# 1. How to prove an expression is irreducible?

#### Secret

Registered Senior Member
Some months ago I came across the "solving equation with indeterminate answers" and this bit interested me:

This looks like a fairly straightforward problem but it's been bugging me for a while

$$y=\frac{1-\frac{1}{e^{bx}}}{x}$$

How do I show that when x = 0 , y = b ?

It is a transcendental expression, transcendental expressions aren't reducible, did you pass basic algebra? An example of reducible expression is $$\frac{x^2-y^2}{x-y}$$. Does the expression in your exercise look like a ratio of two polynomials?

So is it true that for an expression to be reducible, it must be a ratio of polynomials ( where x, and y can be a function) or is there more subtle cases?
If there is more subtle cases, then in general, how to prove that an arbitrary expression f(x)/g(x) is irreducible?

So is it true that for an expression to be reducible, it must be a ratio of polynomials ( where x, and y can be a function) or is there more subtle cases?
If there is more subtle cases, then in general, how to prove that an arbitrary expression f(x)/g(x) is irreducible?

It doesn't have to be a ratio of polynomials. You just have to be able to decompose the numerator into a product of the denominator and something else. Tach's example that you quoted works because $$x^2-y^2=(x-y)(x+y)$$ and $$x-y$$ is the denominator. A non-polynomial example would be $$\frac{\tan(x)}{\csc(x)}=\frac{\cos(x)\csc(x)}{\csc(x)}=\cos(x)$$. (In case you've forgotten, $$\csc(x)=\frac{1}{\sin(x)}$$; I wrote it in this form to make it clearer what's in the numerator and what's in the denominator.) I don't know of an algorithmic way to decide whether an expression is irreducible, but if you have a polynomial in one part of the fraction and an exponential in the other, that's generally a bad sign.

Also, I don't know if this was mentioned in the thread you quoted, but the irreducible example can still be solved by taking the limit as x approaches zero and applying L'Hospital's rule.

ok, so I guess another user might be able to tackle the "an algorithmic way to decide whether an expression is irreducible". I'll also take note about the "but if you have a polynomial in one part of the fraction and an exponential in the other, that's generally a bad sign." for my future studies

As for the "Also, I don't know if this was mentioned in the thread you quoted, but the irreducible example can still be solved by taking the limit as x approaches zero and applying L'Hospital's rule", yes I knew of that, and yes it has mentioned in that thread

Maybe I should have been more clear in my OP. By reducible, I mean "it is possible to simplify the expression, as long the stuff being cancelled does not involve division by zero"

So to refine my question it is as follows:
Q: In general, is there a way to check whether an arbitrary expression, given division by zero is not involved, can be simplified

P.S. Solving this question might make me understand more clearly why expressions such as x=sin(x) (i.e. x-sin(x)=0) cannot be solved symbolically

If there is more subtle cases, then in general, how to prove that an arbitrary expression f(x)/g(x) is irreducible?

You need a concept of a class of all functions that meet a certain property, like $$C^{\tiny \infty}$$.

Then f(x)/g(x) is "reducible" if there is a unique admissible function h(x) such that whenever g(x) is not zero f(x) = h(x) g(x) and whenever g(x) is zero, then f(x) is also zero and the limit of f(x)/g(x) at the zeros of g(x) is equal to the limit of h(x).

$$\textrm{divides}(g,f) \leftrightarrow \exists ! h \left[ \left( \forall x f(x) = h(x) g(x) \right) \; \textrm{and} \; \forall x \left( g(x) = 0 \rightarrow \left( f(x) = 0 \; \textrm{and} \; \lim_{y\to x} \frac{f(y)}{g(y)} = \lim_{y\to x} h(y) \right) \right) \right]$$

So if the class of "all functions" are polynomials in $$\mathbb{C}$$ this conforms to the standard definition of polynomial division.

Likewise if the class is "all smooth functions in $$\mathbb{R}$$ with isolated singularities" we have "sin(2x)/cos(x) = 2 sin(x)". And "tan(x)/sec(x) = sin(x)". But not "sin(x)/cos(x) = tan(x)", because cos(x) has zeros where sin(x) is finite, so the domain of f is not the domain of h.