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Proof. $S(n):$ Given any set of numbers, if one of the numbers is prime then all the numbers in the set are prime.

Starting with $S(1)$, means that we are starting with one prime number in the set, and the number in the set is prime. Which proves that $S(1)$ is true. Now, we if we make an induction hypothesis that S(n) is true, and we consider a set consisting of $n + 1$ numbers and at least one of them is prime. Let $P_1,P_2,P_3,...,P_{n+1}$ be the numbers in the set with $P_1$ being the number that is prime. If we consider the subset {${P_1,P_2,...,P_{n}}$}, we see it is the set with $n$ numbers and one of them is prime, so by induction hypothesis, numbers $P_1,P_2,...P_{n}$ are primes. Next the subset {${P_1,P_2,...,P_{n+1}}$} is also a subset of $n$ numbers. So, by induction hypothesis, we can say that they are all prime. Hence, it is proved that $S(n+1)$ is true, which completes the proof.

Bill Dubuque
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bwaluvory
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    Perhaps searching on the web for "all horses have the same colour" helps. That's the most common way to exhibit this mistake. – Dermot Craddock Jun 18 '25 at 09:52
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    Just see where your argument fails for ${2,4}$. – Kavi Rama Murthy Jun 18 '25 at 09:54
  • "Next the subset {P1,P2,...,Pn+1} is also a subset of n numbers" Um.... no it isn't. It is a subset of $n+1$ numbers. – fleablood Jun 18 '25 at 15:00
  • I think what you meant to write was ${P_2,P_3,....,P_{n+1}}$ is a subset of $n$ numbers. Here the error is that if $n=1$ then ${P_{n+1}}$ is a subset of $n$ numbers.... but you have no reason to assume it has a prime in it. – fleablood Jun 18 '25 at 15:03
  • Basically you induction step makes the tacit assumption: $n > 1$ which is not satisfied in your base case. If an induction step involves a property that isn't satisfied by the base case then the induction isn't valid to apply. You can "fix" this by making your base case $n=2$ (but then it just isn't true). Or you can see where it fails as @KaviRamaMurthy suggests (${2}$ is subset of 1 with a prime and all are prime. Then ${4}$ is a subset of 1 with...hey wait, if $2$ isn't in there then I don't know it has at least one prime!) – fleablood Jun 18 '25 at 15:11

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