Ways to arrive at the existence of irrational numbers I know about a way Greeks arrived at the existence of irrational numbers by showing that sometimes two line segments can be incommensurable.
And about a simple way by which it can be shown that some numbers are irrational, for example, as is usually shown that $\sqrt2$ is irrational.
Also, it can be shown that some numbers are transcendental and because all rationals are algebraic that shows that there are some non-rational, that is, irrational numbers.
And also there is a way that shows that all rational numbers have periodic expansion in every base and since there are non-periodic expansions that also shows the existence of irrationals.
And there is countability/uncountability way.
Are there some other ways?
 A: Any infinite simple continued fraction is irrational because any rational number has a finite simple continued fraction.
A: Your third statement is not worded correctly.  In point of fact, to show that some numbers are transcendental is not particularly easy.  The first proof for a well-known number is for $e$.  The simplest argument depends on a series of approximations (the Taylor series for $e$ that converges "too quickly" to be algebraic, but it is nowhere near as simple as the proof that $\sqrt{2}$ is irrational.
There are many ways to show irrationallity, and several to show trancendentallity. For example, a number expressed as a repeating continued fraction is algebraic.
A: From my perspective, the key to all such proofs is the completeness of the real numbers.
At the most basic level, the defining property of $\mathbb{R}$ is that it is a complete ordered field.  Now $\mathbb{Q}$ is also an ordered field,  so you are not going to be able to prove the existence of irrationals by pure algebra or inequalities.  Somewhere you will have to use completeness.  Indeed, the existence of irrationals is equivalent to the fact that $\mathbb{R}$ is complete but $\mathbb{Q}$ isn't.
It's interesting to look where completeness is used in the other usual proofs:


*

*$\sqrt{2}$ is irrational:  How do you know that $2$ has a square root at all?  There are plenty of number systems in which lots of numbers just don't have square roots (e.g. integers, modular arithmetic; even in $\mathbb{R}$ the negative numbers don't have square roots).  So what's so special about $\mathbb{R}$ to guarantee that $2$ has a square root?  Completeness, of course.  You can construct a Cauchy sequence of rationals such that if it has a limit, the square of the limit must be 2.  Or you can look at the Dedekind cut $\{x : x^2 < 2\}$.  Or you can prove the intermediate value theorem (using completeness!) and consider the continuous function $x^2$ which takes values both above and below 2.

*$\log_2 3$: Likewise, how do you know that $3$ has a log base 2?  There are lots of other number systems in which logs don't exist, e.g. there just doesn't exist any number $x$ such that $2^x = 3$.

*Continued fractions: how do you know that the continued fraction actually converges to something?

*$\mathbb{Q}$ is countable but $\mathbb{R}$ is not: you have to use completeness somewhere in proving that $\mathbb{R}$ is uncountable.  (The Baire category theorem is a neat way to see it made explicit.)

*0.1010010001...: why does every sequence of decimal digits actually represent a number?

*$e$: You have to first give a rigorous definition of $e$.  Whatever definition you choose, proving that there exists a number satisfying that definition will require using completeness.
A: Logorithms are an easy "in" for finding irrationals. Suppose $\log_2 3=\frac{a}{b}$. Then $3=2^{a/b}$, giving $3^b=2^a$, which is impossible unless $a=b=0$, which wouldn't be allowed by the original setup because $\log_2 3=\frac{a}{b}$ would lead to $\frac{0}{0}$. 
