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I'm working on a task I dont really understand. It says

"Define the relation $R\subset \mathbb{ℕ} × \mathbb{ℕ}$ by: $R = \{(a,b) \in \mathbb{N} : a-2 \le b \le a+2\}$.

Draw the graph of the relation, restricted on the natural numbers between $0$ and $10$. Expected answer (motivated): A graph, with $11$ vertices, and an arrow pointing from $V_1$ and $V_2$ if $(V_1, V_2) \in R$."

So $R\subset \mathbb{ℕ} × \mathbb{ℕ}$. $\mathbb{N} \times\mathbb{N}$ is $\{(0,0),(0,1),(0,2),...\}$ and so on all the way to $10$, then $\{(1,0), (1,1), ...\}$ all the what to $10$ agan, and again and so on. U get the point. So I now what kind of numbers I am suppose to work with. When I get to the $R = \{(a,b) ∈ ℕ : a-2 ≤ b ≤ a+2\}$ part, it's kind of more confusing. $(a,b) \in\mathbb{N}$ is fair enough. "$a$" and "$b$" could be a number between $0-10$. The last part is okey too, just choose a number for $a$ and $b$ from set $R$. If true it is in the relation. But how should I draw a graph out of this? I don't kind of see the connection between.
Thanks for all answer! :D

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    $\begingroup$ The basic idea is that you write the numbers $0, 1, 2, ..., 10$ out on a piece of paper and then connect the two numbers $a$ and $b$ if $(a, b) \in R$. Each number is a vertex and $R$ represents the edge list. $\endgroup$ – Noble Mushtak Jun 9 '16 at 19:43
  • $\begingroup$ Are you familiar with the notion of a combinatorial graph? See en.wikipedia.org/wiki/Graph_(discrete_mathematics). $\endgroup$ – Jakob Hansen Jun 9 '16 at 19:44
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Would the graph look something like this? : enter image description here

My drawing skills is not at a top level today, but I think this is how the graph should look like if I did understand u guys right. NOTE: The both way arrows are suppose to be two separated arrow, I just draw both way arrows to save some space. I dont know if it matters.

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  • $\begingroup$ This is correct. Great job! $\endgroup$ – Noble Mushtak Jun 9 '16 at 20:09
  • $\begingroup$ Okey, nice! Thanks 4 the help! :D $\endgroup$ – Christian Iversen Jun 9 '16 at 20:10
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They want you to create a graph where $0, 1, 2, .., 10$ are the verticies and the vertices are connected if and only if $(a, b) \in R$. The relation $R$ is basically an edge list for this graph. We know that: $$(a, b) \in R \iff a-2 \leq b \leq a+2$$

However, this inequality can be more succinctly written as $\lvert a-b \rvert \leq 2$. Thus, two vertices are connected if they have a difference less than or equal to $2$. If we examine each vertex individually, we find:

  • $0$ will be connected to $0, 1, 2$
  • $1$ will be connected to $0, 1, 2, 3$
  • $2$ will be connected to $0, 1, 2, 3, 4$
  • $3$ will be connected to $1, 2, 3, 4, 5$
  • $4$ will be connected to $2, 3, 4, 5, 6$
  • $5$ will be connected to $3, 4, 5, 6, 7$
  • $6$ will be connected to $4, 5, 6, 7, 8$
  • $7$ will be connected to $5, 6, 7, 8, 9$
  • $8$ will be connected to $6, 7, 8, 9, 10$
  • $9$ will be connected to $7, 8, 9, 10$
  • $10$ will be connected to $8, 9, 10$

Thus, by using this rule of connected the vertices of numbers $0, 1, 2, ..., 10$, we can represent $R$ using a graph.

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  • $\begingroup$ Yeah, that's what I have done, but I have that all the vertices is also connected to them self, since it is true with this statement: a−2≤b≤a+2 $\endgroup$ – Christian Iversen Jun 9 '16 at 20:01
  • $\begingroup$ @ChristianIversen That is correct. All of the vertices are connected to themselves. $\endgroup$ – Noble Mushtak Jun 9 '16 at 20:05
  • $\begingroup$ So the picture I added as an answer is how the graph should look like, right ? $\endgroup$ – Christian Iversen Jun 9 '16 at 20:06
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Start with $0.$ (0,0)(0,1),(0,2) are in R.

Make nodes for $0, 1$ and $2$ and draw your arrows from $0 \to 1$ and $0 \to 2$ and a little loop from 0 to itself.

What is $1$ related to? (1,0),(1,1,),(1,2),(1,3).

Make a node for $3$ that is somewhere convenient. Double end the existing arrow $0\iff 1$, make your arrows for $1\to2$ and $1\to3$, and the loop from 1 to 1.

etc.

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