Does a different inequality automatically contradict a given inequality in a proof

Question

I’m working through a proof in Baby Rudin and I had a general question about proofs and their givens:

Let’s say there’s a known true statement like “for any positive integer n” and then some proposition that needs to be proven.

Now I assume the opposite of the proposition and a few lines later a result like n > 1 pops out. Does this automatically imply contradiction with the given above because it excludes the case where n = 1, or is it merely “new” information that further constrains n? My gut tells me it’s a contradiction because it’s ignoring the n=1 case that was guaranteed to be valid by the given, but I have no professor to ask.

I’m sorry if this is a bit vague but it’s a general question about proofs and I don’t want anyone to accuse me of asking for Homework help (I’m self-studying for my own curiosity for what it’s worth).

Answer 1

No, it's not a contradiction. $n>1$ is perfectly compatible with $n$ being a positive integer.

If, say, $n=7$, then it is both true that $n >1$ and that $n$ is a positive integer.

So it is indeed more like new information we get about $n$, and hence we now know more about $n$, but we do not obtain contradictory information.

And when you say that the case $n=1$ is 'valid', please know that all you really know is that initially it was possible that $n=1$, i.e. that it was an option that $n=1$, but now that you found out that $n>1$, that particular option is ruled out, but there are of course still plenty of options open compatible with the claim that $n$ is a positive integer. Indeed, by saying that $n$ is a positive integer, you were not forced to say that $n=1$, so in that sense it wasn't 'valid' at all to say that $n=1$.

Answer 2

No, it's no contradiction at all. It would be a contradiction if you had got $0<n<1$ (since there's no integer in $(0,1)$), but since there are integers greater than $1$, there is nothing to deduce.

Answer 3

No. That proves that it can't be false/must be true for $n=1$. You still need to prove that it can't be false/must be true for all other positive integers.

Let's supposed I want to be prove that for all positive integers $n$ that $n^2 \le n$. (This is false but I'm trying to prove it is true.)

I try a proof by contradiction. Assume $n^2 > n$. As $n > 0$ we know $\frac {n^2}n > \frac nn$ so $n > 1$.

This is the same situation as you have described. This is not a contradiction except for $n = 1$. So I can conclude that if $n = 1$ then $n^2 \le n$. And that is true. But I can't conclude $n^2 \le n$ for all $n$ as $n > 1$ is not a contradiction for all $n$.