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How to calculate float type precision and does it make sense?

I have a problem understanding the precision of float type.
The msdn writes that precision from 6 to 9 digits. But I note that precision depends from on the size of the number:
float smallNumber = 1.0000001f;
Console.WriteLine(smallNumber); // 1.0000001
bigNumber = 100000001f;
Console.WriteLine(bigNumber); // 100000000
The smallNumber is more precise than big, I understand IEEE754, but I don't understand how MSDN calculate precision, and does it make sense?
Also, you can play with the representation of numbers in float format here. Please write 100000000 value in "You entered" input and click "+1" on the right. Then change the input's value to 1, and click "+1" again. You may see the difference in precision.

The MSDN documentation is nonsensical and wrong.
Bad concept. Binary-floating-point format does not have any precision in decimal digits because it has no decimal digits at all. It represents numbers with a sign, a fixed number of binary digits (bits), and an exponent for a power of two.
Wrong on the high end. The floating-point format represents many numbers exactly, with infinite precision. For example, “3” is represented exactly. You can write it in decimal arbitrarily far, 3.0000000000…, and all of the decimal digits will be correct. Another example is 1.40129846432481707092372958328991613128026194187651577175706828388979108268586060148663818836212158203125•10−45. This number has 105 significant digits in decimal, but the float format represents it exactly (it is 2−149).
Wrong on the low end. When “999999.97” is converted from decimal to float, the result is 1,000,000. So not even one decimal digit is correct.
Not a measure of accuracy. Because the float significand has 24 bits, the resolution of its lowest bit is about 223 times finer than the resolution of its highest bit. This is about 6.9 digits in the sense that log10223 is about 6.9. But that just tells us the resolution—the coarseness—of the representation. When we convert a number to the float format, we get a result that differs from the number by at most ½ of this resolution, because we round to the nearest representable value. So a conversion to float has a relative error of at most 1 part in 224, which corresponds to about 7.2 digits in the above sense. If we are using digits to measure resolution, then we say the resolution is about 7.2 digits, not that it is 6-9 digits.
Where do these numbers came from?
So, if “~6-9 digits” is not a correct concept, does not come from actual bounds on the digits, and does not measure accuracy, where does it come from? We cannot be sure, but 6 and 9 do appear in two descriptions of the float format.
6 is the largest number x for which this is guaranteed:
If any decimal numeral with at most x significant digits is within the normal exponent bounds of the float format and is converted to the nearest value represented in the format, then, when the result is converted to the nearest decimal numeral with at most x significant digits, the result of that conversion equals the original number.
So it is reasonable to say float can preserve at least six decimal digits. However, as we will see, there is no bound involving nine digits.
9 is the smallest number x that guarantees this:
If any finite float number is converted to the nearest decimal numeral with x digits, then, when the result is converted to the nearest value representable in float, the result of that conversion equals the original number.
As an analogy, if float is a container, then the largest “decimal container” guaranteed to fit inside it is six digits, and the smallest “decimal container” guaranteed to hold it is nine digits. 6 and 9 are akin to interior and exterior measurements of the float container.
Suppose you had a block 7.2 units long, and you were looking at its placement on a line of bricks each 1 unit long. If you put the start of the block at the start of a brick, it will extend 7.2 bricks. However, if somebody else chooses where it starts, they might start it in the middle of a brick. Then it would cover part of that brick, all of the next 6 bricks, and and part of the last brick (e.g., .5 + 6 + .7 = 7.2). So a 7.2-unit block is only guaranteed to cover 6 bricks. Conversely, 8 bricks can cover the 7.2-unit block if you choose where they are placed. But if somebody else chooses where they start, the first might cover just .1 units of the block. Then you need 7 more and another fraction, so 9 bricks are needed.
The reason this analogy holds is that powers of two and powers of 10 are irregularly spaced relative to each other. 210 (1024) is near 103 (1000). 10 is the exponent used in the float format for numbers from 1024 (inclusive) to 2048 (exclusive). So this interval from 1024 to 2048 is like a block that has been placed just after the 100-1000 ends and the 1000-10,000 block starts.
But note that this property involving 9 digits is the exterior measurement—it is not a capability that float can perform or a service that it can provide. It is something that float needs (if it is to be held in a decimal format), not something it provides. So it is not a bound on how many digits a float can store.
Further Reading
For better understanding of floating-point arithmetic, consider studying the IEEE-754 Standard for Floating-Point Arithmetic or a good textbook like Handbook of Floating-Point Arithmetic by Jean-Michel Muller et al.

Yes number of digits before rounding errors is a measure of precision but you can not asses precision from just 2 numbers because you might be just closer or further from the rounding threshold.
To better understand the situation then you need to see how floats are represented.
The IEEE754 32bit floats are stored as:
bool(1bit sign) * integer(24bit mantisa) << integer(8bit exponent)
Yes mantissa is 24 bit instead of 23 as it's MSB is implicitly set to 1.
As you can see there are only integers and bitshift. So if you are representing natural number up to 2^24 you are without rounding completely. Fro bigger numbers binary zero padding occurs from the right that causes the difference.
In case of digits after decimal points the zero padding occurs from the left. But there is another problem as in binary you can not store some decadic numbers exactly. For example:
0.3 dec = 0.100110011001100110011001100110011001100... bin
0.25 dec = 0.01 bin
As you can see the sequence of 0.3 dec in binary is infinite (like we can not write 1/3 in decadic) hence if crop it to only 24 bits you lose the rest and the number is not what you want anymore.
If you compare 0.3 and 0.125 the 0.125 is exact and 0.3 is not but 0.125 is much smaller than 0.3. So your measure is not correct unless explored more very close values that will cover the rounding steps and computing the max difference from such set. For example you could compare
1.0000001f
1.0000002f
1.0000003f
1.0000004f
1.0000005f
1.0000006f
1.0000007f
1.0000008f
1.0000009f
and remember the max difference of fabs(x-round(x)) and than do the same for
100000001
100000002
100000003
100000004
100000005
100000006
100000007
100000008
100000009
And then compare the two differences.
On top of all this you are missing one very important thing. And that is the errors while converting from text to binary and back which are usually even bigger. First of all try to print your numbers without rounding (for example force to print 20 decimal digits after decimal point).
Also the numbers are stored in binary base so in order to print them you need to convert to decadic base which involves multiplication and division by 10. The more bits are missing (zero pad) from the number the bigger the print errors are. To be as precise as you can a trick is used and that is to print the number in hex (no rounding errors) and then convert the hex string itself to decadic base on integer math. That is much more accurate then naive floating point prints. for more info see related QAs:
my best attempt to print 32 bit floats with least rounding errors (integer math only)
How do libraries/programming languages convert floats to strings
How do I convert a very long binary number to decimal?
Now to get back to number of "precise" digits represented by float. For integer part of number is that easy:
dec_digits = floor(log10(2^24)) = floor(7.22) = 7
However for digits after decimal point is this not as precise (for first few decadic digits) as there are a lot rounding going on. For more info see:
How do you print the EXACT value of a floating point number?

I think what they mean in their documentation is that depending on the number that the precision ranges from 6 to 9 decimal places. Go by the standard that is explained on the page you linked, sometimes Microsoft are a bit lazy when it comes to documentation, like the rest of us.
The problem with floating point is that it is inaccurate. If you put the number 1.05 into the site in your link you will notice that it cannot be accurately stored in floating point. It's actually stored as 1.0499999523162841796875. It's stored this way to do calculations faster. It's not great for money, e.g. what if your item is priced at $1.05 and you sell a billion of them.

The smallNumber is more precise than big
Incorrect compare. The other number has more significant digits.
1.0000001f is attempting N digits of decimal precision.
100000001f attempts N+1.
I have a problem understanding the precision of float type.
To best understand float precision, think binary. Use "%a" for printing with a C99 or later compiler.
float is stored base 2. The significand is a Dyadic rational, some integer/power-of-2.
float commonly has 24 bits of binary precision. (23-bit explicitly encoded, 1 implied)
Between [1.0 ... 2.0), there are 223 different float values.
Between [2.0 ... 4.0), there are 223 different float values.
Between [4.0 ... 8.0), there are 223 different float values.
...
The possible values of a float are not distributed uniformly among powers-of-10. The grouping of float values to power-of-10 (decimal precision) results in the wobbling 6 to 9 decimal digits of precision.
How to calculate float type precision?
To find the difference between subsequent float values, since C99, use nextafterf()
Illustrative code:
#include<math.h>
#include<stdio.h>
void foooo(float b) {
float a = nextafterf(b, 0);
float c = nextafterf(b, b * 2.0f);
printf("%-15a %.9e\n", a, a);
printf("%-15a %.9e\n", b, b);
printf("%-15a %.9e\n", c, c);
printf("Local decimal precision %.2f digits\n", 1.0 - log10((c - b) / b));
}
int main(void) {
foooo(1.0000001f);
foooo(100000001.0f);
return 0;
}
Output
0x1p+0 1.000000000e+00
0x1.000002p+0 1.000000119e+00
0x1.000004p+0 1.000000238e+00
Local decimal precision 7.92 digits
0x1.7d783ep+26 9.999999200e+07
0x1.7d784p+26 1.000000000e+08
0x1.7d7842p+26 1.000000080e+08
Local decimal precision 8.10 digits

Convert float to double

I'm trying to convert Single to Double while maintaining the original value. I've found the following method:
Single f = 5.2F;
Double d1 = f; // 5.19999980926514
Double d2 = Double.Parse(f.ToString()); // 5.2 (correct)
Is this practice recommendable? I don't need an optimal method, but the intended value must be passed on to the double. Are there even consequences to storing a rounded value in a double?

You could use "decimal" instead of a string.
float f = 5.2F;
decimal dec = new decimal(f);//5.2
double d = (double)dec; //5.2

The conversion is exact. All the Single values can be represented by a Double value, because they are "built" in the same way, just with more possible digits. What you see as 5.2F is in truth 5.1999998092651368. If you go http://www.h-schmidt.net/FloatConverter/IEEE754.html and insert 5.2 you'll see that it has an exponent of 2^2 (so 4) and a mantissa of 1.2999999523162842. Now, if you multiply the two numbers you'll get 5.1999998092651368.
Single have a maximum precision of 7 digits, so .NET only shows 7 digits. With a little rounding 5.1999998092651368 is 5.2

If you know that your numbers are multiples of e.g. 0.01, I would suggest that you convert to double, round to the nearest integer, and subtract that to get the fractional residue. Multiply that by 100, round to the nearest integer, and then divide by 100. Add that to the whole-number part to get the nearest double representation to the multiple of 0.01 which is nearest the original number.
Note that depending upon where the float values originally came from, such treatment may or may not improve accuracy. The closest float value to 9000.02 is about 9000.019531, and the closest float value to 9000.021 is about 9000.021484f. If the values were arrived at by converting 9000.020 and 9000.021 to float, the difference between them should be about 0.01. If, however, they were arrived at by e.g. computing 9000f+0.019531f and 9000f+0.021484f, then the difference between them should be closer to 0.02. Rounding to the nearest 0.01 before the subtract would improve accuracy in the former case and degrade it in the latter.

Why does Convert.ToDecimal(Double) round to 15 significant figures?

I have a double with 17 digits after the decimal point, i.e.:
double myDouble = 0.12345678901234567;
If I convert this to a decimal like this:
decimal myDecimal = Convert.ToDecimal(myDouble);
then the value of myDecimal is rounded, as per the Convert.ToDecimal documentation, to 15 digits (i.e. 0.0123456789012345). My question is, why is this rounding performed?
I understand that if my original number could be accurately represented in base 10 and I was trying to store it as a double, then we could only have confidence in the first 15 digits. The final two digits would be subject to rounding error. But, that's a base 10 biased point of view. My number may be more accurately represented by a double and I wish to convert it to decimal while preserving as much accuracy as possible.
Shouldn't Convert.ToDecimal aim to minimise the difference between myDouble and (double)Convert.ToDecimal(myDouble)?

From the documentation of Double:
A Double value has up to 15 decimal digits of precision, although a
maximum of 17 digits is maintained internally
So, as the double value itself has a maximum of 15 decimal places, converting it to Decimal will result in a Decimal value with 15 significant figures.

The behavior of the rounding guarantees that conversion of any Decimal which has at most fifteen significant figures to double and back to Decimal will yield the original value unchanged. If values were rounded to sixteen figures rather than fifteen, such a guarantee would not only fail to hold for number with sixteen figures, but it wouldn't even hold for much shorter values. For example, the closest Double value to 9000.04 is approximately 9000.040000000000873115; rounding that to sixteen figures would yield 9000.040000000001.
The choice of rounding one should use depends upon whether regards the best Decimal value of double value 9000.04 as being 9000.04m, 9000.040000000001m, 9000.0400000000008731m, or perhaps something else. Microsoft probably decided that any representation other than 9000.04m would be confusing.

The following is from the documentation of the method in question.
http://msdn.microsoft.com/en-us/library/a69w9ca0(v=vs.110).aspx
"The Decimal value returned by this method contains a maximum of 15 significant digits. If the value parameter contains more than 15 significant digits, it is rounded using rounding to nearest.

Every terminating binary fraction is exactly representable as a decimal fraction, so the minimum possible difference for a finite number is always 0. The IEEE 754 64-bit representation of your number is exactly equal to 0.1234567890123456634920984242853592149913311004638671875
Every conversion from binary floating point to decimal or decimal string must embody some decision about rounding. In theory, they could preserve all the digits, but that would result in some very long outputs with most of the digits having little meaning.
One option, used in Java's Double.toString, is to stop at the shortest decimal representation that would convert back to the original double.
Most set some fairly arbitrary limit. 15 significant digits preserves most meaningful digits.

How to check if a double can fit into a float without conversion to infinity

Is there a standard way to check if a 64 bit floating point number can be converted to a 32 bit one without being converted to +/- Infinity?
I know it can be checked afterwards for +/- Inf.
Convert.ToSingle() won't do this.

In order to test whether a double value d will convert to float without yielding infinity, you can test whether d's absolute value is strictly below the double written as 3.4028235677973366e+38 (that is, 0x1.ffffffp+127 in hexadecimal). This number is the exact limit between doubles that can be converted to float and those that cannot.
To expand a little, the last finite float is:
0x1.fffffep+127 (3.4028234663852886e+38 in decimal)
What would be the next float number if type float had more exponent bits is:
0x2.000000p+127 (3.4028236692093846e+38 in decimal)
The numbers that are closer to the former than to the latter are rounded down to the former. The limit is the midpoint of the two numbers:
0x1.ffffffp+127 (3.4028235677973366e+38 in decimal)

Storing a double with 2 digit precision

How do I convert a double to have precision of 2 places?
For ex:
double x = 1.00d;
Console.WriteLine(Math.Round(x,2,MidpointRounding.AwayFromZero));
//Should output 1.00 for me, but it outputs just 1.
What I'm looking for is to store 1.00 in a double variable which when printed to console prints 1.00.
Thanks,
-Mike

"x" stores the number - it doesn't pay attention to the precision.
To output the number as a string with a specific amount of precision, use a format specifier, ie:
Console.WriteLine(x.ToString("F2"));

double is an IEEE-754 double-precision floating point number. It always has exactly 53 bits of precision. It does not have a precision that can be exactly represented in "decimal places" -- as the phrase implies, "decimal places" only measures precision relative to the base-10 number system.
If what you want is simply to change the precision the value is displayed with, then you can use one of the ToString() overloads to do that, as other answers here point out.
Remember too that double is an approximation of a continuous real value; that is, a value where there is no implied discrete quantization (although the values are in fact quantized to 53 bits). If what you are trying to represent is a discrete value, where there is an implicit quantization (for example, if you are working with currency values, where there is a significant difference between 0.1 and 0.099999999999999) then you should probably be using a data type that is based on integer semantics -- either a fixed-point representation using a scaled integer or something like the .NET decimal type.

This is not a question of storing, it is a question of formatting.
This code produces 1.00:
double x = 1.00d;
Console.WriteLine("{0:0.00}", Math.Round(x, 2, MidpointRounding.AwayFromZero));

The decimal data type has a concept of variable precision. The double data type does not. Its precision is always 53 bits.
The default string representation of a double rounds it slightly to minimize odd-looking values like 0.999999999999, and then drops any trailing zeros. As other answers note, you can change that behavior by using one of the type's ToString overloads.