Much less than, what does that mean? What exactly does $\ll$ mean?
I am familiar that this symbol means much less than. 
...but what exactly does "much less than" mean? (Or the corollary, $\gg$)
On Wikipedia, the example they use is that $1\ll 9999999999$
But my thought on that is that $10^{10^{10^{11}}}\ll 10^{10^{10^{11}}}+9999999999$, based on the same logic. But I am confused because the numbers are comparitively. 
Logically, I am really kinda spent dealing with this issue of mine. Thanks for any input.
 A: It has no precise meaning.
It's just a shorthand for saying "sufficiently small" or "easily seen to be so much smaller that the following argument will work" -- and shouldn't be expected to have a more precise technical content than it would have to write those fuzzy descriptions out in words.
A: A (too) simple definition would be
$$
a \ll b ~ \Leftrightarrow ~ \frac{a}{b} \approx 0
$$
There may be some context implied: a variable tending toward $0$ or $\infty$, or a certain imprecision level that the user is willing to accept.
A: Much less than, much more than is useful in mathematical formulas applying to everyday life.
If I must find a life saving value using a formula, which has a part that doesn't contribute significantly to the end result, there it has a meaning.(<< or >>). Otherwise you might lose precious life saving time
It doesn't have to be precisely defined theoretically(and imo it can't); practically speaking though it can. 
Would you use relativistic equations for computing physics and engineering in everyday tasks(γ= 1 / SQR 1- (v2/c2)  )  ??
Why not? Because C is much much bigger than every day life speed that it's only a waste of time to compute it in this formula, because v2/c2  would be <<1, meaning it wouldn't appreciably change the end result. 
So, it doesn't matter if it can't be defined theoretically or not, as long as it has a practical value.
A: "Much less than" is a qualitative assessment of comparative inequality.
$a\ll b$ means that $a$ is not only less than $b$, but muchly so by some convention.
That convention is fairly arbitrary, or rather contextual; there is no fixed quantitative basis.   Sometimes it is based on additive position, sometimes on multiplicative magnitude.   It's mostly based on utility and context; it's a declaration that $a$ is less enough than $b$ to be useful in the discussion (such as the case where certain terms involving $a$ become vanishingly small enough to conveniently ignore).   To know what sufficiently less means, you have to understand what the context of the comparison is.
A: In certain contexts (primarily in Number Theory), $a(x) \ll b(x)$ is an alternative notation to  $a(x) = O(b(x))$.  See e.g. this question.
A: The entire point is that it's NOT very clear what "much less than" is. What is "much less than"? In some contexts $1\ll 2$ and in others $ 1\not \ll 2$ but $1 \ll 100000$ and in others, even $1\not\ll 100000$. What precisely makes something "much less than"?
In most contexts, $\ll$ is used in approximation. For example: for $0<x\ll 1$, $(1+x)^n\approx 1+nx$.
So how small does $x$ have to be? Well small enough for the approximation to be "good enough"! What is good enough? It seems like I'm just pushing off the question, and that's indeed what I'm doing!
The entire point isn't whether something is "small enough" (big enough), or if the approximation is "good enough", the point is control. Can you control the error in a manageable way.
For the example $0<x \ll 1, (1+x)^n \approx 1+nx$. You can make $(1+x)^n$ as close as you'd like to $1+nx$ by making $x$ small enough! I think this is a more rigorous meaning. 
$\ll$ is imprecise in the sense that you don't know how "small" something is. But it is precise in the sense that it implies that there is some control. There is some way of making the error in the approximation, argument, etc. as small as you'd like, provided that you make $x$ small enough.
A: Let me chip in from a physicists vantage... It's all about orders of magnitude. Are the observed effects so different in order that your simplified model is sufficiently accurate? In this case, sufficient depends on how closely you need to look and how feasible it is to even look deeper.
You'll rarely if ever see the $\ll$ sign or its counterpart unless the numbers are at least an order of magnitude apart. I love avid19's answer on this question, as the part on control is very intriguing. I can see that in many contexts, but I don't find that the notion of control is universally applied, in that we cannot choose the numbers that we measure from reality. We can control our required levels of tolerance in modelling reality, though.
Physically, then, there is a benchmark, and that is "Does the model work?" We approximate because the real world is a mess of particle interactions. Additionally, many effects that we observe are non-linear, that is, not too friendly to effectively compute. The harmonic oscillator is a great example.

To add to this quandry, let me reference Schroeder's book An Introduction to Thermal Physics. As a preface, this discussion is in regards to systems involving many, many elements, such as modelling atoms in a gas. You're either going to love or hate this excerpt:

There are three kinds of numbers that commonly occur in statistical mechanics: small numbers, large numbers, and very large numbers...
You already know how to manipulate small numbers.
Large numbers are much larger than small numbers... The most important property of large numbers is that you can add a small number to a large number without changing it. For example, $10^{23} + 23 = 10^{23}$. The only exception is when you plan to eventually subtract off the same large number: $10^{23} +42 - 10^{23} = 42$.
Very large numbers are even larger than large numbers... [they] have the amazing property that you can multiply them by large numbers without changing them. For instance, $10^{10^{23}} \times 10^{23} = 10^{10^{23} +23}= 10^{10^{23}}$.

A: Par example: Choose a priori positive number $N>1$ such that $0<x\ll y$. If $\frac{y-x}{x}>N$ for then $y>x\cdot(1+N)$. With repect to N, your 2nd example does not fit this requirement: $10^{10^{10^{10^{11}}}}\ll10^{10^{10^{10^{11}}}}+9999999999$. I had in mind $\frac{N-O}{O}\times 100\%$, the relative change has to overcome a number greater than one. If my profits this month is USD10,000 and next month it is USD20,000 then it is much greater if you have choose a priori p.e. $N=\frac{3}{2}$. One can play with the apriori choosen N to figure out a bit more. 
