# Unknown maths topic, does the numbers hold

Let $A=\{a,b,c,d,e,f\}$ and let $R\subseteq A\times A$ 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: $$R(e,a),\quad R(b,f),\quad R(e,c),\quad R(d,a),\quad R(b,a)\;.$$ Given all the above information does $R(c,f)$ hold or not? Explain your answer using the information you have been provided with above.

Hey i have the questions

Could someone please tell me if this is correct

$R(b,f)$ holds therefore $R(c,b)$ holds
$R(e,c)$ holds, $R(b,a)$ holds and $R(a,e)$ holds therefore $R(b,e)$ holds
therefore $R(c,f)$ holds

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• You can upload images to stackexchange, although it would be preperafble to transcribe it, it helps others and isn't that hard. – Jorge Fernández-Hidalgo Dec 2 '13 at 22:17
• How are you getting from $R(b,f)$ to $R(c,b)$? Your steps don't seem to be in order. – Henry Swanson Dec 2 '13 at 22:19

Judging by the rest of your answer, I suspect that $R(c,b)$ in the first line is a typo for $R(f,b)$. If so, you have many of the necessary pieces, but you’ve not explained how they fit together, and some of your steps don’t make sense. Here’s how I would do it if I keep as much as possible of what you’ve written.

$R(e,a)$ holds, so by symmetry $R(a,e)$ holds. $R(b,a)$ also holds, so $R(b,e)$ holds by transitivity. $R(e,c)$ holds, so by transitivity $R(b,c)$ holds, and by symmetry $R(c,b)$ holds. Finally, $R(b,f)$ holds, so $R(c,f)$ holds by transitivity.

first note we have simmetry, so the order of the elements in the pairs matters not.

using this we have:

$R(f,b)$ and $R(b,a) \rightarrow R(f,a)$.

$R(a,e)$ and $R(e,c) \rightarrow R(a,c)$

using this we have

$R(f,a)$ and R(a,c)$\rightarrow R(fc)$