Finding the error in this induction proof

Question

Claim: If $n$ belongs to $\mathbb{N}$, and $p$ and $q$ are natural numbers with maximum $n$, then $p=q$.

Let $S$ be the subset of the natural numbers for which the claim is true. $1$ belongs to $S$, since if $p$ and $q$ belong to $N$ and their maximum is $1$, then $p=q=1$. Now assume $k$ belongs to $S$, and that the maximum of $p$ and $q$ is $k+1$. Then the maximum of $p-1$, $q-1$ is $k$. But $k$ is in $S$, so $p-1=q-1$, thus $p=q$ and $k+1$ is in $S$, so the assertion is true for all $n$ in $N$.

Subtracting 1 can take p and q out of the natural numbers (which I gather you are starting at 1). — lulu, Jul 19 '15 at 12:00
This question is also the same as one of the answers provided here on the thread Fake Induction Proofs. — Daniel W. Farlow, Jul 19 '15 at 16:13

score 5 · Answer 1 · edited Jul 19 '15 at 14:08

5

By natural number I assume you mean positive integer. The error in the proof occurs when $k+1=2,p=2,q=1$. Then $q-1 =0$, not a natural number, so there is no case with $p-1$ and $q-1$.

edited Jul 19 '15 at 14:08

Race Bannon

1,829

answered Jul 19 '15 at 13:06

DanielWainfleet

57,985

Finding the error in this induction proof

1 Answers1