cantor's diagonalization argument (multiple sizes of infinities) Yes, many people have asked the same question, and I have tried by best to get my head around it, but I still can't understand.
Every time I look at the proofs, they seem like a sleight of hand to me..
Kindly point out the faults in my arguments:
Argument - 1

To map every Real Number between 0 and 1, to a unique Natural Number,
  simply remove all trailing zeros, and then remove the decimal
  point! so,
    0.1                        maps to                1
    0.254                    maps to             254
    0.2540000           maps to             254               (because we are removing the trailing zeros)
    0.31415926...       maps to             31415926...    (if a real number can have infinite precision, then most probably a natural number can be
  infinitely large? Am I wrong?)

Argument - 2

Every pair of Natural Numbers can be mapped to a unique Natural
  Number. (for eg, Cantor's Pairing Function). Every Rational Number
  'r' can be mapped to a pair of Natural Numbers (p,q) such
  that        r = p/q Since for every rational number 'r', we have an
  infinite number of such pairs       {(p,q),(2p,2q),(3p,3q)...} the
  cardinality of Natural Numbers must be greater than the cardinality of
  Rational Numbers. So, Irrational Number can be mapped to Natural
  Numbers. So, Real Numbers can be mapped to Natural Numbers.

 A: Your first Argument is invalid because there exists no such Number $n\in \mathbb{N}$ with (f. ex.)
$$\pi = 10^k\cdot n$$
which is essentially what you require.
Your second Argument has the same error. There only exists a bijection from
$$\mathbb{N} \to \mathbb{Q}$$
but none from $\mathbb{N} \to \mathbb{R} \setminus \mathbb{Q}$ nor from $\mathbb{N} \to \mathbb{R}$
A: The problem with argument 1 is that no, natural numbers cannot be infinitely long, and so your mapping has no natural number to which $\frac{\pi}{10}$ maps.
The (Well, one, at least) problem with argument 2 is that you assume that there being an infinite number of pairs of naturals that represent each rational means that there are more natural numbers than rationals.
You have established that there is an injection from $\mathbb{N}^2 \to \mathbb{Q}$, and so the cardinality of the rationals is at least the cardinality of $\mathbb{N}^2$, which is the same as the cardinality of $\mathbb{N}$.
However, there are also injections from $\mathbb{Q}$ to $\mathbb{N}^2$ (For example, consider the function that takes the fraction $\frac{p}{q}$ to $(p,q)$, where $p$ and $q$ are relatively prime.), and so rather than $|\mathbb{N}|$ being greater than $|\mathbb{Q}|$, we can see that they are, in fact, equal.
A: There is one other point to consider.  It is quite common among those who do not understand diagonalization that they do what you did: construct a map they believe works.  The answers so far do a good job of showing what is wrong with your attempts.  But you need to understand that, despite you not yet understanding diagonalization, it is correct nonetheless.  Any mathematician knows immediately that attempts like yours have to be wrong because diagonalization is correct.
So what would probably be more useful in your studies is to figure out exactly what parts of Cantor don't make sense to you, and then learn why they are correct.  Who knows--not all proofs are perfect, and mathematicians do find errors in proofs.  Diagonalization is very well studied, so you aren't likely to find an error. But the point is that the procedure that might convince a mathematician is to show where a seemingly correct proof is wrong, not to exhibit an example that a correct proof says does not exist.
