I recognize we start with the base case, where, if we speak to the above equation $P(n)$, $P(0)$, because that $0$ lines would be $0$. But I really have actually no idea how to begin the inductive step. How do we know what $k+1$ we"re claimed to come at?

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edited Feb 16 "14 in ~ 16:37

amWhy

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asked Mar 15 "13 in ~ 17:30

Muath05Muath05

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Let $P(n)$ be the statement that the variety of straight heat segments figured out by $n$ points in the plane no three of i m sorry lie ~ above the same straight heat is $\fracn^2-n2$.

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When over there is $1$ point, there room $0=\frac1^2-12$ line segments. Thus we have $P(1)$.

Now suppose we have actually $P(k)$ for some optimistic integer $k$. Then $k$ points determine $\frack^2-k2$ heat segments. Once we include another point, we attach this allude to each of the present $k$ lines. So now we have actually $\frack^2-k2+k=\frack^2-k+2k2=\frack^2+k2=\frac(k+1)^2-(k+1)2$ heat segments. This way that we have actually $P(k+1)$.

Hence we have actually $P(n)$ for all hopeful integers $n$.

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reply Mar 15 "13 in ~ 17:38

user4594user4594

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Say you"ve obtained $k$ points and the variety of unordered pairs of them---hence the variety of lines---is $\dfrack^2-k2$.

Add a $(k+1)$th point. You deserve to pair the brand-new point with any of the old $k$ points, gaining $k$ brand-new lines.

So the number of lines is now$$\frack^2-k2 + k.$$

So currently all you require is to prove the that"s the same as $\dfrac(k+1)^2-(k+1)2$.

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answer Mar 15 "13 in ~ 17:46

Michael HardyMichael durable

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P(n): For all $n$, the number of straight heat segments identified by $n$ clues in the plane, no 3 of i beg your pardon lie on the very same straight line,is: $\large \fracn^2 - n2$.

Inductive hypotheses: offered $n = k$ points, assume $P(k)$ is true: $P(k) = \dfrack^2 - k2$.

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Proving $P(k+1)$ would require proving the for $n = k+1$ points, making use of your inductive hypothesis, the number of lines passing v $k + 1$ clues is same to

$$P(k+1) = \dfrac(k+1)^2 - (k+1)2$$That is, $P(k+1)$ is the amount of $P(k)$, the number of lines figured out by $k$ points, add to the number of additional heat segments result from the extr point: the $(k+1)$th point. Since there space $k$ original points, the variety of line segments that can attach with the $(k+1)$st point is specifically $k$, one heat segment connecting each of the $k$ initial points v $k+1$th point.

That is, our amount is:

$$\beginalignP(k) + k &= \dfrac(k^2 - k)2 + k = \dfrac(k^2 - k)2 + \dfrac 2k2 \\ \\& = \dfrac k^2 + 2k - k2 \\ \\ & = \frack^2 + 2k +1 - k - 12 \\ \\& = \frac(k+1)^2 - (k + 1)2 \\\endalign$$

Hence, native the fact of the base case, and the truth that $P(k+1)$ follows from assuming $P(k)$, we have actually thus verified by induction on $n$ the $P(n) = \dfracn^2 - n2$