In real numbers, the number of rational numbers is more or the number of irrational numbers is more? In real numbers, the number of rational numbers is more or the number of irrational numbers is more ?
 A: By the Cantor Diagonalization argument, we can prove that $\mathbb{R}$ uncountable. Also, we know that the set of rational numbers $\mathbb{Q}$ is countable. This part is not difficult. At first, find a bijection between $\mathbb{Z}$ and $\mathbb{N}$. Then use famous Diagonal arguments to show that $\mathbb{Q}$ and $\mathbb{Z}$ has a set bijection. Hence, we have a set bijection between $\mathbb{N}$ and $\mathbb{Q}$. This proves that $\mathbb{Q}$ is countable.
Since the irrationals, $\mathbb{Q^c}$ is the complement of $\mathbb{Q}$, and hence $\mathbb{Q^c}$ must be uncountable. Suppose $\mathbb{Q^c}$ countable, and then get a contradiction. Therefore, $\mathbb{Q^c}$ is uncountable. Hence, the cardinality of $\mathbb{Q^c}$ is greater than that of $\mathbb{Q}$. In fact, by the Continum Hypothesis, we can say that the cardinality of $\mathbb{Q^c}$ is exactly equal to that of $\mathbb{R}$. For more details  
For more details please have a look:
https://en.wikipedia.org/wiki/Uncountable_set
The following link has an excellent answer that gave proof of my last argument.
Cardinality of the Irrationals
Finally, by the Continum Hypothesis (https://en.wikipedia.org/wiki/Continuum_hypothesis) we can readily conclude the assertion that the cardinality of $\mathbb{Q^c}$ is greater than that of $\mathbb{Q}$.
A: There are more irrationals than rationals by far. This all goes back to the famous Set Theory grandfather Georg Cantor (yes, it is spelled "Georg", not "George").
Let $\aleph_0 := |\mathbb{N}|$ (The symbol $\aleph$ is pronounced ah-lehf or ahl-ehf). From this we can start to express the cardinalities of other sets in terms of $\aleph_0$.
To compare different cardinalities we consider if there exists a bijection between the sets. Bijections are where a function from one set to another is both one-to-one (also known as "injective", where each input value maps to a unique output value) and onto (also known as "surjective", where every output value is mapped an input value). E.g. the sets $A = \{1, 3, 5, 7\}$ and $B = \{0, 2, 4, 6\}$ have equal cardinalities because there exists a function $f(n) = n - 1 = m$, s.t. $n \in A$ and $m \in B$, so that every element of $A$ is paired a unique element of $B$ and vice-versa. The same can be done with one set of all uppercase letters and one set of all lowercase letters in English: simply let each letter map to its upper/lower version of itself.
Intuitively speaking, we would say that $|\mathbb{Z}| = 2 \cdot |\mathbb{N}_1| + 1 > |\mathbb{N}_1|$ since the integer set has two "versions" of each positive integer (itself and its negative) and $0$. However, Cantor found that, using a bijection, $|\mathbb{Z}| =  |\mathbb{N}|$. To prove so, write $\mathbb{Z}$ as
$$\mathbb{Z} = \{0, 1, -1, 2, -2, ...\}$$
Then what we find is that we can assign a natural number to each element of $\mathbb{Z}$ starting at $1$ or $0$ (arbitrary) then incrementing each element by $1$. I.e. $f(0) = 0$, $f(1) = 1$, $f(2) = -1$, $f(3) = 2$, $f(4) = -2$, etc. This means that for any positive integers $n$ and $m$, it is true that $n \cdot \aleph_0 + m = \aleph_0$, despite how absurb that may seem.
Now consider the rational set $\mathbb{Q}$. Intuitively speaking, we would suggest that $|\mathbb{Q}| < |\mathbb{Z}| \cdot |\mathbb{N}_1|$ since every rational number has an integer as its numerator and a non-zero natural as its demoninator and there are repeats (e.g. $10/2 = 5/1$). Yet again, Cantor showed that $|\mathbb{Q}| = |\mathbb{N}|$. To prove so express $\mathbb{Q}$ as
$$\mathbb{Q} =
\begin{Bmatrix} 0, & 1/1, & -1/1, & 2/1, & \cdots,\\
            & 1/2, & -1/2, & 2/2, & \cdots,\\
            & 1/3, & -1/3, & 2/3, & \cdots,\\
            & \vdots & \vdots & \vdots & \ddots
\end{Bmatrix}$$
If we draw a line starting at $0$ going through $1/1$ to $-1/1$, then down-left to $1/2$ down to $1/3$ then up-right through $-1/2$ to $2/1$ and so on, we can then index each rational via a natural number. Ergo, a bijection between $\mathbb{Q}$ and $\mathbb{N}$ exists meaning that $|\mathbb{Q}| = |\mathbb{N}|$. From this it can be proven that for any positive integer $n$, ${\aleph}_0^n = \aleph_0$.
Moving onto the real set, Cantor showed that a contradiction occurs when attempting to assume that $|\mathbb{R}| = |\mathbb{N}|$. Starting simple, suppose such a bijection exists between $(0, 1)$ and $\mathbb{N}_1$ then let's write $(0, 1)$ as
$$\begin{array}{c|c}
\mathbb{N_1} & \mathbb{R}\\
1 & 0.550501 \cdots\\
2 & 0.101011 \cdots\\
3 & 0.999099 \cdots\\
4 & 0.876123 \cdots\\
\vdots & \vdots
\end{array}$$
Now lets make a real number by carefully choosing its digits based on the above table: let the integer part be $0$, the first decimal place not be whatever the first decimal place of $f(1)$ is (i.e. $5$), the second decimal place not be the second decimal place of $f(2)$ is (i.e. $0$), etc. Now notice that our new real number is an element of $(0, 1)$, isn't $f(1)$ because the first decimal place differs, isn't $f(2)$ because the second decimal place differs and so on so forth. Hence, our new number isn't in our table above. You may ask "Well, can't we just put the new number in there?" and we certainly can; however, we can simply repeat the process above to find yet another number not in our table. In fact, even if we could repeat this process an $\aleph_0$ number of times we still wouldn't have all of them. Ergo, $|\mathbb{N}| < |(0, 1)|$, which obviously means $|\mathbb{N}| < |\mathbb{R} \setminus \mathbb{Q}|$.
I encourage you to further research this "transfinite number theory", as I personally call it, since you will find many counter-intuitive, yet sometimes very simple to prove, theorems. E.g. $|(0, 1)| = |\mathbb{R}| = 2^{\aleph_0} = 3^{\aleph_0} = \cdots = {\aleph_0}^{\aleph_0}$. There are even wierder and increasingly abstract infinities such as $\aleph_1$, $\aleph_2$, etc. This becomes much more complicated and abstract when we start using "transfinite ordinal numbers" which is where Cantor's work starts really to become fascinating.
