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Define a graph $G$ such that $V(G) = \{2,3,4,5,11,12,13,14\}$ and two vertices $s$ and $t$ are adjacent if and only if $\gcd\{s,t\} = 1$. Draw a diagram of $G$ and find its size $e(G)$.

I can understand V(G) = {2,3,4,5,11,12,13,14} but what are "two vertices $s$ and $t$ are adjacent if and only if $\gcd\{s,t\} = 1$"?

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A graph is a bunch of dots on your paper, called "vertices". Each dot has a label, which is its name. In your example there are eight dots, named 2, 3, 4, 5, 11, 12, 13, and 14.

Sometimes two vertices are connected by a line, and sometimes they aren't. When two dots are connected, we say they are "adjacent". The line is called an "edge".

This question is asking about a graph where two vertices are connected whenever their labels (which are numbers) have no common divisor bigger than one. For example, vertices 4 and 14 are not connected because the numbers 4 and 14 are both divisible by 2. But vertices 5 and 14 are connected because there is no number bigger than 1 that divides both 5 and 14. We write $\gcd(4, 14)$ for the greatest number, 2, that divides both 4 and 14, and $\gcd(5,14)$ for the greatest number, 1, that divides both 5 and 14.

Your job is to draw all the connections between the dots. You should decide if 2 and 3 are connected, and then draw an edge between them if so. Then decide if 2 and 4 are connected, and so on.

$e(G)$ is just the total number of edges in the graph.

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Thank you! it was simple! – Intellectual_ Jul 22 '12 at 19:23
I'm glad I could help. – MJD Jul 22 '12 at 21:07
+1 That's a great, simple, clear, elementary explanation! – M Turgeon Jul 22 '12 at 21:20

You have $8$ vertices, labelled $2$, $3$, and so on. Now we need to determine the edges.

Look for example at the vertices $2$ and $3$. Are they joined by an edge? They are to be joined precisely if $\gcd(2,3)=1$. The greatest common divisor of $2$ and $3$ is indeed $1$, so draw an edge joining $2$ and $3$.

Are vertices $2$ and $4$ joined by an edge? Well, $\gcd(2,4)=2\ne 1$, so no edge.

Are vertices $2$ and $5$ joined by an edge? Yes, because $\gcd(2,5)=1$. Continue.

Vertex $2$ will be joined to $3$, $5$, $11$, $13$. Now we have produced all the edges that involve $2$.

In addition, $3$ is joined to $4$, $5$, $11$, $13$, $14$.

In addition, $4$ is joined to $5$, $11$, and $13$.

In addition, $5$ is joined to $11$, $12$, $13$, $14$.

In addition, $11$ is joined to $12$, $13$, $14$.

In addition, $12$ is joined to $13$.

And finally, $13$ is joined to $14$.

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oh! this was simple!!! thank you! – Intellectual_ Jul 22 '12 at 19:00

$G=G(V,E)$ is a graph, where $V$ is the set of vertices - in your case the vertices "have names'' or correspond to natural numbers.

Now $E$ is the set of edges, in your case $E:=\{(u,v):gcd(u,v)=1\}$ and by $gcd(u,v)$ I mean the gcd of the natural numbers corresponding to $u,v$.

For example there is an edge between the vertices corresponds to the numbers $1$ and $2$ i.e. $(1,2)\in E$ since $gcd(1,2)=1$ but $(2,4)\not\in E$ since $gcd(2,4)=2\neq1$.

Note:$gcd$ - is the greatest common divisor

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Here's a drawing of the graph in question.

A drawing of the graph

The $7$ dashed lines represent non-edges, and the $\binom{7}{2}-7=14$ solid lines are the present edges.

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