I believe I have seen almost every elementary bogus proof of $0=1$ or $1=2.$ I am wondering if there are any lesser-known "proofs" where the error is harder to spot.
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
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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 $1-1+1-1+\cdots$?
Let $x=1-1+1-1+\cdots.$
1.
$$x=1-1+1-1+\cdots=1-(1+1-1+\cdots)=1-x.$$Thus, $x=\dfrac{1}{2}.$
2.
$$x=1-1+1-1+\cdots=(1-1)+(1-1)+\cdots=0+0+\cdots=0.$$Thus, $x=0.$
3.
$$x=1-1+1-1+\cdots=1+(-1+1)+(-1+1)+\cdots=1+0+0+\cdots=1.$$Thus, $x=1.$
Therefore,$$\dfrac{1}{2}=0=1?$$

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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