Where is the mistake?

Statement: For $n\in \mathbb{N}_{0}$ is $2n=0$.

Bais: Show that the basis holds for $n=0$ : $2*0=0$.

Assumption: The statement is valid for all $k\leq n$ : $2*k=0$ for all $k \leq n$

Inductive Step: For $k = n+1$ is $k=a+b$ for two natural numbers $a,b \leq n$. It is $2(n+1) = 2a + 2b = 0+0=0$.

It is obviously wrong. I have a few ideas but I would like to be sure where the exact mistake is.

The first thing I noticed was that in this equoation $2(n+1) = 2a + 2b = 0+0=0$ the left side $2(n+1)$ is always positive but the right side isn't.

I cant find a mathematical mistake here so I thought the Assumption is wrong. I always thought the Assumption is valid for a specific n. It is is indeed freely selectable but specific. Is that right and is that the problem with this induction?

If it isn't - maybe it might be the fact that there aren't two natural numbers below 1 that $ a+ b =k$ and thet you cant use the assumption two times?


2 Answers 2


You use the fact that $2\times1 = 0$, therefore your inductive step does not work for the first step, since you claim that : $$2\times1 = 2\times0+2\times1 = 0+0=0$$ which is not in your induction hypothesis.

Simply put : $1$ is not the sum of two natural numbers $a+b$ where $a\lt 1$ and $b \lt 1$ thus your indutive step does not hold for $n=0$.

  • $\begingroup$ Nicely explained. +1 It's a very similar logical flaw as in the "inductive proof" that all persons alive are 5 years old, or that we all are women, etc. $\endgroup$
    – DonAntonio
    Oct 30, 2016 at 11:01
  • $\begingroup$ @As Exactly the same idea, indeed. $\endgroup$
    – DonAntonio
    Oct 30, 2016 at 11:07

Inductive Step: For $k = n+1$ is $k=a+b$ for two natural numbers $a,b \leq n$. It is $2(n+1) = 2a + 2b = 0+0=0$.

This seems to be a variant of the alternative inductive step $$ (\forall m \le n: S(m)) \Rightarrow S(m+1) $$

Which would complete your statement to

Inductive Step: For $k = n+1$ is $k=a+b$ for two natural numbers $a,b \leq n$.
[$2k = 0$ holds for all $k \le n$, therefore it holds for $a$ and $b$ ]
It is $2(n+1) = 2a + 2b = 0+0=0$.

However only $S(0)$ is true and $S(m)$ is false for $m \in \mathbb{N}$, where

$$ S(m) = ( 2m = 0) $$

so the premise of the implication in square brackets above is wrong and indeed your equation $$ 2a + 2b = 0 + 0 $$ is wrong for $a \in \mathbb{N}$ or $b \in \mathbb{R}$.

  • $\begingroup$ I don't think this answers the question, OP obviously knows that it is not true than $\forall n\in \Bbb N, 2n=0$ (who doesn't ?) but here he is asking why his induction proof does not work. (or did I missunderstand your answer ?) $\endgroup$
    – Astyx
    Oct 30, 2016 at 11:15
  • $\begingroup$ Yes all the quantor are (even though badly put out). His induction hypothesis at rank $n$ is $P_n : \forall k\le n, 2k=0$ which is fine, and his inductive step is that $P_n \implies P_{n+1}$ which is also true for all $n \in \Bbb N^*$ (but not $n=0$ and this is where it fails) $\endgroup$
    – Astyx
    Oct 30, 2016 at 11:20
  • $\begingroup$ $P_n$ is only true as $P_0$ and otherwise false. So it is not fine. $P_0 \Rightarrow P_1$ is false (because $P_0$ is true and $P_1$ is false), while indeed $P_n \Rightarrow P_{n+1}$ for $n \in \mathbb{N}$ is true (just because $P_n$ is false). $\endgroup$
    – mvw
    Oct 30, 2016 at 12:13
  • $\begingroup$ Yes, exactly, this is what I wrote $\endgroup$
    – Astyx
    Oct 30, 2016 at 12:18

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