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I love teaching, learning, and that wonderful feeling you get when you really understand something.


3h
comment Are there mathematical contexts where “finite” implicitly means “nonzero?”
@anorton The definition you're referencing from the dictionary is Dedekind-infiniteness. One way of proving that the set of rational numbers is infinite is to find a bijection from the rationals to a subset of the rationals. In other words, the definition that you were referencing does correctly let you conclude that the rationals are infinite, though it doesn't let you conclude just how big they are.
3h
comment Are there mathematical contexts where “finite” implicitly means “nonzero?”
@anorton That's correct, which is why the set of all rational numbers is infinite. :-)
6h
comment Are there mathematical contexts where “finite” implicitly means “nonzero?”
@JonasMeyer Yep, that part was fine. The issue is basically whether 0 is considered finite in mathematics, and the answers here seem to suggest "yes, unless you're doing applied math."
6h
comment Are there mathematical contexts where “finite” implicitly means “nonzero?”
I agree about dictionary definitions not being valid in math. The main reason I was asking was to check whether there actually was some part of math where 0 was not considered finite, since it wouldn't surprise me in the slightest if there was some validity to the students' concerns that just happen to be in a branch of math I'm not familiar with.
6h
comment Are there mathematical contexts where “finite” implicitly means “nonzero?”
When I asked this question, I specifically did so to point out how some definitions end up not describing anything. I totally agree about the root of the paradox.
7h
asked Are there mathematical contexts where “finite” implicitly means “nonzero?”
Sep
30
awarded  Explainer
Sep
24
awarded  Autobiographer
Sep
19
comment What's the parametric equation for the general form of an ellipse rotated by any amount? Thanks
@HighPerformanceMark I was answering the question with the intent that it would then get migrates to the math site. My intent was to help the OP out, but also suggest a better site for questions like this in the future. I was under the impression that this was considered appropriate, and I'm sorry if this wasn't the right approach.
Sep
19
answered What's the parametric equation for the general form of an ellipse rotated by any amount? Thanks
Sep
11
comment Can't find mistake in an easy proof.
@Taemyr I actually don't think that statement is a problem. Suppose that B is empty. Then the statement "if $a \in A$ and $b \in B$, then $(a, b) \in A \times B$" is true because the antecedent is false.
Sep
10
comment Can't find mistake in an easy proof.
@Taemyr Interesting - I was taught that to prove a universally quantified statement, you can start with "choose an arbitrary x" even in the case where no such x actually exists. Is this not correct? Also, is there a reason why the flaw is with the second arbitrary choice (pick b) not having a caveat of the form "assuming B is not empty" and not with the first choice (pick a) for the same reason?
Sep
10
awarded  Nice Answer
Sep
9
comment Can't find mistake in an easy proof.
There's nothing wrong per se with saying to choose an arbitrary element from a set if you're trying to prove a universally-quantified statement. I think that the real issue is that the proof is a valid proof of a result other than the one that's trying to be proven.
Sep
9
comment Can't find mistake in an easy proof.
I see what you're saying, but I disagree that the problem in the proof is that $B$ isn't asserted to be nonempty. None of the actual steps in the proof are actually wrong - it's more that the result shown isn't the right result. Think of it this way - suppose you want to prove that $A \cap B \subset A$. One way to do that is to consider an arbitrary element $x \in A \cap B$, then to say that this element must belong to $A$ because it belongs to $A \cap B$. This proof is valid even if $A \cap B$ is empty, even though "choose an arbitrary $x \in A \cap B$" isn't qualified. (continued...)
Sep
9
awarded  Custodian
Sep
9
reviewed Approve suggested edit on Can't find mistake in an easy proof.
Sep
9
comment Can't find mistake in an easy proof.
+1 This is an excellent problem - it took me a while to figure out what was wrong with it!
Sep
9
comment Can't find mistake in an easy proof.
@Frunobulax I don't actually think that this is the problem with the proof. There's nothing wrong with saying "let $b$ be an arbitrary element of $B$" without having to stipulate that such a $b$ actually exists. Typically, if you do something like that, you'll end up with a result that's true, but may be vacuously true if there isn't a choice of $b$ to make. I think that the issue is with the overall logical structure of the proof; can you check my answer and see if it makes sense?
Sep
9
answered Can't find mistake in an easy proof.