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If you take n dots (or vertices) and connect them together with lines, you'll have n−1 lines. But why n−1? Please assist me with this triviality. Other than the obvious visual proof, can you prove this more rigorously?

If there's n dots, only the first line requires two unique dots and every subsequent one requires only 1 additional dot. But I still don't see the $n-1$ in that.

Let ⬤ be a dot.

Then ⬤ ------------------------ ⬤ ------------------------ ⬤ denotes 3 dots, 2 lines.

In general, $n$ dots are separated by $n-1$ lines.

For example, I can easily see how $n$ disconnected dots give $n/2$ lines, because there must be $m$ groups of 2 dots or $m$ lines.

NKLost
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  • Can you explain more about what exactly you are talking about? I can't tell. – Mariano Suárez-Álvarez Nov 21 '11 at 16:59
  • If you take $n$ vertices and connect them together with lines, you'll have $n-1$ lines, why $n-1$? Other than the obvious visual proof, is there something that could be more, I don't know, substantial? – NKLost Nov 21 '11 at 17:02
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    Explain that in the body of the question, so that people do not have to read through all comments. – Mariano Suárez-Álvarez Nov 21 '11 at 17:06
  • @NKLost You'll also have to be more clear about how you are connecting the dots. In my mind, connecting 3 dots yields three lines, and 4 dots yield 6 lines. I also see ways in which 234987 dots yield one line. – mboratko Nov 21 '11 at 17:11
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    @NKLost That said, once you sufficiently rigorize the question, induction will work. – mboratko Nov 21 '11 at 17:17
  • Dot, then line, dot, then line, ..., dot, then line, dot. There's one more dot than line. Isn't that enough of a proof? – Dimitrije Kostic Nov 21 '11 at 17:23
  • (A remark from @Pete Clark's notes on induction.) Some scholars have suggested that what is essentially an argument by mathematical induction appears in the later middle Platonic dialogue Parmenides, lines 149a7-c3. But this argument is of mostly historical and philosophical interest. The statement in question is, very roughly, that if $n$ objects are placed adjacent to another in a linear fashion, the number of points of contact between them is $n − 1$. [contd] – Srivatsan Nov 21 '11 at 19:57
  • [contd] (Maybe. To quote the lead in the wikipedia article on the Parmenides: "It is widely considered to be one of the more, if not the most, challenging and enigmatic of Plato’s dialogues.") There is not much mathematics here! Nevertheless, for a thorough discussion of induction in the Parmenides the reader may consult [Ac00] and the references cited therein. – Srivatsan Nov 21 '11 at 19:57
  • Ultimately it doesn't have to be "in a linear fashion" - this generalizes to [finite] trees. (basically, you get to attach the "then line, dot" anywhere instead of at the dot you added most recently) – Random832 Nov 21 '11 at 21:58

3 Answers3

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If you are in $1$ town and you want to visit $n$ towns, then you have to walk to $n-1$ towns.

TROLLHUNTER
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See Fencepost Error. This is really just a matter of counting carefully, and of having a precise understanding of what you are counting.

Here's what I think you mean: You have $n$ vertices $v_1, v_2, v_3, \ldots, v_n$, and you want to count the line segments connecting ($v_1$ and $v_2$), ($v_2$ and $v_3$), $\ldots$, all the way to ($v_{n-1}$ and $v_n$).

We can label each line segment by taking the smaller of the indices of the two vertices $v_i$ and $v_j$ it connects. Then we have $n-1$ line segments, $\ell_1, \ell_2, \ldots, \ell_{n-1}$.

Notice that if we connected the vertices in a loop, we would have one more line segment, connecting ($v_n$ and $v_1$), to make $n$ total line segments. If we have a single chain (not a loop) with $n$ interior vertices, then we have $n+1$ line segments. (That is, we add vertices $v_0$ on one end and $v_{n+1}$ on the other end, making $n+2$ vertices total, but only $n$ interior vertices.)

In counting problems like this, you have to think carefully about the correct way to count.

Jonas Kibelbek
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Instead of counting lines, how about counting left ends of lines? Every dot except the last one is adjacent to exactly one left end, and the last one is adjacent to none.

There are $n-1$ dots-except-the-last-one (which is just how subtraction works, nothing to prove there), so that is also the number of left ends, and therefore also the number of lines.

hmakholm left over Monica
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