Every interval in real numbers has a rational and irrational How can you prove the following: where $a\not= b$, show every $[a,b]$ of R has a rational and irrational number.
The context for my question is as follows. In my intro calculus class, we were showing that the Dirichlet function is not integrable. This involved showing that the least upper bound of all upper estimates, U, and the greatest lower bound of all lower estimates, L, were not the same (on any partition of any interval.) This involved showing that L=0 because any subinterval $[x_{i-1},x_i]$ contains both a rational and irrational, so $m_i=o$ and thus every lower estimate will be 0. But the fact that every subinterval has a rational and irrational member, though intuitively obvious, was merely stated.
 A: Let $a<b$ be real numbers.
Then we can see by dividing $\mathbb{R}$ into intervals of size ${1\over 2}\left(\left[{1\over b-a}\right]+1\right)^{-1}$ that some rational number with this denominator is in the interval, because they are equally spaced at the given length.
For irrational numbers, you can use cardinality arguments, since $(a,b)$ is uncountable, all the numbers within cannot be rational.
Edit (addition): to leapfrog off of @user99680's answer:
Let
$$a=\sum_{n=-N}^\infty {a_n\over 10^n}, b=\sum_{n=-N}^\infty {b_n\over 10^n}$$
with $0\le a_n,b_n\le 9$ as digits of the decimal expansion.
Then since $b>a$ we can find a smallest $n_0$ such that $b_{n_0}>a_{n_0}$. Then the number
$$a<\sum_{n=-N}^{n_0+1}a_n(10)^{-n}+10^{-n_0-1}<b$$
is a rational number between the two.
Similarly, we may use the fact that numbers are rational if and only if the decimal expansion is eventually periodic to show that if we write $\pi=3.c_1c_2\ldots$ where the $c_i$ are digits, that
$$a<\sum_{n=-N}^{n_0+1}a_n(10^{-n})+10^{-n_0-2}\sum_{m=1}^\infty c_m10^{-m}<b$$
is an irrational between them.
Still further:  we can see using ideas from Liouville's original proof that transcendental numbers exist to see that
$$a<\sum_{n=-N}^{N}a_n10^{-n}+\sum_{m=N}^\infty{1\over m!}<b$$ is transcendental so long as $N$ is sufficiently large, so choose $N$ so large that this is so and that $N>n_0$ and we get another one.
A: You can use decimal expansions : let $a:=a.a_0a_1.... ; b:=b.b_0b_1....$ .To construct a Rational number  in-between , make sure it has a periodic expansion , to construct an irrational, in-between, make sure its decimal expansion is not periodic.  You can find a term a_j in the expansion of a so that you can repeat the string .a_1a_2....a_j as a period. Then, to construct an irrational , as Adam suggested, you can go far-enough in the expansion of a and append the digits o $\pi$ to it and get an irrational in-between a,b.  
