Stuck with infinities I have heard this "some infinities are bigger than others" . How can this be ? 
The context was that the cardinality of the set of integers is less than that of the cardinality of th real numbers , but how do we prove this . The more I think about this it becomes more absurd . 
 A: For the particular results that the cardinality of the real numbers is greater than the cardinality of the integers, there are many proofs. Perhaps the most famous one is Cantor's diagonal argument for that fact, which actually exploits a typographical fact about the representation of real numbers as strings of digits. There are other proofs that involve more analytical facts about the real numbers, but they all require a bit of preparation. 
To answer your question it is perhaps better to consider Cantor's Theorem: For any set $X$, its cardinality is strictly smaller than the cardinality of $\mathcal P (X)$, the set of all subsets of $X$. This result then actually says not just that some infinite sets are larger than other sets, but that for any set, you can always find a set with strictly larger cardinality. The proof requires first an agreed upon way to measure which cardinality is bigger. The meaning of "The cardinality of a set $A$ is strictly smaller then the cardinality of a set $B$" is that there exists a injective function $f:A\to B$, and that there does not exist a surjective function $A\to B$. If you agree to that definition (which, I hope, makes sense at least), then here is a proof of Cantor's Theorem.
Let $A$ be a set and let $B=\mathcal P (A)$. The function $f:A\to B$ given by $f(a)=\{a\}$ is clearly injective. So, we just need to verify there is no surjection $g:A\to B$, so let's assume we are given such a surjection and we'll reach a contradiction. Consider the set $C=\{a\in A\mid a\notin g(a)\}$. Now, clearly $C\in \mathcal P (A)$, and thus $C=g(c)$ for some $c\in A$ (since we assumed $g$ is surjective). But then, try to answer the question "does $c\in g(c)$?". If it does, then $c\in g(c)=C$ and thus $c\notin g(c)$. If it does not, then $c\notin g(c)$, and thus $c\in C=g(c)$. A contradiction. 
A: The standard result here is that if $S$ is an arbitrary set, then the cardinality of $S$ is strictly smaller then the cardinality of the power set $\mathcal P(S)$, i.e., there is no surjection $\varphi: S \twoheadrightarrow \mathcal P(S)$.
The proof is Cantor's classic diagonal argument: suppose $\varphi$ were any such map, and consider the set $B = \{s \in S: s \notin \varphi(s)\}$. Then by definition, $s \in B \iff s \notin \varphi(s)$, so $B \neq \varphi(s)$ for any $s$, which implies $\varphi$ is not onto.
This implies you can give an infinite list of strictly increasing infinite cardinalities by $\mathbb Z, \mathcal P(\mathbb Z), \mathcal P^2(\mathbb Z), \mathcal P^3(\mathbb Z), \cdots$. The second term in the list, $\mathcal P(\mathbb Z)$, can be shown to have the same cardinality as $\mathbb R$.
A: The classic proof that the cardinality of the set of real numbers is greater than that of the set of integers is Cantor's diagonalization argument.
In general, two infinite sets are considered the same "size" if there exists a one-to-one correspondence between them. There does not exist a one-to-one correspondence between the set of integers and the set of real numbers. This means that the cardinality of $\mathbb{Z}$ is not equal to the cardinality of $\mathbb{R}$.
Clearly, the set of integers is a subset of the set of real numbers, so the cardinality of $\mathbb{Z}$ cannot be greater than that of $\mathbb{R}$, so it must be less.

You may want to read about aleph numbers.
A: The reason we know that there are more real numbers than there are natural numbers, in terms of cardinality, is the following:
Assume we have a complete, enumerated list of all the real numbers (which would allow a bijection between $\Bbb N$ and $\Bbb R$).  Let's get a sample of that list:
$$\begin{align}&.123789714361\ldots\tag{1}\\
&.134897313982\ldots\tag{2}\\
&.274329379725\ldots\tag{3}\\
&.796972438937\ldots\tag{4}\end{align}\\\vdots$$
Now, we're going to make a real number that is not in our list.  How do we do that, you might ask?  Great question!
We're going to consider the $n^{th}$ digit in the $n^{th}$ number.  If that digit is $1$, our number will have a $2$ in that place.  Otherwise, it will have a $1$.  So, based on our first four numbers, our number would begin with $.2111$.  Since our magical number is different from every number in our enumerated list in at least one digits place, our number is not on the list, and so we do not have a complete bijection onto the real numbers.
A: The answer is simple. Mathematics works with precise definitions.
To say that two sets have the same cardinality is to say that these two sets satisfy that certain definition. To say that they don't have the same cardinality means that they don't satisfy the definition. That is all.
We define two sets $A$ and $B$ to have the same cardinality when there is a function $f$ whose domain is $A$, its range is exactly $B$, and it has the properties that two distinct elements of $A$ are mapped to distinct elements of $B$. Note that a function in mathematics is an abstract idea, it doesn't have to be expressed with a formula like $\sin(x^2)+3\ln(x)-1$.
Finally, to show that two infinite sets don't have the same cardinality we need to show that there is no such function. How do you show that something doesn't exist? You can start with assuming it exists and reach a contradiction. Or you can just show, sometimes, directly from the definitions that it cannot exist.
We can show directly that any function from the integers to the real numbers misses at least one real number. So it is impossible to find any function from the integers whose range is exactly the real numbers, it will always be a subset thereof. This shows that the definitions of "have the same cardinality" does not apply for the two sets "the integers" and "the real numbers".
