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I have recently used induction in solve a problem. But I am told that the proof is wrong. Below is the problem and solution.

I don't see any issue with this proof. What are the issues with this approach?

The population of every city in US is the same

The above statement is equivalent to: “For every n, if S is any set of n cities in US, then the population of all cities from S is the same”.

Base Case: Size of S is 1. There is only one city in S. Thus population of every city in S is the same.

Inductive Hypothesis:

Let S be a set of m cities and assume that population of every city in S is the same.

Induction Step:

We will prove that if S is a set of m+ 1 cities, then the population of every city in S is the same.

Let S = {c1, c2, · · · , cm, cm+1}. Consider the following two subsets of S: S1 = {c1, · · · , cm},

S2 = {c2, · · · , cm+1}

Note that both S1 and S2 are of size m. Thus by induction hypothesis:

population(c1) = polulation(c2) = · · · = population(cm),
population(c2) = polulation(c2) = · · · = population(cm+1).

Since c2 appears in both sets, we have

population(c1) = polulation(c2) = · · · = population(cm) = population(cm+1).

Thus for every set of m + 1 cities, their population is the same. By induction principle, every city in US has the same population.

Are there any rules to decide whether to use Induction or not? What are the mistakes in the above approach.

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    The problem with the above is that you can't get from $n=1$ to $n=2$ this way. – lulu Sep 07 '17 at 13:14
  • I don't know of any general rule that determines when Induction is the way to go. You get a feel for it after you work a lot of cases, but of course that is not always reliable. – lulu Sep 07 '17 at 13:14
  • @lulu not sure I understood your comment. Can you explain further? Thanks. –  Sep 07 '17 at 13:16
  • In the step "Induction Hypothesis", there is a condition $m \geq 1$. In your case, the statement is false for $m+1=2$ – Krish Sep 07 '17 at 13:16
  • Which comment, the $n=1$ to $n=2$? Just write it out. You'll see that the "induction" depends on choosing common elements in the sets of size $n-1$, not possible when $n=2$. – lulu Sep 07 '17 at 13:17
  • See https://en.wikipedia.org/wiki/All_horses_are_the_same_color – Barry Cipra Sep 07 '17 at 14:23

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Check what happens when you take $m=1$ in the inductive step. You get $S_1 = \{ c_1 \}$ and $S_2 = \{ c_2 \}$. So, you can no longer conclude that $c_2$ belongs to both sets.

It isn't that induction is not the way to go in this problem. It is just that the base case here is $m=2$, not $m=1$. If you can show that for every set of two cities, the population of both cities is the same, then the inductive step works as you have written it. Of course, the fact that the base case cannot be shown now should be clear.