Let $R$ be a relation from $N$ to $N$ defined by $R = \{(a, b) : a, b \in N \text{ and } a = b^2\}$. Is the following statement true?
$(a, b) \in R, (b, c) \in R$ implies $(a, c) \in R$
$(a, b) \in R, (b, c) \in R$ implies $(a, c) \in R$
(N/A) The relation is defined as $R = \{(a, b) : a, b \in N \text{ and } a = b^2\}$.
To check if $(a, b) \in R$ and $(b, c) \in R$ implies $(a, c) \in R$,we test with counterexamples.
Consider $a = 16, b = 4, c = 2$.
Since $16 = 4^2$,we have $(16, 4) \in R$.
Since $4 = 2^2$,we have $(4, 2) \in R$.
Now,we check if $(16, 2) \in R$.
For $(16, 2)$ to be in $R$,it must satisfy $a = b^2$,which means $16 = 2^2$.
Since $16 \neq 4$,$(16, 2) \notin R$.
Therefore,the statement is false.
To check if $(a, b) \in R$ and $(b, c) \in R$ implies $(a, c) \in R$,we test with counterexamples.
Consider $a = 16, b = 4, c = 2$.
Since $16 = 4^2$,we have $(16, 4) \in R$.
Since $4 = 2^2$,we have $(4, 2) \in R$.
Now,we check if $(16, 2) \in R$.
For $(16, 2)$ to be in $R$,it must satisfy $a = b^2$,which means $16 = 2^2$.
Since $16 \neq 4$,$(16, 2) \notin R$.
Therefore,the statement is false.
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