So this is going to sound insane to some of you... I found out recently that my math education was worthless. Meaning, if you were to compare my math education to those of people at a decent math school, my education didn't cover everything. For example, I never took topology, diff eq., discrete math, and the abstract algebra and analysis courses I had were "introductory" and Rudin and Dummit & Foote make my education look like a joke.

I'm trying to prep for the GRE subject test, and the last time I took it, I studied my calculus well and scored in the 18th percentile. What do you recommen…

that's a bit old "As far as I can tell--and I have actually read some of the stuff Mochizuki has written, unlike many people who casually dole out harsh criticism of this dear mathematician--as far as I can tell this man has been extremely polite throughout this whole episode. Even describing him as "angry" is unsupported by any facts I have seen. If you read this article, he is quoted as saying he is "a bit disappointed" -- extremely mild, perfectly polite. When I read Mochizuki's latest update, he seemed to understand why mathematicians don't have the motivation to learn his work. I would…

@DanielFischer Is it right so far?

We want to show that $\forall k<n+m$ there is a $b \in X \cup Y$ such that $h(b)=k$.

We know that $X \cap Y=\varnothing$

We will show that $\forall k<n$ there is a $x \in X$ such that $h(x)=k$.

For $x \in X$: $h(x)=k \Rightarrow f(x)=k$

Since $f$ is surjective, we conclude that $\forall k<n, \exists x$ such that $f(x)=k$.


Now we will show that $\forall n \leq k <n+m$ there is a $y \in Y$ such that $h(y)=k$.

For $y \in Y: h(y)=k \Rightarrow n+g(b)=k$

I haven't defined subtraction on that set before.. Could you explain me further what we have to do?