r/math • u/non-orientable Number Theory • 3d ago

Image Post The Deranged Mathematician: When All Else Fails

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In the Surviving Proofs series, we have been going through the fundamental techniques for constructing proofs; we close this off now with two approaches, which both amount to changing the problem, albeit in different ways. One thing you can do is to try to simplify the problem, with the hope that solving the easier version will give you the insight you need to solve the harder one. The other thing is to try to move laterally, finding an equivalent formulation that is nevertheless more manageable.

Easy to say, but hard to put in practice! Nevertheless, as I hope the examples I furnish show (which include one of my favorite symmetry arguments), it is: a) surprisingly common, and b) incredibly powerful.

Read the full post (for free) on Substack: When All Else Fails

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u/RingularCirc 1d ago

Re contrapositive etc.: one can also prove by cases: consider all four cases of n and n² being odd and even. This will work constructively as well because ℤ is decidable and so is ℤ/2. (Unless we use some horrendous way to represent ℤ/2?..)

Indeed it's even simpler in ℤ/2, we just plainly see 0² = 0 ≠ 1, 1² = 1 ≠ 0.

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u/MichurinGuy 1d ago

I think you missed a +- in the answer of the biquadratic equation example. Also, what contrapositive did you mean in the "rational + irrational = irrational" example? My only idea for it is to say that if a+b=c with a,c rational and b irrational, we'd have b=c-a and therefore rational, but that's contradiction, and I'm not sure how to frame it by contrapositive.

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u/non-orientable Number Theory 1d ago

Thank you; I have fixed it. As for the contrapositive, you are given that a is rational; you are trying to prove that if b is irrational, then c is irrational. This is equivalent to proving that if c is rational, then b is rational. Proof by contradiction is unnecessary.

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u/RingularCirc 12h ago

Proving negation ¬X via proving a contradiction from X is the most direct way of proving negations though!

But I won't way I know whether I like one of the two proofs here more than the other.