Which one in this problem is A and which is B?

Question

This following homework problem is proof by contradiction (that is: $\neg B \Rightarrow \neg A$): Problem: Let $m=a^n-1$ for $a>1$ and $n>1$. Show that $\operatorname{ord}_m a = n$.

Solution: by supposition $m= a^n - 1 $, $a^n \equiv 1 \pmod m$ holds. If $x \in N$, $x < n$ and $a^x \equiv \pmod m$ holds. Then $m \mid a^x -1 $, so $a^n - 1 \leq a^x - 1$. Contradiction, for $a^x -1 < a^n - 1 = m$. Therefore $\operatorname{ord}_m a = n$.

Now is A="$a^n-1 < a^x -1$" and B= "$x> n$" ~ "$\operatorname{ord}_m a =n$"? if not then what is A and what is B?

Answer 1

First let’s try to clear up the difference between proving the contrapositive of an implication and proving an implication by contradiction. Suppose that you want to prove $A\to B$. This implication is logically equivalent to its contrapositive, $\lnot B\to\lnot A$, so one way to prove $A\to B$ is to prove $\lnot B\to\lnot A$; this is not a proof by contradiction.

In a proof of $A\to B$ by contradiction, you begin by assuming that both $A$ and $\lnot B$ are true, and you derive a contradiction. That is, you prove $A\land\lnot B\to\bot$, where $\bot$ stands for any false statement.

For this result you don’t need a proof by contradiction. Suppose that $\operatorname{ord}_ma=r$. Then $m\mid a^r-1$, and $m=a^n-1$, so $a^n-1\mid a^r-1$. It follows that $a^n\le a^r$ and hence that $n\le r$. On the other hand, we know that $a^n\equiv 1\pmod m$, so $r\le n$. Combining results, we see that $r=n$, i.e., that $\operatorname{ord}_ma=n$.

Now, you can cast this as a proof by contradiction, as you’re trying to do. Suppose that $r<n$ and $a^r\equiv 1\pmod m$. Then $m\mid a^r-1$, so $a^n-1\mid a^r-1$, and therefore $a^n\le a^r$ and $n\le r$, which contradicts the assumption that $r<n$.

Note that although the mechanics of the two arguments are very similar, the logic is quite different. In the first I proved directly that the order of $a$ is $n$. In the second I assumed that it was less than $n$ and showed that some number was then simultaneously less than and not less than $n$, which is impossible. In other words, I assumed the negation of what I wanted to prove, as well as the various hypotheses of the result, and derived a false statement. In the first argument I merely showed that the order of $a$ had to be both $\le n$ and $\ge n$ and therefore had to be equal to $n$: no false statement in sight.

Answer 2

$A$ is the premise, $B$ is the conclusion/consequent.

The premise $A$ is that "$m=a^n-1$ with $a\gt 1$ and $n\gt 1$."

The consequent, $B$, is that "$\mathrm{ord}_m(a) = n$."

Now, you are apparently confused as to what is a proof by contradiction, because you parenthetically claim that a proof by contradiction of $A\implies B$ is a proof of $\neg B\implies \neg A$. That is incorrect. A proof that $A\implies B$ done by showing $\neg B\implies \neg A$ is a proof by contrapositive, not by contradiction. A proof by contradiction has the form $(A\land \neg B)\implies (P\land \neg P)$.

Note that it is very commonly (and in my opinion, badly) done to claim that a proof that is essentially a proof by contrapositive is in fact a proof by contradiction. If you have a proof of $\neg B\implies \neg A$, then one can turn it into a "fake proof by contradiction" by simply adding the premise $A$. Because if $\neg B\implies \neg A$, then $(A\land \neg B)\implies (A\land\neg A)$, which gives a proof by contradiction. But one never uses the assumption $\neg A$ until the very end, to deduce the contradiction.

Added. A similar way to do a "fake proof by contradiction" is when you have a direct proof $A\implies B$; adding $\neg B$ as a premise leads to $A\land \neg B\implies B\land\neg B$". Note that I call them "fake proof by contradiction" not because it's a fake proof, but rather because these are often proofs by "fake contradictions". As Bill Dubuque notes in the comment, what makes them "fake" is that one never uses the hypothesis $\neg B$ until the very end, when we have successfully concluded $B$, and then simply add "And since we assumed $\neg B$, we get a contradiction, which shows $B$ is true." If you delete the first line that says "Assume $\neg B$", and you delete the final line that says "which contradicts $\neg B$", you get a perfectly fine proof of $A\implies B$, so that the contradiction in the other argument is "artificial" or "fake". A classical example of such "fake proof by contradiction" is the common way of casting the proof that there are infinitely many primes. What we have is a perfectly fine direct proof that given any finite set of primes, there is a prime not in the set; we don't need to add the assumption that our set contains all primes only to later deduce that our set does not contain all primes.

For this proof you don't need a proof by contradiction, or by contrapositive, or a fake proof by contradiction: since $m|a^n-1$, it follows that $a^n\equiv 1\pmod{m}$, so the order of $a$ divides $n$. On the other hand, if $a^r\equiv 1\pmod{m}$, then $m|a^r-1$, which implies that $a^n-1|a^r-1$. In particular, either $a^r-1=0$ or $a^n\leq a^r$. Since $a\gt 1$, $a^r=1$ if and only if $r=0$; and $a^n\leq a^r$ if and only if $n\leq r$. Thus, the smallest positive integer $r$ such that $a^n\equiv 1\pmod{m}$ is $n$, so $\mathrm{ord}_m(a) = n$, as claimed.