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QS: Indicate if the relation on the given set are reflexive on a given set, which are symmetric, and which are transitive.

$\not = \text{on } \Bbb N$

So for this problem I am trying to comprehend why this question is not transitive.

$(i)$ Reflexive: No because $\forall x \in \Bbb N$ thus $x \not = x $.

$(ii)$ Symmetric : Yes because if $ x \not = y \; \Rightarrow y\not = x.$ Thus $xRy \rightarrow yRx.$

$(iii)$ Transitive: $x \not = y , y\not=z, x\not = z.$

At least this is what I think, but this is wrong and I want to understand my mistake. One last question are $(x,y) \in \Bbb N$?

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    $\begingroup$ Take $x = 1, y = 2, z = 1$. $\endgroup$ – Parth Kohli Jul 8 '16 at 18:21
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    $\begingroup$ $x\in \Bbb N$ and $y\in \Bbb N$ would imply $(x,y)\in\Bbb N^2$ but $(x,y)$ is a tuple, not a single number and so it not itself an element of $\Bbb N$. For disproving the various relations, take a specific counterexample. Show why the counterexample satisfies the premises but does not satisfy the conclusion. $\endgroup$ – JMoravitz Jul 8 '16 at 18:22
  • $\begingroup$ Then this is not transitive? $\endgroup$ – Jon Jul 8 '16 at 18:23
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    $\begingroup$ Your phrasing is a bit off too. "No because $\forall x \in \Bbb N$ thus $x\neq x$" does not make grammatical sense. Suggest replacing with the phrase "No (it is not reflexive) because $\forall x\in \Bbb N$ you have $x=x$, thus $x\neq x$ is false." Alternatively, just a single counterexample will work. Let $x=1$. Then it is not true that $x \neq x$. $\endgroup$ – JMoravitz Jul 8 '16 at 18:24
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A simple counter example will suffice. Take $x=z=1$, $y=2$. Then $x \neq y$ and $y\neq z$, but $x = z$.

As to your last question, we use $x,y \in \mathbb{N}$ to mean that both $x$ and $y$ are natural numbers. This differs from $(x, y) \in \mathbb{N^2}$ where $(x,y)$ is an ordered pair.

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  • $\begingroup$ When should you use an ordered pair? Would it be better like this $x \in \Bbb N \; and \; y \in \Bbb N$. $\endgroup$ – Jon Jul 8 '16 at 18:36
  • $\begingroup$ This is what I meant sorry. $(x,y) = 1 $ $\endgroup$ – Jon Jul 8 '16 at 18:37
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    $\begingroup$ $(x, y) = 1$ dosen't really make sense (unless you interpret it as a $gcd$, but I suspect that's not what you're after). That is an ordered pair is exactly that, a pair of values, so assigning it one single value doesn't make sense. You could say $(x, y) = (1, 1)$ to denote $x=1, y=1$ if you'd like. My answer was pointing out the difference between $x, y \in \mathbb{N}$ to mean that both $x$ and $y$ are natural numbers and $(x, y) \in \mathbb{N}$ which dosen't make sense as an ordered pair is a pair and cannot be a single number in $\mathbb{N}$. $\endgroup$ – Zain Patel Jul 8 '16 at 18:50
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The two things you wanted to know:

  1. THE MISTAKE:

$x \not= y \land y \not= z \not \implies x \not =z$

Take any counter-example, as in Parth Kohli's comment.

  1. $(x,y)$ is an ordered pair, so essentially $(x,y) \in \mathbb{N} \times \mathbb{N}$.
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