Any real number has at most two decimal representations Here,  Tao says that

any real number has at most two decimal representations.

Is this really true? I always thought $\pi$ has only one decimal representation.
 A: $\pi$ does indeed have only one decimal representation. However, some numbers have two! For example, $$1.0000...=0.9999....$$ Based on this, it's reasonable to ask how many decimal representations a number can have. E.g., is there a number with three decimal representations?
Tao's comment is that the answer is no: every number has either one or two decimal representations, that is, at most two.
A: 
any real number has at most two decimal representations.



*

*to dispute the truth of this, you need to find a real number with more than two decimal representations.

A: Most real numbers, including all the irrationals like $\pi$, have only one decimal expansion.  The only ones that have two are rationals with terminating decimals like $\frac 12$.  You can write $\frac 12=0.50000\ldots =0.499999\ldots $
A: As many others have said 'all have at most two' is no contradiction to 'this specific one has exactly one' not even to 'all have exactly one.' 
But I want to stress something else, namely where your surprise likely comes from.
Many expositions indeed adopt a convention for "decimal representations" that makes it so  that every real number has a unique decimal representation. And against this background the remark, while not in itself a contradictory to it, seems peculiar. Indeed, there a different convention seems to be adopted. 
This has however the slight drawback that the definition of decimal representation is such that it is slightly less natural. 
Let us get to the bottom of this by starting at the start, that is by considering what do we even mean by a decimal representation. For simplicity I will only consider non-negative numbers less than $1$. 
The decimal expansion is $0.d_1d_2d_3 \dots$ where each $d_i \in \{0,\dots, 9\}$. 
So for a start we could say naively the decimal expansion of a number $0 \le x \lt 1$ is 'the' sequence $(d_i)_{i \in \mathbb{N}}$ such that $x = \sum_{i=1}^{\infty} d_i 10^{-i}$. 
That seems all good an well at first but if we consider, say, $x=1/2$ then we find that besides the familiar $(5,0,0,\dots) $ also the sequence $ (4,9,9,9,9, \dots)$ would be 'the' decimal expansion. This is not good there are two ways out. 
Version 1 (slight ambiguity):

A decimal expansion of a number $0 \le x \lt 1$ is a sequence $(d_i)_{i \in \mathbb{N}}$ such that $x = \sum_{i=1}^{\infty} d_i 10^{-i}$.   

Then one can show that each number has at least one and at most two decimal representations. Those that have two are numbers of the form $a/10^n$ with $a$ a non-zero integer, i.e., those that whose decimal representations 'end.'
Version 2 (restrict the sequences):

The decimal expansion of a number $0 \le x \lt 1$ is a sequence $(d_i)_{i \in \mathbb{N}}$ that is not eventually constant $9$ such that $x = \sum_{i=1}^{\infty} d_i 10^{-i}$.   

Version 2' (restrict the sequences differently):

The decimal expansion of a number $0 \le x \lt 1$ is a sequence $(d_i)_{i \in \mathbb{N}}$ that is not eventually constant $0$ such that $x = \sum_{i=1}^{\infty} d_i 10^{-i}$.   

Both 2 and 2' allow to show that every number has a unique decimal representation. 
The most common one is in my experience V2, although restriction is not always made explicit. In the source you mention V1 is adopted.
A: "All real numbers have at most two decimal representations" means that all real numbers have either zero, one, or two decimal representations. 
Of course, all real numbers have at least one decimal representation, so we might have instead said:

Each real number has either one or two decimal representations— never more, never less.


We can divide all decimal expansions into three categories:


*

*If the decimal expansion of $x$ terminates, you can get a second representation by decrementing the last digit and appending infinite 9s. For example, $0.128 \equiv 0.127999\overline{9}$.

*Conversely, if the decimal expansion of $x$ eventually results in an infinite sequence of 9s: $\ldots 999\overline{9}$, you can get a second representation by deleting the sequence of 9s and incrementing the final digit. For example, $0.127999\overline{9} \equiv 0.128$.  

*There is only one other case: if the decimal expansion of $x$ continues forever without resulting in infinite 9s, it is the one and only representation of $x$.

