9
$\begingroup$

My concern is with choosing specific conditions within a proof to arrive at a general result. As an example, I'll use the proof that $\mathbb{Q}$ is dense in $\mathbb{R}$. The proof I know goes as follows:

Claim: Let $x,y \in \mathbb{R}$ such that $y > x$. Then there exists a rational number $\frac{m}{n}$ such that $y > \frac{m}{n} > x$. Since $y > x \implies y - x > 0$, by the Archimedean property there exists a positive natural number $n$ such that $n(y - x) > 1$ so $ny > 1 + nx$. If $m$ is the smallest integer such that $m > nx$ (also note this inequality is the same as $\frac{m}{n} > x$), then notice that $m - 1 \leq nx$, so from this we have $m \leq nx + 1$. Combining, we have $m \leq nx + 1 < ny$, so taking the left and right sides of this inequality, we have $y > \frac{m}{n}$ and $\frac{m}{n} > x$ and so $y > \frac{m}{n} > x$.

1) So first, by hypothesis we have that $y > x$, and note that this is the same as $y - x > 0$. Then, the Archimedean property is invoked by stating there exists a positive natural number $n$ such that $n(y - x) > 1$. While this is certainly true, why isn't it an issue that the number $1$ was specifically chosen as opposed to some arbitrary number? Why doesn't it affect the end result that $1$ in particular is being used instead of some arbitrary real number?

2) Next in the proof, we choose an integer $m$ such that $m$ is the smallest integer that satisfies $m > nx$. Here we are imposing a specific condition on the integer $m$ (that it is the smallest integer), as opposed to assuming an arbitrary $m$ without any special qualities. Why does this not "ruin" the generality of the proof?

In general, how do we know that by putting special conditions in parts of this proof, that the end result of the proof that we get ($y > \frac{m}{n} > x$) is not just a coincidence because we forced specific restrictions to make the proof appear that way?

Aren't there "false proofs" of results that are not true in general that appear to make the fake result true, but in fact are not valid because they assumed too many specific conditions to "force" the result, when the result in actuality is not true in general?

$\endgroup$
  • 1
    $\begingroup$ It's natural that you had to force specific restrictions on $m$ and $n$. After all, you are trying to produce a specific rational number $m/n$ that lies between $x$ and $y$. Most $m/n$ won't lie between $x$ and $y$! $\endgroup$ – tracing Dec 23 '14 at 15:10
1
$\begingroup$

In part (1), you imply that for an arbitrary positive integer $q$ there exists a positive integer $n$ such that $n(y−x)>q.$ If this is true for arbitrary $q$ then it is certainly also true for the case $q=1.$

In part (2), given that there is a set of integers such that for any integer $r$ in the set, $r > nx,$ then it is certainly true that $m > nx$ where $m$ is the smallest integer in that set, provided of course that the set has a smallest member.

One problem that can occur when you take a general statement and use a specific example of it in a proof is that the specific example is not enough to prove the thing that was to be proved, whereas if you had kept the statement more general then you would have been able to finish. Another problem that can occur is that you implicitly assume some restriction (for example, $q > 1$), make a step in your proof that works only under that assumption, and then assert that a result is true for arbitrary $q$ (ignoring the fact that you had to restrict $q$ to get the result). The "all horses are the same color" fake proof is an example of this.

Where the restrictions are explicitly stated and are not later ignored, however, the restrictions do not invalidate anything that was deduced from them.

If the proof at some point had needed to restrict to $x>1$ in order to do one of its steps, and later claimed that the result of that step applied to all $x$, then the proof would indeed be invalid. That did not happen here.

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

The reason we can choose $m$ and $n$ as we do in this proof is that we want the result to hold for every $x, y \in \mathbb{R}$, not every $m, n$. So we can impose any arbitrary condition we want on the $m$ and $n$ without affecting the result. In general, it's important to consider which part of the proof need to be general for the result to hold. Here we just need to construct one specific rational for it to hold, so we can impose any conditions we want to on that rational.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.