A Question about Lebesgue measure Cover every rational number on [0,1] with an interval of radius $\epsilon/2^n$, the total length of these intervals is $\epsilon$, so the measure of rational numbers on [0,1] is 0. My question is that, can you prove there exist a irrational number on [0,1] that is not covered by these intervals without the use of prove by contradiction?  
 A: To do what you ask more concretely, one needs to be more specific.
Let's consider the enumeration of the rationals according to the following sequence:
$$\tag0 0, 1, \frac12, \frac13, \frac23,\frac14,\frac34,\frac15,\frac25,\frac35,\frac45,\frac16,\frac56,\ldots, \frac{p_n}{q_n},\ldots $$
i.e. all reduced frections ordered first by denominator, then by numerator.
I claim that $\phi=\frac{\sqrt 5-1}2$, which is a root of $f(X)=X^2+X-1$, is not covered.
Assume on the contrary that the $n$th element $\frac {p_n}{q_n}$ of the above sequence covers $\phi$, i.e. we have 
$$\tag1\left|\phi-\frac {p_n}{q_n}\right|<2^{-n}.$$
Then $q_n^2f\left(\frac{p_n}{q_n}\right)=p_n^2+p_nq_n-q_n^2$ is a nonzero integer, hence $$\tag2\left|f\left(\frac{p_n}{q_n}\right)\right|\ge \frac1{q_n^2}.$$
On the other hand $f\left(\frac{p_n}{q_n}\right)=f\left(\frac{p_n}{q_n}\right)-f(\phi)=\left(\frac{p_n}{q_n}-\phi\right)f'(\xi)=\left(\frac{p_n}{q_n}-\phi\right)(2\xi+1)$ for some $\xi$ between $\phi$ and $\frac{p_n}{q_n}$.
From $(1)$, $(2)$ and $\xi\in[0,1]$ we conclude
$$\tag3 \frac1{q_n^2}<\frac{3}{2^n}.$$
For each denominator $q>2$, there appear at least two fractions in $(0)$, namely $\frac1q$ and $\frac{q-1}q$. Taking the initial exceptions for $q\le2$ into account, we see that $q_n\le \frac n2+1$. Together with $(3)$ this implies
$$ 2^n<3\left(\frac n2+1\right)^2,$$
 which holds only for $n\le 5$. Checking the first five fractions explicitly, we find that $(1)$ does not hold for them either.
Remark: Formally, this is a proof by contradiction (though it can be converted to a direct one), but presumably not the kind you really wanted to exclude (i.e. simply using that measure $<1$ implies proper subset). Also note that I actually used $\epsilon=1$. Of course, smaller values of $\epsilon$ will also fail to cover $\phi$.
A: My friend told me another method: get an enumeration of all rational numbers, like the one given by Hagen, denote it by {$q_1,q_2,q_3...$},  there is an interval of radius $\epsilon/2^i$ centered at each $q_i$. Choose an irrational number $s$, it can be either algebraic or transcendental, now look at $q_1$, if the distance $d_1$ between $q_1$ and $s$ is less than $\epsilon$/2, then make $q_1$ the $n$th term of this sequence such that $n=inf\{n\in\mathbb{N^+}:\epsilon/2^n<d_1\}$, if $d_1>\epsilon/2$, leave $q_1$ alone and look at $q_2$. Proceed like this, it is easy to show that this sequence will finally converge, because if you move the $n$th term of this sequence, you will not change the order of sequence before n (which is apparent that you can only move a term forward). The intervals which cover the final sequence of all rational numbers will not cover the irrational number $s$.
