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My course problem booklet (Mathematics BSc, 2nd year module in analysis, unpublished) has,

Let $x_1=2.5$ and $x_{n+1}=\frac{1}{5}(x_n^2+6)$ for $n>1$. Show that each $2\leq x_n\leq 3$. (Hint: Try a proof by contradiction.)

The solution booklet has,

Suppose that $x_n \geq 2$ for all $1\leq n \leq m$ but $x_{m+1}<2$. Then $$x_{m+1} = \frac{1}{5}(x_m^2+6) \geq \frac{1}{5}(4+6) = 2$$ and we have a contradiction. So $x_n \geq 2$ for all $n$.

and similarly for $x_n \leq 3$.

But why can we "Suppose that $x_n\geq2$ for all $1\leq n \leq m$", and why would we think of doing that? Shouldn't a proof by contradiction begin "Suppose there exists $N\in\mathbb{N}$ such that $x_N<2$"?

It seems to me that this is actually a proof by induction in disguise, with $x_1=2.5$ implicitly taken as the base case. Am I right?

Edit: In the statement of the problem I originally wrote $2\leq x_n\leq x_3$ instead of $2\leq x_n\leq 3$. Sorry for any confusion!

mjc
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    It’s a proof by contradiction wrapped inside a proof by induction. (In particular, they are using contradiction to prove the induction hypothesis.) These proof methods aren’t mutually exclusive. – symplectomorphic Aug 11 '21 at 03:53
  • @symplectomorphic That's interesting. Would you agree that the contradiction here is redundant? – mjc Aug 11 '21 at 03:56
  • Yes: the work that is given provides a direct proof of the induction hypothesis. Strictly speaking there is nothing logically wrong with what the solution book does, but it is silly. (To be clear, the silliness lies in not noticing that the inequality that is proven is precisely what we need to show in the induction step, so that we don’t need to go a further step and contradict anything.) – symplectomorphic Aug 11 '21 at 04:37

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The proof you have written certainly uses induction ("suppose that $x_n \geq 2$ for $1 \leq n \leq m$..."). It also uses contradiction ("suppose $x_{m+1}<2$", then derive a contradiction).

However, the contradiction part isn't necessary, since $x_m \geq 2$ directly implies $x_{m+1} \geq 2$. Similarly, if $2 \leq x_m \leq 3$, then $x_{m+1} = \frac{1}{5}(x_m^2 + 6) \leq \frac{1}{5}(9 + 6) = 3$. It's generally better style to avoid a proof by contradiction where it is not necessary; in this case, the assumption that $x_{m+1} < 2$ isn't really used at all.

J.E.
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You're right, that solution is odd! They are supposed to be using strong induction and "contradiction" (actually, isn't), but it can be done with the usual induction:

Is clear that $2 \leq x_1 \leq 3$. Now let $n \geq 1$, and assume that $2 \leq x_n \leq 3$. Then $4 \leq x_n^2 \leq 9$ implies $2 = \tfrac15(4+6) \leq \tfrac15(x_n^2+6) \leq \tfrac15(9+6) = 3$, that is, $2 \leq x_{n+1} \leq 3$.

The true proof by contradiction here is as follows:

You assume that there exists $m \geq 1$ such that $2 \leq x_n \leq 3$ for all $1 \leq n \leq m$, but $x_{m+1}<2$ or $x_n > 3$ (the negation of "for every $m \geq 1$ such that $2 \leq x_n \leq 3$ for all $1 \leq n \leq m$, we then have $2 \leq x_{m+1} \leq 3$"). Then, using this (not only mentioning it), you arrive to an absurdity.

azif00
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    @symplectomorphic Oh, that's right, I hadn't read that. But still, that's not contradiction, since they are not using the hypothesis at all (it's not enough to assume something is false and immediately prove its true). For example, if you want to prove some statement of the form $A \to B$, so you assume $A$ and $\neg B$, and then you prove $\neg A$ without using $A$, that's really not a proof by contradiction, its by contrapositive: you showed $\neg B \to \neg A$. – azif00 Aug 11 '21 at 04:41
  • Yes, I deleted my comment because I was referring only to your point about the proper negation of the inequality, not to your point about how the contradiction worked. The example you give in your comment regarding contraposition isn’t what’s going on here, though. We want to show $A\to B$. The quoted solution assumes $\lnot B$ and then proves $B$ using only $A$. That is indeed proof by contradiction… it’s just very silly, because it contains a direct proof of $A\to B$ within it. The contradiction step is just literally redundant. – symplectomorphic Aug 11 '21 at 05:45
  • @symplectomorphic I didn't said that my comment regarding the contrapositive proof is what is going here, its just an example of fake contradiction. Indeed, if we want to prove by actual contradiction some implication $A \to B$ we assume that $\neg (A \to B)$ and (somehow) we derive an absurd. Assuming $A \wedge \neg B$ and then proving $B$ without $\neg B$ is just a silly way of writting the proof, that's not a proof by contradiction. – azif00 Aug 11 '21 at 06:08
  • @symplectomorphic By the way, this is related, maybe you want check the discussion there. – azif00 Aug 11 '21 at 06:13
  • I’m aware of all these things, but you are not literally correct when you say “that’s not a proof by contradiction.” A proof that assumes $A$ and $\lnot B$ and derives a contradiction — any contradiction — is indeed a “proof by contradiction” of the implication $A\to B$. In the special case that the contradiction involves deriving $B$ directly from $A$ to contradict $\lnot B$, you are right that this is “silly,” as I have said in comments to the OP. But it’s not, strictly speaking, not a “proof by contradiction.” – symplectomorphic Aug 11 '21 at 15:06