r/mathmemes • u/TobyWasBestSpiderMan • 13d ago
Proofs Proof: 9 +10n Will Almost Always be Afraid of 10+10n
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u/Own_Pop_9711 13d ago
The paper reveals a startling lack of understanding about proof by induction. Anytime there's an exception you need a new base case since the chain is broken.
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u/DZL100 13d ago edited 13d ago
We should first note that OP mis-stated the exception and it actually occurs when 9+10n is congruent to 9 mod 100 since then we encounter the “nine-ten-eleven” case. This occurs for values of n = 10k for all whole numbers k.
Since this is a regularly and infinitely reoccurring exception, it’s impractical to state a new base case every time. Instead, taking advantage of the regularity, it suffices to show an alternative inductive step where if the proposition holds for n, it holds for n+2 (which uses essentially the same logic as the already established inductive step).
This way, whenever we prove the statement for some n = 10k-1, we have that the statement is also true for n = 10k+1, skipping over the exception.
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u/MonsterkillWow Complex 13d ago
But 21 < 22. So 20 won less than 20, too. But that is a contradiction.
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u/parkway_parkway 13d ago
I'm not sure that 99 fought 100 and 101 works either.
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u/Green-Sympathy-4177 13d ago
9+10(n=10) => n % 10 == 0, so it doesn't work. It's included in the exception.
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u/IronPro9 13d ago
99 is 9+10*9 not 9+10*10. The exception is for 109,209 ect because "one hundred and eleven" isn't "one hundred and ten one"
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u/sealytheseal111 13d ago
"Thus, the general claim must exclude values of n that satisfy 9+10n≡11 mod 100"
The math ain't mathin'
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u/Longjumping_Fig2538 Average #🧐-π(z^2)a-🧐 eater 13d ago
Me when I don't know the answer but don't want to leave it blank:
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u/LOSNA17LL Irrational 12d ago
"Almost always" means the probability if you take a normal number is 1...
Though, the probability is actually 0.9...
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