Can all mathematical proof be represented visually? Can all mathematical proof and all mathematical be represented visually?
When defining visually I mean "could potentially be expressed visually", such as graphically or geometrically so the layman could still potentially understand the proof. I arrived at this idea when this video on topology which was understandable to me: a high school student that has not studied topology. Obviously there could be some problems when extending to higher dimensions but nonetheless we can represent higher dimensional objects on a piece of paper.
As I research higher problems and topics in calculus and analysis I feel detached from visual proofs as there seems to be no visual intuition grounding my problems, such as one I was working on  yesterday.
$$\int \arcsin(x) \ln(x) dx$$
However this problem can still be grounded in visual intuition if you graph it and understand Riemann sums visually.
I feel as if most students, when studying maths, rely on "seeing the problem". Does this way of problem solving fade when studying higher maths?
So could all problems and proofs be represented visually or can some proofs only be expressed and understood through definitions? Is there a universal visualisation?
 A: It's very hard to prove a negative. What do you expect an answer of "no, you can't possibly visualize topic _____ in sesquicontravariant cotopology theory" to look like?
I don't expect that there are any topics that people only work with through rigorous application of definitions. That sort of thing could be done by computer. Whenever we try to solve a problem that computers can't solve yet, we must rely on intuitive guesses, which means having an intuition for what the rigorous definitions are meant to do.
(See also: There’s more to mathematics than rigour and proofs)
But that intuition doesn't have to be visual for everyone; people have different ways they understand math. Some people can visualize complicated geometric objects when they close their eyes, other people think that's complete nonsense. Some people will look for visualizations of the math they're doing; other people will find their intuition someplace else, and possibly this will affect the kind of math they prefer to do.
A: From my experience, whenever you are proficient enough in a part of mathematics, you can always generate a sort of intuition of objects and their properties. You may call this a visualization, but this "visualization" is very hard to convey to anyone who doesn't posses some knowledge of the subject already.
Taking cat theory as an example. I usually visualize functors as a continuous deformation of a net into another. But this break downs at times: As a well-known example, the image of a functor is not necessarily a category. Also, it won't help you at other times: An algebra is defined as a morphism $FA \to A$, where $F$ is an endofunctor. Now a better imagery of the functor would be to consider the special case of polynomial functors, e.g. $FX = X + 1$.
In some sense, your statement is correct, because learning math is more or less about building up intuition -- or visualization -- about different concepts; Theorems and definitions on their own are just meaningless symbols strung together, and only when we create an imagery out of them do they make sense.
A: I will tell you, I think this question is too broad. I suggest asking the same question in Mathematics Educators Stack Exchange.
In my opinion and it's exactly that, that yes, every math problem can be presented visually and my guess is that if you "can't" represent it visually, you either don't understand or just don't have the tools to do it (e.g. it's very difficult to create diagrams).
Your example (although a little odd with the $\ln$) is a good one:
$$
\int\arcsin(x)dx
$$
Draw a picture:

Then this integral can be solved by integration by parts.
Edit: Here's an odd mix of visualization and numbers
I am obsessed with perfect numbers (because of a PBS program that introduced me to them--in the 90's). So I spent an inordinate amount of time thinking about perfect numbers. At some point, I realized that perfect numbers could be generated by Mersennes Primes ($2^n - 1 = \text{a prime number}$). I think this is related to Fermat's Little Theorem...but a little of an aside.
Alright, so we need to find $n$ such that $2^n - 1 = \text{a prime number}$. We can immediately rule out all non-prime $n$. Why? Easy visualization if you look at the number in binary!
What does $2^n - 1$ look like as a binary integer?
$$
11111111...n\text{-times}
$$
If $n$ is is not prime, then I can group the $1$'s and easily create a composite number. Example: $2^{2\cdot3\cdot5} - 1 = 2^{30} - 1$. In binary, this is:
$$
111111111111111111111111111111
$$
Because I know there are $30$ $1$'s, I can "organize" this into:
\begin{align*}
(11)(11)(11)(11)(11)(11)(11)(11)(11)(11)(11)(11)(11)(11)(11) \\
(111)(111)(111)(111)(111)(111)(111)(111)(111)(111)\\
(11111)(11111)(11111)(11111)(11111)(11111)
\end{align*}
And there are more ways. The key thing, is that I can say, $2^n - 1$ is definitely not prime if $n$ isn't prime because if $n$ isn't prime, I can do this decomposition and find a multiple (i.e. it's not prime).
Note: This does not prove that "if $n$ is prime, then $2^n - 1$ is prime" and in fact it's not true: $2^{11} - 1$ is not prime (among numerous others). But it does help in narrowing the search for perfect numbers (only test prime exponents).
