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


Nov
6
awarded  Nice Question
Oct
31
accepted Are there mathematical contexts where “finite” implicitly means “nonzero?”
Oct
31
comment Are there mathematical contexts where “finite” implicitly means “nonzero?”
@MarkBennet From experience, that part of the definition didn't actually seem to trip anyone up. It was just the finite part.
Oct
23
comment Are there mathematical contexts where “finite” implicitly means “nonzero?”
@Joshua My idea with the question was to give a formal definition to "a lot of rocks," then have people write a proof about that definition. My hope was that under that interpretation, the English meaning of "a lot of rocks" would not come into play. To the best of my knowledge, that aspect of the question wasn't really the problem; the "finite" part is what was confusing people.
Oct
23
awarded  Popular Question
Oct
22
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.
Oct
22
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. :-)
Oct
22
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."
Oct
22
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.
Oct
22
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.
Oct
22
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?
@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?
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...)