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I'm having some trouble identifying where specifically the following proof is incorrect:

Statement: Given any positive integer $n$, every n people have the same name.

"Proof:" We prove by induction. When $n = 1$, the statement is trivially true. Now, assume the statement is true when $n = m$. In other words, every $m$ people have the same name. We show the statement is true when $n=m+1$ in the following way. Consider the following list of people: $p_{1}, p_{2},..., p_{m+1}$. By the inductive hypothesis, the individuals $p_{1},..,p_{m}$ have the same name. But we also see that the list $p_{2},...,p_{m+1}$ consists of $m$ individuals, so they all have the same name as well. It follows that $p_{1}, p_{2},..., p_{m+1}$ all have the same name. Thus, since we have shown the statement holds when $n=m+1$, the statement holds for all positive integers.

I'm given the following options:

  • The basis step is incorrect
  • The inductive step is wrong when $m = 1$
  • The inductive step is incorrect for every $m$
  • The inductive hypothesis is incorrect.

Any help would be greatly appreciated!

Michael
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  • The problem is hidden here: Given that $p_1,\dots, p_m$ all have the same name and $p_2,\dots,p_{m+1}$ all have the same name, it actually doesn't necessariily follow that $p_1,\dots,p_{m+1}$ all have the same name. Think carefully about why it seems that it does follow... – David C. Ullrich Jul 30 '18 at 16:58
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    The correct answer is : "The inductive step is wrong when $m=1$". The claim is true for $m=1$ , but wrong for $m=2$. It it were true for $m=2$, it would be true for every $m$. – Peter Jul 30 '18 at 17:02
  • @Peter I knew something was wrong with the base case but couldn't work out exactly what! An interesting question which arises is what constitutes a good base case? – packetpacket Jul 30 '18 at 17:04
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    @packetpacket Actually, this proof is tricky. The base case is correct just because if we only have one person we cannot compare the names, so mathematically "all the persons have the same name". In the case $m=2$, what we actually have is only $p_1=p_1$ and $p_2=p_2$ by induction hypothesis which of course does not imply $p_1=p_2$ – Peter Jul 30 '18 at 17:10
  • I'm a little bit formal to make the argument of the principle of induction appear clearer. Let $P(n)$ is the statement every $n$ people have the same name. As you pointed out, $P(1)$ is trivially true. And you are trying to prove $$P(n)\implies P(n+1)$$ to finish the proof. But It's clear that $P(1)$ does NOT implies $P(2)$. Thus the statement $$P(n)\implies P(n+1)$$ is not true. And this is where your proof falls. – Akira Jul 31 '18 at 05:46

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