Andrew Li

I've been at this simple proof for hours and I have no idea how to prove it except setting up many cases... is there any better way? $|a+b|+|a-b|=|a|+|b| \implies |a|=|b|$. Is there something obvious I'm missing? I've tried applying the triangle inequality but it's an equality I want.

@MikeMiller Sorry I had a terrible brainfart, thanks

Wait a sec, how is it guaranteed that X is a subset of N?

Oh so it's just $X$? But what about the inverse? I used that to prove surjection

$\mathbb N - (\mathbb N - X) = \mathbb N \cap X$

@TobiasKildetoft What do you mean? Apply difference again?

Exactly

I've proved $f$ is bijective given $f : \mathcal P(\mathbb N) \rightarrow \mathcal P(\mathbb N)$ and $f(X) = \mathbb N - X$ but how am I supposed to find an inverse? How could you invert a set difference?

And nothing else

If I need to write a formal proof that involves an implication where the hypothesis is false, can I just say the hypothesis is false and that the implication is thus always true?

Wait, I had the symbols messed up...

Okay, thanks!

I have a quick question. If I'm proving an iff statement (say $A \Longleftrightarrow B$), can I use information I gained from $A \Longrightarrow B$ in proving $A \Longleftarrow B$?

So it's by definition and I don't have to prove it.

Is any set a subset of the universe?

If I want to say $x$ is an element of A, B, and C, can I chain it like $x \in A \in B \in C$ or is that ambiguous/unclear?

@Ryan What about them?

Heh

Okay, just shot my prof an email. Thanks

Yeah, I'll just ask my prof

Not from a book. Also, I heard that if you're in the US, it's usually that natural numbers include 0

So then I guess $\mathcal N$ excludes 0 since it holds for 1.

Which would kind of imply $n \in \mathcal N$ includes 0?

It's awkward because other questions explicitly mention $n\geq 1$

Problem.

I'm trying to establish a base case for an induction problem $\forall n \in \mathcal{N}, (n+1)\cdot(n+2)\cdots 2n = 2^n\cdot 1\cdot 3\cdot 5\cdots (2n-1)$. How would I interpret the case $n=0$? Would it just be $2\cdot0 = 2^0$ (which isn't true)?

@Semiclassical Okay, thanks again. I'll go with conjecture.

@Semiclassical Okay thanks. I'm new to formal proofs and all. Can I use a theorem (like in latex) to describe a statement I later disprove?

Can the noun "proof" refer to some statements that disprove as well as prove?

Thanks!

Nevermind, Wikipedia answered my question: adverbial clauses

Such as "Although X, Y" or "X, who is a Y, ...", etc.

I see a dependent clause as a part of sentence that can't stand on its own

So it's almost like a conditional thing?

Oh, I see

So the example should have three independent right? "I want to do well in school", "I want to go to college", and "I have a lot of trouble with math"

@Mitch Sure. So a compound-complex sentence is a sentence with 2 or more independent clauses and at least one dependent

^^ That's a good idea

What don't you understand

Also notice that my answer states ln|1-x| not ln(1-x). But in the interval of convergence for the series (|x| < 1), they are the same thus the series is still applicable

No. If you want to reindex you're looking for \sum_{n=1} \frac{x^n}{n}

Hello? I don't have much time to chat @tienlee

@tienlee Yes, but since we have $n + 1$ in the denominator and a power of $n+1$, we can change that to just $n$ and start from $n=1$ rather than $n=0$.

@tienlee If you want just one summation, you can reindex to $-\sum_{n = 1}^\infty {x^n\over n}$

@tienlee Yes, that is correct, in the interval of convergence.

@JohnDoe Really? I'll edit it then.

@tienlee Did you use the hint?

Can I get some assistance flagging this: