I believe I have seen almost every elementary bogus proof of $0=1$ or $1=2.$ I am wondering if there are any lesserknown "proofs" where the error is harder to spot.

Do you mean lesserknown or more difficult to identify? – For the love of maths Jul 04 '18 at 02:35

https://math.stackexchange.com/questions/348198/bestfakeproofsamseaprilfoolsdaycollection – JavaMan Jul 04 '18 at 02:37

From the link above: Proof that a dog has 9 legs: No dog has 5 legs, A dog has 4 more legs than no dog. A dog has 9 legs. – For the love of maths Jul 04 '18 at 02:45

Unfortunately I have already seen all the given proofs so far. Perhaps I'll have to wait a while to get a satisfactory answer. – Display name Jul 05 '18 at 05:51
3 Answers
There's the classic example of the proof that all horses are the same color.
We prove by induction on $ n $ that for any set of $ n $ horses, all horses in that set have the same color.
Base case: If $ n = 1$ then certainly that horse has the same color as itself.
Inductive step: Suppose that for all sets of $ n $ horses that the result holds. Consider a set of $ n + 1 $ horses, say $ A = \{h_1, \ldots, h_{n+1}\} $. Then every horse in the set $ \{ h_1, \ldots, h_n \} $ has the same color by the inductive hypothesis. Thus, $ h_1 $ has the same color as all horses in $ \{h_2, \ldots, h_n \} $, as does $ h_{n+1} $. Thus, all the horses in $ A $ are the same color.
EDIT: To show that $ h_{n+1} $ has the same color as all horses in $ \{h_2, \ldots, h_n \} $, use the inductive hypothesis on $ A  \{h_1\} $.
Hence, by induction, all horses have the same color, as there are only finitely many horses.
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What argument are you using to show that $h_{n+1}$ has the same colour? – qwertz Jul 04 '18 at 04:45

I should have mentioned this explicitly, but the same argument as for showing $ h_1 $ has the same color, remove an element (in this case we remove $ h_1 $) and use the inductive hypothesis. – paul blart math cop Jul 04 '18 at 04:47
Let $$x=1+1+1+1+.....$$ Then $$x=1+(1+1+1+.....)=1+x$$
Now subtract $x$ from bout sides of the $$x = 1+x$$ to get $0=1$ and add $1$ to this new equality to get $1=2$
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What's the sum of $11+11+\cdots$?
Let $x=11+11+\cdots.$
1.
$$x=11+11+\cdots=1(1+11+\cdots)=1x.$$Thus, $x=\dfrac{1}{2}.$
2.
$$x=11+11+\cdots=(11)+(11)+\cdots=0+0+\cdots=0.$$Thus, $x=0.$
3.
$$x=11+11+\cdots=1+(1+1)+(1+1)+\cdots=1+0+0+\cdots=1.$$Thus, $x=1.$
Therefore,$$\dfrac{1}{2}=0=1?$$
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