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Edit: For an example of a proof where we are led to false expectations in a proof by contradiction, consider Euclid's proof that there are infinitely many primes. In a common proof by contradiction, one assumes that p₁, ..., p_n are all the primes. It follows that since none of them divide the product-plus-one p₁...p_n+1, that this product-plus-one is also prime. This contradicts that the list was exhaustive. Now, many beginner's falsely expect after this argument that whenever p₁, ..., p_n are prime, then the product-plus-one is also prime. But of course, this isn't true, and this would be a misplaced instance of attempting to extract greater information from the proof, misplaced because this is a proof by contradiction, and that conclusion relied on the assumption that p₁, ..., p_n were all the primes. If one organizes the proof, however, as a direct argument showing that whenever p₁, ..., p_n are prime, then there is yet another prime not on the list, then one is led to the true conclusion, that p₁...p_n+1 has merely a prime divisor not on the original list.

Edit: For an example of a proof where we are led to false expectations in a proof by contradiction, consider Euclid's proof that there are infinitely many primes. In a common proof by contradiction, one assumes that p₁, ..., p_n are all the primes. It follows that since none of them divide the product-plus-one p₁...p_n+1, that this product-plus-one is also prime. This contradicts that the list was exhaustive. Now, many beginners falsely expect after this argument that whenever p₁, ..., p_n are prime, then the product-plus-one is also prime. But of course, this isn't true, and this would be a misplaced instance of attempting to extract greater information from the proof, misplaced because this is a proof by contradiction, and that conclusion relied on the assumption that p₁, ..., p_n were all the primes. If one organizes the proof, however, as a direct argument showing that whenever p₁, ..., p_n are prime, then there is yet another prime not on the list, then one is led to the true conclusion, that p₁...p_n+1 has merely a prime divisor not on the original list.

Edit: For an example of a proof where we are led to false expectations in a proof by contradiction, consider Euclid's proof that there are infinitely many primes. In a common proof by contradiction, one assumes that p₁, ..., p_n are all the primes. It follows that since none of them divide the product-plus-one p₁...p_n+1, that this product-plus-one is also prime. This contradicts that the list was exhaustive. Now, many beginner's falsely expect after this argument that whenever p₁, ..., p_n are prime, then the product-plus-one is also prime. But of course, this isn't true, and this would be a misplaced instance of attempting to extract greater information from the proof, misplaced becasue this is a proof by contradiction, and that conclusion relied on the assumption that p₁, ..., p_n were all the primes. If one organizes the proof, however, as a direct argument showing that whenever p₁, ..., p_n are prime, then there is yet another prime not on the list, then one is led to the true conclusion, that p₁...p_n+1 has merely a prime divisor not on the original list.

Edit: For an example of a proof where we are led to false expectations in a proof by contradiction, consider Euclid's proof that there are infinitely many primes. In a common proof by contradiction, one assumes that p₁, ..., p_n are all the primes. It follows that since none of them divide the product-plus-one p₁...p_n+1, that this product-plus-one is also prime. This contradicts that the list was exhaustive. Now, many beginner's falsely expect after this argument that whenever p₁, ..., p_n are prime, then the product-plus-one is also prime. But of course, this isn't true, and this would be a misplaced instance of attempting to extract greater information from the proof, misplaced because this is a proof by contradiction, and that conclusion relied on the assumption that p₁, ..., p_n were all the primes. If one organizes the proof, however, as a direct argument showing that whenever p₁, ..., p_n are prime, then there is yet another prime not on the list, then one is led to the true conclusion, that p₁...p_n+1 has merely a prime divisor not on the original list.

With reductio, in contrast, a proof of (p implies q) by contradiction seems to carry little of this extra value. We assume p and ¬q, and argue r₁, r₂, and so on, before arriving at a contradiction. The statements r₁ and r₂ are all deduced under the contradictory hypothesis that p and ¬q, which ultimately does not hold in any mathematical situation. The proof has provided extra knowledge about a nonexistant, contradictory land. (Useless!) So these intermediary statements do not seem to provide us with any greater knowledge about the p worlds or the q worlds, beyond the brute statement that (p implies q) alone.

Edit: For an example of a proof where we are led to false expectations in a proof by contradiction, consider Euclid's proof that there are infinitely many primes. In a common proof by contradiction, one assumes that p₁, ..., p_n are all the primes. It follows that since none of them divide the product-plus-one p₁...p_n+1, that this product-plus-one is also prime. This contradicts that the list was exhaustive. Now, many beginner's falsely expect after this argument that whenever p₁, ..., p_n are prime, then the product-plus-one is also prime. But of course, this isn't true, and this would be a misplaced instance of attempting to extract greater information from the proof, misplaced becasue this is a proof by contradiction, and that conclusion relied on the assumption that p₁, ..., p_n were all the primes. If one organizes the proof, however, as a direct argument showing that whenever p₁, ..., p_n are prime, then there is yet another prime not on the list, then one is lead to the true conclusion, that p₁...p_n+1 has merely a prime divisor not on the original list.

With reductio, in contrast, a proof of (p implies q) by contradiction seems to carry little of this extra value. We assume p and ¬q, and argue r₁, r₂, and so on, before arriving at a contradiction. The statements r₁ and r₂ are all deduced under the contradictory hypothesis that p and ¬q, which ultimately does not hold in any mathematical situation. The proof has provided extra knowledge about a nonexistent, contradictory land. (Useless!) So these intermediary statements do not seem to provide us with any greater knowledge about the p worlds or the q worlds, beyond the brute statement that (p implies q) alone.

Edit: For an example of a proof where we are led to false expectations in a proof by contradiction, consider Euclid's proof that there are infinitely many primes. In a common proof by contradiction, one assumes that p₁, ..., p_n are all the primes. It follows that since none of them divide the product-plus-one p₁...p_n+1, that this product-plus-one is also prime. This contradicts that the list was exhaustive. Now, many beginner's falsely expect after this argument that whenever p₁, ..., p_n are prime, then the product-plus-one is also prime. But of course, this isn't true, and this would be a misplaced instance of attempting to extract greater information from the proof, misplaced becasue this is a proof by contradiction, and that conclusion relied on the assumption that p₁, ..., p_n were all the primes. If one organizes the proof, however, as a direct argument showing that whenever p₁, ..., p_n are prime, then there is yet another prime not on the list, then one is led to the true conclusion, that p₁...p_n+1 has merely a prime divisor not on the original list.