Find a Mistake in Proof: 0,1,2,3 are All Even

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The proof claiming that 0, 1, 2, and 3 are all even is flawed due to a misunderstanding in the induction step. The base case correctly identifies 0 as even, but the assumption that both k and 1 are even when k = 0 is incorrect. This leads to the erroneous conclusion that k + 1 (which equals 1) is also even. The key mistake lies in assuming the inductive hypothesis holds for k = 0, which fails since it does not apply to the case of k = 1. Ultimately, the induction step must be independent of the specific value of k, which this argument does not satisfy.
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Basically I need to find a mistake in this "proof".

I claim that 0,1,2,3...are all even.

I will use induction to prove that 'n is even' for n = 0,1,2,3...
Base case is n = 0, which is true, 0 is even. I assume that the statement is true for
n = 0,1,2,3...,k and consider n = k+1. By assumption, 1 and k are both even, and thus k+1 is even as well. This means that n = 0,1,2,3... are all even.

I can't seem to find a hole in the proof. I know that 1 is not even and when we add 1 to an even number, we get an odd number. But, by assumption 1 is even, so, what do I do know ?:-p
 
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Try your argument when getting from k = 0 to k = 1.
 
If k is even, then k+1 is odd...
 
There is nothing wrong in you proof...wat you are doing is dat you are assigning a subscript for each even number so A0=0, A1=2...so by induction you r getting this peculiar result which is for the subscript...:smile:
 
As LCKurtz said, the problem is between 0 and 1. You seem to be tacitly assuming that k is greater than 1 when you say that 1 is even by the inductive hypothesis. But for k=0, this isn't true since k+1=1.
 
spamiam said:
As LCKurtz said, the problem is between 0 and 1. You seem to be tacitly assuming that k is greater than 1 when you say that 1 is even by the inductive hypothesis. But for k=0, this isn't true since k+1=1.

More accurately, his argument fails because he can't choose 1 and k > 1 to add together in the induction step because there is no such k.

The real lesson in this example is that in induction arguments, the induction step must be independent of the value of k, which it isn't in this argument.
 
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