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How can doubles represent higher numbers than decimals if they can't hold as many significant figures?

I may very well have not the proper understanding of significant figures, but the book
C# 6.0 in a Nutshell by Joseph Albahari and Ben Albahari (O’Reilly).
Copyright 2016 Joseph Albahari and Ben Albahari, 978-1-491-92706-9.
provides the table below for comparing double and decimal:
Is it not counter-intuitive that, on the one hand, a double can hold a smaller quantity of significant figures, while on the other it can represent numbers way bigger than decimal, which can hold a higher quantity of significant figures ?

Imagine you were told you can store a value, but were given a limitation: You can only store 10 digits, 0-9 and a negative symbol. You can create the rules to decode the value, so you can store any value.
The first way you store things is simply as the value xxxxxxxxxx, meaning the number 123 is stored as 0000000123. Simple to store and read. This is how an int works.
Now you decide you want to store fractional numbers, so you change the rules a bit. Now you store xxxxxxyyyy, where x is the integer portion and y is the fractional portion. So, 123.98 would be stored as 0001239800. This is roughly how a Decimal value works. You can see the largest value I can store is 9999999999, which translates to 999999.9999. This means I have a hard upper limit on the size of the value, but the number of the significant digits is large at 10.
There is a way to store larger values, and that's to store the x and y components for the formula in xxxxxxyyyy. So, to store 123.98, you need to store 01239800-2, which I can calculate as . This means I can store much bigger numbers by changing 'y', but the number of significant digits is basically fixed at 6. This is basically how a double works.

The answer lies in the way that doubles are encoded. Rather than just being a direct binary representation of a number, they have 3 parts: sign, exponent, and fraction.
The sign is obvious, it controls + or -.
The fraction part is also obvious. It's binary fraction that represents a number in between 0 and 1.
The exponent is where the magic happens. It signifies a scaling factor.
The final float calculation comes out to (-1)^$sign * (1 + $fraction) * 2 ^$exponent
This allows much higher values than a straight decimal number because of the exponent. There's a lot of reading out there on why this works and how to do addition and multiplication with these encoded numbers. Google around for "IEEE floating point format" or whatever topic you need. Hope that helps!

The Range has nothing to do with the precision. Double has a binary representation (base 2). Not all numbers can be represented exactly as we humans know them in the decimal format. Not to mention and accumulated rounding errors of addition and division. A larger range means a greater MAX VALUE and a smaller MIN VALUE than decimal.
Decimal on the other side is (base 10). It has a smaller range (smaller MAX VALUE and larger MIN VALUE). This has nothing to do with precision, since it is not represented using floating binary point representation, it can represent numbers more precisely and though is recommended for human-made numbers and calculations.

Bitwise representation of division of floats - how division of floats works

A number can have multiple representations if we use a float, so the results of a division of floats may produce bitwise different floats. But what if the denominator is a power of 2?
AFAIK, dividing by a power of 2 would only shift the exponent, leaving the same mantissa, always producing bitwise identical floats. Is that right?
float a = xxx;
float result = n/1024f; // always the same result?
--- UPDATE ----------------------
Sorry for my lack of knowledge in the IEEE black magic for floating points :) , but I'm talking about those numbers Guvante mentioned: no representation for certain decimal numbers, 'inaccurate' floats. For the rest of this post I'll use 'accurate' and 'inaccurate' considering Guvante's definition of these words.
To simplify, let's say the numerator is always an 'accurate' number. Also, let's divide not by any power of 2, but always for 1024. Additionally, I'm doing the operation the same way every time (same method), so I'm talking about getting the same results in different executions (for the same inputs, sure).
I'm asking all this because I see different numbers coming from the same inputs, so I thought: well if I only use 'accurate' floats as numerators and divide by 1024 I will only shift the exponent, still having an 'accurate' float.
You asked for an example. The real problem is this: I have a simulator producing sometimes 0.02999994 and sometimes 0.03000000 for the same inputs. I thought I could multiply these numbers by 1024, round to get an 'integer' ('accurate' float) that would be the same for those two numbers, and then divide by 1024 to get an 'accurate' rounded float.
I was told (in my other question) that I could convert to decimal, round and cast to float, but I want to know if this way works.

A number can have multiple representations if we use a float
The question appears to be predicated on an incorrect premise; the only number that has multiple representations as a float is zero, which can be represented as either "positive zero" or "negative zero". Other than zero a given number only has one representation as a float, assuming that you are talking about the "double" or "float" types.
Or perhaps I misunderstand. Is the issue that you are referring to the fact that the compiler is permitted to do floating point operations in higher precision than the 32 or 64 bits available for storage? That can cause divisions and multiplications to produce different results in some cases.

Since people often don't fully grasp floating point numbers I will go over some of your points real quick. Each particular combination of bits in a floating point number represent a unique number. However because that number has a base 2 fractional component, there is no representation for certain decimal numbers. For instance 1.1. In those cases you take the closest number. IEEE 754-2008 specifies round to nearest, ties to even in these cases.
The real difficulty is when you combine two of these 'inaccurate' numbers. This can introduce problems as each intermediate step will involve rounding. If you calculate the same value using two different methods, you could come up with subtly different values. Typically this is handled with an epsilon when you want equality.
Now onto your real question, can you divide by a power of two and avoid introducing any additional 'inaccuracies'? Normally you can, however as with all floating point numbers, denormals and other odd cases have their own logic, and obviously if your mantissa overflows you will have difficulty. And again note, that no mathematical errors are introduced during any of this, it is simply math being done with limited percision, which involves intermittent rounding of results.
EDIT: In response to new question
What you are saying could work, but is pretty much equivalent to rounding. Additionally if you are just looking for equality, you should use an episilon as I mentioned earlier (a - b) < e for some small value e (0.0001 would work in your example). If you are looking to print out a pretty number, and the framework you are using isn't doing it to your liking, some rounding would be the most direct way of describing your solution, which is always a plus.

Why can't c# calculate exact values of mathematical functions

Why can't c# do any exact operations.
Math.Pow(Math.Sqrt(2.0),2) == 2.0000000000000004
I know how doubles work, I know where the rounding error is from, I know that it's almost the correct value, and I know that you can't store infinite numbers in a finite double. But why isn't there a way that c# can calculate it exactly, while my calculator can do it.
Edit
It's not about my calculator, I was just giving an example:
http://www.wolframalpha.com/input/?i=Sqrt%282.000000000000000000000000000000000000000000000000000000000000000000000000000000001%29%5E2
Cheers

Chances are your calculator can't do it exactly - but it's probably storing more information than it's displaying, so the error after squaring ends up outside the bounds of what's displayed. Either that, or its errors happen to cancel out in this case - but that's not the same as getting it exactly right in a deliberate way.
Another option is that the calculator is remembering the operations that resulted in the previous results, and applying algebra to cancel out the operations... that seems pretty unlikely though. .NET certainly won't try to do that - it will calculate the intermediate value (the root of two) and then square it.
If you think you can do any better, I suggest you try writing out the square root of two to (say) 50 decimal places, and then square it exactly. See whether you come out with exactly 2...

Your calculator is not calculating it exactly, it just that the rounding error is so small that it's not displayed.

I believe most calculators use binary-coded decimals, which is the equivalent of C#'s decimal type (and thus is entirely accurate). That is, each byte contains two digits of the number and maths is done via logarithms.

What makes you think your calculator can do it? It's almost certainly displaying less digits than it calculates with and you'd get the 'correct' result if you printed out your 2.0000000000000004 with only five fractional digits (for example).
I think you'll probably find that it can't. When I do the square root of 2 and then multiply that by itself, I get 1.999999998.
The square root of 2 is one of those annoying irrational numbers like PI and therefore can't be represented with normal IEEE754 doubles or even decimal types. To represent it exactly, you need a system capable of symbolic math where the value is stored as "the square root of two" so that subsequent calculations can deliver correct results.

The way calculators round up numbers vary from model to model. My TI Voyage 200 does algebra to simplify equations (among other things) but most calculators will display only a portion of the real value calculated, after applying a round function on the result. For example, you may find the square root of 2 and the calculator would store (let's say) 54 decimals, but will only display 12 rounded decimals. Thus when doing a square root of 2, then do a power of that result by 2 would return the same value since the result is rounded. In any case, unless the calculator can keep an infinite number of decimals, you'll always have a best approximate result from complexe operations.
By the way, try to represent 10.0 in binary and you'll realize that you can't represent it evenly and you'll end up with (something like) 10.00000000000..01

Your calculator has methods which recognize and manipulate irrational input values.
For example: 2^(1/2) is likely not evaluated to a number in the calculator if you do not explicitly tell it to do so (as in the ti89/92).
Additionally, the calculator has logic it can use to manipulate them such as x^(1/2) * y^(1/2) = (x*y)^1/2 where it can then wash, rinse, repeat the method for working with irrational values.
If you were to give c# some method to do this, I suppose it could as well. After all, algebraic solvers such as mathematica are not magical.

It has been mentioned before, but I think what you are looking for is a computer algebra system. Examples of these are Maxima and Mathematica, and they are designed solely to provide exact values to mathematical calculations, something not covered by the CPU.
The mathematical routines in languages like C# are designed for numerical calculations: it is expected that if you are doing calculations as a program you will have simplified it already, or you will only need a numerical result.

2.0000000000000004 and 2. are both represented as 10. in single precision. In your case, using single precision for C# should give the exact answer
For your other example, Wolfram Alpha may use higher precision than machine precision for calculation. This adds a big performance penalty. For instance, in Mathematica, going to higher precision makes calculations about 300 times slower
k = 1000000;
vec1 = RandomReal[1, k];
vec2 = SetPrecision[vec1, 20];
AbsoluteTiming[vec1^2;]
AbsoluteTiming[vec2^2;]
It's 0.01 second vs 3 seconds on my machine
You can see the difference in results using single precision and double precision introduced by doing something like the following in Java
public class Bits {
public static void main(String[] args) {
double a1=2.0;
float a2=(float)2.0;
double b1=Math.pow(Math.sqrt(a1),2);
float b2=(float)Math.pow(Math.sqrt(a2),2);
System.out.println(Long.toBinaryString(Double.doubleToRawLongBits(a1)));
System.out.println(Integer.toBinaryString(Float.floatToRawIntBits(a2)));
System.out.println(Long.toBinaryString(Double.doubleToRawLongBits(b1)));
System.out.println(Integer.toBinaryString(Float.floatToRawIntBits(b2)));
}
}
You can see that single precision result is exact, whereas double precision is off by one bit

C#: Storing percentages, 50 or 0.50?

When holding percentage values in variables is there a preference between holding them as whole numbers vs fractions. That is should the variable hold numbers between 0 and 100 or between 0.00 and 1.00? In either case, the variable holding the values is of decimal type.
The database I'm interacting with happens to store them as whole numbers 0 to 100.
Note: I'm using the word "whole number" to denote values in the range 0 to 100 though those values may contain fractional components (e.g., 25.75). I don't know how else to describe the difference between the two ranges of percentage values

I would be inclined to store them as 0 to 1, then there is no conversion required when using the number in a calculation.
Using a number between 0 and 100 is more of a display / readability way of looking at the number. When you are displaying the percentage you can use String.Format("{0:P},percentage)
which will expect the number to be between 0 and 1 but will display as 0 to 100.

If you are using floating point fields (float, decimal, double), then 50.0 and 0.50 will have the same degree of precision. So, from this point, I would make decisions based on what similar fields in similar tables do, to give a sense of design unity.

If you are doing a lot of math with fractions, you want to store the numerator and denominator separately as whole numbers. When using fractions to represent repeating or non-repeating decimals you can get rounding errors that grow the more operations you have.
If you are just storing and presenting the number, then it really doesn't matter.
What math do you plan to do with this number, if any?

You can use doubles to represent both percentages and 0-1. But if you choose to store percentage as "whole number" I would advise against using integers, which would limit you if you ever want to represent 12.3%, for example.

I think this is trivial... both are the same
I mean in terms of calculations/displaying info...
00.50 is cool for calculations
50.00 is cool for displaying

Is a double really unsuitable for money?

I always tell in c# a variable of type double is not suitable for money. All weird things could happen. But I can't seem to create an example to demonstrate some of these issues. Can anyone provide such an example?
(edit; this post was originally tagged C#; some replies refer to specific details of decimal, which therefore means System.Decimal).
(edit 2: I was specific asking for some c# code, so I don't think this is language agnostic only)

Very, very unsuitable. Use decimal.
double x = 3.65, y = 0.05, z = 3.7;
Console.WriteLine((x + y) == z); // false
(example from Jon's page here - recommended reading ;-p)

You will get odd errors effectively caused by rounding. In addition, comparisons with exact values are extremely tricky - you usually need to apply some sort of epsilon to check for the actual value being "near" a particular one.
Here's a concrete example:
using System;
class Test
{
static void Main()
{
double x = 0.1;
double y = x + x + x;
Console.WriteLine(y == 0.3); // Prints False
}
}

Yes it's unsuitable.
If I remember correctly double has about 17 significant numbers, so normally rounding errors will take place far behind the decimal point. Most financial software uses 4 decimals behind the decimal point, that leaves 13 decimals to work with so the maximum number you can work with for single operations is still very much higher than the USA national debt. But rounding errors will add up over time. If your software runs for a long time you'll eventually start losing cents. Certain operations will make this worse. For example adding large amounts to small amounts will cause a significant loss of precision.
You need fixed point datatypes for money operations, most people don't mind if you lose a cent here and there but accountants aren't like most people..
edit
According to this site http://msdn.microsoft.com/en-us/library/678hzkk9.aspx Doubles actually have 15 to 16 significant digits instead of 17.
#Jon Skeet decimal is more suitable than double because of its higher precision, 28 or 29 significant decimals. That means less chance of accumulated rounding errors becoming significant. Fixed point datatypes (ie integers that represent cents or 100th of a cent like I've seen used) like Boojum mentions are actually better suited.

Since decimal uses a scaling factor of multiples of 10, numbers like 0.1 can be represented exactly. In essence, the decimal type represents this as 1 / 10 ^ 1, whereas a double would represent this as 104857 / 2 ^ 20 (in reality it would be more like really-big-number / 2 ^ 1023).
A decimal can exactly represent any base 10 value with up to 28/29 significant digits (like 0.1). A double can't.

My understanding is that most financial systems express currency using integers -- i.e., counting everything in cents.
IEEE double precision actually can represent all integers exactly in the range -2^53 through +2^53. (Hacker's Delight, pg. 262) If you use only addition, subtraction and multiplication, and keep everything to integers within this range then you should see no loss of precision. I'd be very wary of division or more complex operations, however.

Using double when you don't know what you are doing is unsuitable.
"double" can represent an amount of a trillion dollars with an error of 1/90th of a cent. So you will get highly precise results. Want to calculate how much it costs to put a man on Mars and get him back alive? double will do just fine.
But with money there are often very specific rules saying that a certain calculation must give a certain result and no other. If you calculate an amount that is very very very close to $98.135 then there will often be a rule that determines whether the result should be $98.14 or $98.13 and you must follow that rule and get the result that is required.
Depending on where you live, using 64 bit integers to represent cents or pennies or kopeks or whatever is the smallest unit in your country will usually work just fine. For example, 64 bit signed integers representing cents can represent values up to 92,223 trillion dollars. 32 bit integers are usually unsuitable.

No a double will always have rounding errors, use "decimal" if you're on .Net...

Actually floating-point double is perfectly well suited to representing amounts of money as long as you pick a suitable unit.
See http://www.idinews.com/moneyRep.html
So is fixed-point long. Either consumes 8 bytes, surely preferable to the 16 consumed by a decimal item.
Whether or not something works (i.e. yields the expected and correct result) is not a matter of either voting or individual preference. A technique either works or it doesn't.