# Hasse Diagram for cube functions [duplicate]

How do you draw a hasse Diagram for the following example

Consider a relation $R$ deﬁned on the set $A = \{-7,-6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6,7\}$. Determine for the $R = \{(a, b) : a^3 = b^3\}$ if the relation is reﬂexive, symmetric, anti -symmetric, transitive, partial orders or an equivalence relation.

I have figured out whether it is reﬂexive, symmetric, anti -symmetric, transitive, partial orders or equivalence relations. But cant draw the hasse diagram.

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## marked as duplicate by MJD, azimut, Davide Giraudo, Amzoti, Ishan BanerjeeApr 8 '13 at 14:32

Isn't $R$ the identity relation (equality) on $A$? Each element is connected with and only with itself. – Berci Apr 7 '13 at 0:17
@Berci do u mean to say its not ant-symmetric? If soo could you explain why? – jack Apr 7 '13 at 0:24
I meant that $R==$. And that is antisymmetric: $(a=b) \land (b=a)\implies a=b$. – Berci Apr 7 '13 at 0:52
I mean how would you draw a hasse diagram in which each element is connected with itself? – jack Apr 7 '13 at 0:59
@jack : You put your $\{\text{curly braces}\}$ OUTSIDE of $\TeX$. Standard size and spacing conventions are not followed when you do that. Look at my recent edit. – Michael Hardy Apr 7 '13 at 1:03

We have a partial order $R$, so a Hasse diagram is possible. As always, we have an upward edge between two vertices $i$ and $j$, with $i \neq j$, whenever $iRj$ and there is no $k$ such that $iRkRj$.
In this case, as noted in the comments, \begin{align*} R &= \{(a,b):a^3=b^3\} \\ &= \{(a,b):a=b\} \\ &= \{(a,a):a \in \{-7,\ldots,7\}\}. \\ \end{align*} So $R$ is the partial order $=$.
So, the edges are between $i,j \in \{-7,\ldots,7\}$, with $i \neq j$, such that $i=j$. This never happens, so there's no edges. So here is the Hasse diagram: