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I am given that R is a relation on the given set X, and I have to show if the relation is

  1. (i) reflexive,
  2. (ii) symmetric,
  3. (iii) transitive,
  4. (iv) asymmetric, and
  5. (v) give an example of an element of the relation.

X= the positive $\mathbf{Z}$ (the positive integers)and R is the relation defined by nRm if and only if there is a nonzero $k \in \mathbf{Q}$ for which $n^k=m$.

I don't even know where to start. Keep in mind that I am in an introductory to higher mathematics course so I may not be entirely familiar with really advanced concepts.

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  • $\begingroup$ What did you get when you tried to work out the definitions? $\endgroup$ – Kaladin Mar 30 '14 at 21:14
  • $\begingroup$ @Kaladin I don't even know how to work out the definitions. $\endgroup$ – KawaiiExpert Mar 30 '14 at 21:23
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I'll start with the definitions. Note that '$\wedge$' is the logical AND operator.

  • R is reflexive on a set $\mathbb{X}$ if and only if: $$\forall a \in \mathbb{X}, aRa$$
  • R is symmetric if and only if: $$aRb \Leftrightarrow bRa$$
  • R is transitive if and only if: $$aRb \wedge bRc \Rightarrow aRc$$
  • R is asymmetric if and only if: $$aRb \wedge bRa \Rightarrow a=b$$

The example relation you defined was: $$R = \left\{(a,b) :a^{k}=b \textit{ for some } k \in \mathbb{Q},k\ne 0\right\}$$

R is reflexive because $\forall a \in \mathbb{Z_{+}}, a^{1} = a$. Setting k=1 will yield the equality $a=a$ which is true for all elements of $\mathbb{Z_{+}}$. Thus every element is related to itself.

R is symmetric because: $$aRb \Rightarrow a^{k}=b \Rightarrow b^{1/k}=a$$

and

$$(k \in \mathbb{Q}) \wedge (k \ne 0) \Rightarrow \frac{1}{k} \in \mathbb{Q}$$ Thus $bRa$ satisfies the definition of the relation and so ultimately, $aRb \Rightarrow bRa$

R is transitive because: $$aRb \Rightarrow a^{k_{1}}=b$$ $$bRc \Rightarrow b^{k_{2}}=c$$ Substituting $b$ in the second expression for $a^{k_{1}}$ in the first expression yields: $$(a^{k_{1}})^{k_{2}}=a^{k_{1} k_{2}}=c$$ and so there exists $k =k_{1} k_{2}$ such that $a^{k} = c$. Note once again that $k$ is a nonzero rational number and so $aRc$ satisfies the definition of the relation.

R is not antisymmetric. A simple counterexample would be to notice that: $$2R4 \wedge 4R2$$ because $$2 = 4^{\frac{1}{2}} \textit{ and } 4 = 2^{2}$$ This contradicts the definition of antisymmetry which says that for two elements to be related to eachother, they must be equal. Thus R is not antisymmetric.

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I will try to explain a approach for part ii) maybe that gives you insight on how to tackle the other parts.

ii) Symmetric: if we have $n\sim m$ then there is a $q\in \mathbb{Q}$ such that $n^q=m$. Now for we want to know if it is symmetric, so then we want to know if $m\sim n$. In other words is there a $p\in \mathbb{Q}$ so that $m^p=n$? (Hint: look at $n^q=m$, then raise both sides to a certain power.)

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  • $\begingroup$ I'm a little confused on what you said in the beginning. The tilde means equivalence right? So you are saying that n is equivalent to m? Sorry if I sound so naive, but what allows you to say that? $\endgroup$ – KawaiiExpert Mar 30 '14 at 21:37
  • $\begingroup$ No problem everyone needs to start somewhere. Symmetric means that if $n\sim m$ (so indeed $m$ equivalent to $n$) then we have also that $m \sim n$. So we assume $n \sim m$ and try to proof(disprove) $m \sim n$. $\endgroup$ – Kaladin Mar 30 '14 at 21:41
  • $\begingroup$ I don't know if I am going about this the right way but as an example to wrap my head around it, I have n=4, m=2, q=1/2, p=2 $4^1/2=2$ and $2^2=4$ $\endgroup$ – KawaiiExpert Mar 30 '14 at 21:44
  • $\begingroup$ Yes, that is correct. How did you do this? $\endgroup$ – Kaladin Mar 30 '14 at 21:46
  • $\begingroup$ Thank you. So for the rest of these, I should start with the assumption of n~m? $\endgroup$ – KawaiiExpert Mar 30 '14 at 21:48

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