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The question says $10\cdot n=0$ for all $n\in\mathbb{Z}$ with $n\geq 0$.

Here is my proof by strong induction:

Base case: $10\cdot0=0$.

Let $k\geq 0$, and suppose that for any $m\leq k$ we have that $10\cdot m=0$.

Consider $10\cdot(k+1)$. The number $k+1$ can be written as $m+l$ for some numbers $0\leq m,l\leq k$. By the induction hypothesis, $10\cdot m=0=10\cdot l$.

But then: $10\cdot(k+1)=10\cdot(m+l) = 10\cdot m+10\cdot l=0+0=0$.

Apparently something went wrong in my proof by strong induction, but I cannot seem to figure out what.

Daniel W. Farlow
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Rohan
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3 Answers3

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You cannot go like this from $k=0$ to $k=1$ (i.e. $k=1$ cannot be expressed in the form $m+l$ as you wrote).

truebaran
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I'll try to sketch out a proof similar to the one you provided, and then I will flesh out why this approach does not work (as an aside, you may find this thread on fake induction proofs to be of use in picking up on faulty reasoning sometimes used in induction proofs).


Claim: For every non-negative integer $n, 10n=0$.

Base case: $10\cdot 0=0$.

Inductive step: Suppose that $10m=0$ for all non-negative integers $m$ with $0\leq m\leq k$. Write $k+1=\ell+m$, where $\ell$ and $m$ are natural numbers less than $k+1$ (I am considering the natural numbers to include $0$). By the induction hypothesis, $10(k+1)=10(\ell+m)=10\ell+10m=0+0=0$.


Can you spot the error truebaran alludes to in his answer? Here it is in more detail:

Error: The error lies in going from the base case $n=0$ to the next case, $n=1$. The number $1$ cannot be written as the sum of two smaller natural numbers; thus, we cannot invoke the inductive hypothesis. In the proof, when $k=0$, we cannot write $0+1=\ell+m$ where $0\leq \ell\leq 0$ and $0\leq m\leq 0$.

Does that make sense?

Daniel W. Farlow
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In general, induction works like this.

Let a statement $\mathcal{S}$ depend on a natural number $n \in \mathbb{N}$. We write the statement as $\mathcal{S}_n$.

Induction means to proof the following two conditions:

$$ \mathcal{S}_0 = \textrm{true} \tag{1}\\ $$ $$ \mathcal{S}_n = \textrm{true} \Longrightarrow \mathcal{S}_{n+1} = \textrm{true}\tag{2} $$

whence

$$ \forall n \in \mathbb{N} : \mathcal{S}_n = \textrm{true} $$


Your statement is given by $$ \mathcal{S}_n = \langle\textit{For any natural number $n$ holds that $10n=0$.}\rangle $$

The second part of induction (2) would imply that $$ \mathcal{S}_n = \textrm{true} \Longrightarrow \mathcal{S}_{n+1} = \textrm{false} $$ Whence induction has failed.

After all we have IF $10n = 0$ THEN $10(n+1)=0$, thus $10=0$ which is false.

johannesvalks
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