44

Here's a worked example of what I think is being asked for. Suppose

$$R5 = \left\{(x,y) \mid \frac{10}{x^2} < \frac{y}{3}\right\}$$

The set of tuples that satisfy this relation is $$\left\{ (3, 6), (4, 3), (4, 4), (4, 5), (4, 6), (5, 3), (5, 4), (5, 5), (5, 6), (6, 3), (6, 4), (6, 5), (6, 6)\right\}$$ .

A relation is reflexive iff $\forall x \in A. R(x,x)$. This fails for R5 because we do not have, for example, $R5(1,1)$.

A relation is symmetric iff $\forall x,y \in A. R(x,y) \leftrightarrow R(y,x)$. This fails for R5 because we have $R5(4,3)$ but not $R5(3,4)$.

A relation is antsymmetric iff $\forall x,y \in A. R(x,y) \leftrightarrow \neg R(y,x)$. This fails for R5 because we have both $R5(3,6)$ and $R5(6,3)$.

A relation is transitive iff $\forall x,y,z \in A. R(x,y) \wedge R(y,z) \rightarrow R(x,z)$. This holds for R5.

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