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Some proofs that aren't "essentially the same": 1. Prove that the interior angles of a triangle sum to 180 degrees. First proof: draw a line parallel to one o

Some proofs that aren't "essentially the same": 1. Prove that the interior angles of a triangle sum to 180 degrees. First proof: draw a line parallel to one of the triangle's sides passing through its opposite vertex. There are three angles on one side of this line, and they obviously add to 180 degrees because it's a line. One of the three angles is directly one of the triangle's interior angles; the other two can be shown to be equal to the triangle's other two interior angles. (Try drawing it out.) Second proof: start at one side of the triangle and walk around it. By the time you return to where you started, you must have turned 360 degrees. Thus the sum of the exterior angles is 360 degrees. Each interior angle is 180 minus the corresponding exterior angle, and there are three of them, so calling the interior angles A, B, C and the exterior angles A', B', C' we have A'+B'+C' = 360 implies (180-A) + (180-B) + (180-C) = 360 implies 540 - A - B - C = 360 implies 180 = A + B + C. 2. Prove that the sum of the first N numbers is N(N+1)/2. First proof: sum the first and last number to get 1 + N, then the second and second-to-last to get 2 + (N-1) = 1 + N, repeating until you get to the middle. There are N/2 such pairs, giving a total of (1 + N)N/2. (This assumed that there were an even number of terms; consider the odd case too.) Second proof: proceed by induction. For the base case, it's true for N=1 because 1*2/2 = 1. For the inductive case, suppose it's true for N-1. Then 1 + 2 + ... + N-1 + N = (1 + 2 + ... + N-1) + N = N(N-1)/2 + N = N(N-1)/2 + 2N/2 = N(N+1)/2.

pkoird 2y ago
I'm responding to your second example simply because it's easy to argue about. I'd say that both proofs that you have presented are equivalent ways of saying that "since when you sum all the numbers from 1 to N you obtain a number that's N(N+1)/2, therefore, it is true that the sum of numbers from 1 to N is N(N+1)/2". Now, this argument may appear trite but do consider that both of your proofs essentially do the same thing with the first one summing the numbers from extremities and the second one summing 1...N-1 first and then the last. I'd argue that if addition were not commutative, you may have obtained different results.
- justinpombrio 2y ago
  If two programs are equivalent, you can typically show that they're equivalent with a sequence of small refactorings. Replace `x + x` with `2 * x`. Inline that function call. Etc. Can you do that with these two proofs? What's a proof that's halfway in between the two? If you can get from one proof to the other with small "refactorings", then I agree that they're fundamentally the same. If you can't---if there's an insurmountable gap that you need to leap across to transform one into the other---then I'd call them fundamentally different. If you insist that two proofs are "essentially the same thing" despite having this uncrossable gap between them, then I suspect you're defining "essentially the same" to mean "proves the same thing", which is a stupid definition because it makes all proofs the same by fiat, and avoids the interesting question.
  - pkoird 2y ago
    But can you not? Assume sum, 1..(N+1)/2..N Have you not actually built the same proof via induction in both cases with one of them starting from the middle and subsequently including left and right terms at a unit away (you actually do it in reverse but the crux still holds)? Such that S[0] gives you (N+1)/2 and S[(N-1)/2] gives you the total sum. The argument would be like S[i] = (2i-1)(N+1)/2 only you'd be proving it using induction i.e. given S[i-1], finding S[i]. All it ever matters for this problem is that to prove it as such, you somehow have to add up all of the numbers. The "different" proofs you presented are actually the same since for addition, the order of operation does not matter due to associative and commutative properties. A good question would be to see if any of the proofs still remain valid when either of these properties are removed from an operation.
    - justinpombrio 2y ago
      You're handwaving, but I think there is a middle-ground in this proof: Sum(i=1..n, i) = Sum(i=1..n/2, i) + Sum(i=1..n/2, n+1-i) = Sum(i=1..n/2, n+1) I'm still interested in the general question, of whether some proofs have big gaps between them. The more complex the proofs, the more obvious this would be; my examples are unfortunately simple. Something like proving the fundamental theorem of algebra using Rouche's Theorem (complex analysis) vs. field theory. But I don't know enough math to compare those.

pkoird 2y ago

I'm responding to your second example simply because it's easy to argue about. I'd say that both proofs that you have presented are equivalent ways of saying that "since when you sum all the numbers from 1 to N you obtain a number that's N(N+1)/2, therefore, it is true that the sum of numbers from 1 to N is N(N+1)/2". Now, this argument may appear trite but do consider that both of your proofs essentially do the same thing with the first one summing the numbers from extremities and the second one summing 1...N-1 first and then the last. I'd argue that if addition were not commutative, you may have obtained different results.

justinpombrio 2y ago
If two programs are equivalent, you can typically show that they're equivalent with a sequence of small refactorings. Replace `x + x` with `2 * x`. Inline that function call. Etc. Can you do that with these two proofs? What's a proof that's halfway in between the two? If you can get from one proof to the other with small "refactorings", then I agree that they're fundamentally the same. If you can't---if there's an insurmountable gap that you need to leap across to transform one into the other---then I'd call them fundamentally different. If you insist that two proofs are "essentially the same thing" despite having this uncrossable gap between them, then I suspect you're defining "essentially the same" to mean "proves the same thing", which is a stupid definition because it makes all proofs the same by fiat, and avoids the interesting question.
- pkoird 2y ago
  But can you not? Assume sum, 1..(N+1)/2..N Have you not actually built the same proof via induction in both cases with one of them starting from the middle and subsequently including left and right terms at a unit away (you actually do it in reverse but the crux still holds)? Such that S[0] gives you (N+1)/2 and S[(N-1)/2] gives you the total sum. The argument would be like S[i] = (2i-1)(N+1)/2 only you'd be proving it using induction i.e. given S[i-1], finding S[i]. All it ever matters for this problem is that to prove it as such, you somehow have to add up all of the numbers. The "different" proofs you presented are actually the same since for addition, the order of operation does not matter due to associative and commutative properties. A good question would be to see if any of the proofs still remain valid when either of these properties are removed from an operation.
  - justinpombrio 2y ago
    You're handwaving, but I think there is a middle-ground in this proof: Sum(i=1..n, i) = Sum(i=1..n/2, i) + Sum(i=1..n/2, n+1-i) = Sum(i=1..n/2, n+1) I'm still interested in the general question, of whether some proofs have big gaps between them. The more complex the proofs, the more obvious this would be; my examples are unfortunately simple. Something like proving the fundamental theorem of algebra using Rouche's Theorem (complex analysis) vs. field theory. But I don't know enough math to compare those.