Let $A={a,b,c,d,e,f}$ and let $R\subset A\times A$ be a relation which is symmetric and transitive. You have been given some partial information about the relation which is that the following are known to be true for the relation:
$$R(e,a), R(b,f), R(e,c), R(d,a), R(b,a)$$
Given all the above information does $R(c,f)$ hold or not? Explain your answer using the information you have been provided with above.
I would argue that $R(c,f)$ does not hold. No two pairs would equal $R(c,f)$ and $R(f,c)$.
Does this seem correct?
Update: Just thought of this.
R(b,f) and R(b,a) would be the same as R(f,a).
R(f,a) and R(e,a) would be the same as R(f,e).
Therefore, R(f,e) and R(e,c) would be the same as R(f,c).
As the relation is symmetric, R(f,c) -> R(c,f).
How's that?
This is a great question to apply some methods from Polya's 'How to Solve it'. Namely, can you draw a picture???
Because the relation $R$ is symmetric and transitive, it makes a lot of sense to draw this like a mathematical graph (that's different from a 'graph' of $f(x)=x^2$, by the way - wiki it).
So if you draw $a, b, c, d, e, f$ as points on your page, and connect them with lines whenever you know they're related, the question becomes - is there a path from $c$ to $f$?
If there was any such path, using, for example, $c\mapsto e\mapsto a\mapsto b\mapsto f$, then you would know, using symmetry: $$R(c, e), R(e, a), R(a, b), R(b, f)$$ Applying transitivity to the first pair, you could 'walk' from $c$ to $a$, concluding $R(c, a)$. Then repeat, until you conclude $R(c, f)$!