We Keep Coding

Should this number be subtracted by 1 when implementing the Mersenne Twister?

I found this snippet online along with this Stackoverflow post which converts it into a TypeScript class.
I basically copy and pasted it verbatim (because I am not qualified to modify this sort of cryptographic code), but I noticed that VS Code has a little underline in the very last function:
/**
* generates a random number on [0,1) with 53-bit resolution
*/
nextNumber53(): number {
let a = this._nextInt32() >>> 5;
let b = this._nextInt32() >>> 6;
return (a * 67108864.0 + b) * (1.0 / 9007199254740992.0);
}
Specifically the 9007199254740992.0
VS Code says Numeric literals with absolute values equal to 2^53 or greater are too large to be represented accurately as integers.ts(80008)
I notice that if I subtract that number by one and instead make it 9007199254740991.0, then the warning goes away. But I don't necessarily want to modify the code and break it if this is indeed a significant difference.
Basically, I am unsure, because while my intuition says that having a numerical overflow is bad, my intuition also says that I shouldn't try to fix cryptographic code that was posted in several places, as it is probably correct.
But is it? Or should this number be subtracted by one?

9007199254740992 is the right value to use if you want Uniform values in [0,1), i.e. 0.0 <= x < 1.0.
This is just the automatics going awry, this value can be accurately represented by a JavaScript Number, i.e. a 64bit float. It's just 253 and binary IEEE 754 floats have no trouble with numbers of this form (it would even be represented accurately with a 32bit float).
Using 9007199254740991 would make the range [0,1], i.e. 0.0 <= x <= 1.0. Most libraries generate uniform values in [0,1) and other distributions are derived from that, but you are obviously free to do whatever is best for your application.
Note that the actual chance of getting the maximum value back is 2-53 (~1e-16) so you're unlikely not actually see it in practice.

Will I possibly loose any decimal digits (precision) when multiplying Number.MAX_SAFE_INTEGER by Math.random()?

Will I possibly loose any decimal digits (precision) when multiplying Number.MAX_SAFE_INTEGER by Math.random() in JavaScript?
I presume I won't but it'd be nice to have a credible explanation as to why 😎
Edited, In layman terms, we're dealing with two IEEE 754 double-precision floating-point numbers, one is the maximal integer (for double-precision), the other one is fractional with quite a few digits after a decimal point. What if (say) I first converted them to quadruple-precision format, then multiplied, and then converted the product back to double-precision, would the result be any different?
const max = Number.MAX_SAFE_INTEGER;
const random = Math.random();
console.log(`\
MAX_SAFE_INTEGER: ${max}, \
random: ${random}, \
product: ${max * random}`);
For more elaborate examples, I use it to generate BigInt random numbers.

Your implementation should be safe - in theory, all numbers between 0 and MAX_SAFE_INTEGER should have a possibility of appearing, if the engine implementing Math.random uses a completely unbiased algorithm.
But an absolutely unbiased algorithm is not guaranteed by the specification - the numbers chosen are meant to be psuedo random, not truly, completely random. (does such a thing even exist? it's debatable...) Modern versions V8 and some other implementations use an algorithm with a period on the order of 2 ** 128, larger than MAX_SAFE_INTEGER (2 ** 53 - 1) - but it'd be completely plausible for other implementations (especially older ones) to have a much smaller period, resulting in certain integers within the range being picked much more often than others.
If this is important for your script (which is pretty unlikely in most situations, I'd think), you might consider using a higher-quality random generatior than Math.random - but it's almost certainly not worth worrying about.

What if (say) I first converted them to quadruple-precision format, then multiplied, and then converted the product back to double-precision, would the result be any different?
It could be in cases where the rounding behaves differently between multiplying two doubles vs converting quadruple to double, but the main problem remains the same. The spacing between representable doubles in the range from 2n to 2n+1 is 2n−52. So between 252 and 253 only whole numbers can be represented, between 251 and 252 only every 0.5 can be represented, etc.
If you want more precision you could try decimal.js. The library is included on that documentation page so you can try these out in your console.
Number.MAX_SAFE_INTEGER*.9
8106479329266892
new Decimal(Number.MAX_SAFE_INTEGER).mul(new Decimal(0.9)).toString()
"8106479329266891.9"

Both answers are correct, but I couldn't help running this little experiment in C#, where double is the same thing as Number in JavaScript (fiddle):
using System;
public class Program
{
public static void Main()
{
const double MAX_SAFE_INT = 9007199254740991;
Decimal maxD = Convert.ToDecimal(MAX_SAFE_INT.ToString());
var rng = new Random(Environment.TickCount);
for (var i = 0; i < 1000; i++)
{
double random = rng.NextDouble();
double product = MAX_SAFE_INT * random;
// converting via string to workaround the "15 significant digits" limitation for Decimal(Double)
Decimal randomD = Decimal.Parse(String.Format("{0:F18}", random));
Decimal productD = maxD * randomD;
double converted = Convert.ToDouble(productD);
if (Math.Floor(converted) != Math.Floor(product))
{
Console.WriteLine($"{maxD}, {randomD, 22}, products: decimal {productD, 32}, converted {converted, 20}, original {product, 20}");
}
}
}
}
As far as I'm concerned, I'm still getting the desired distribution of the random numbers within the 0 - 9007199254740991 range.
Here is a JavaScript playground code to check for possible recurrences.

Javascript scientific notation floats vs integers explained

I am working with js numbers and have lack of experience in that. So, I would like to ask few questions:
2.2932600144518896
e+160
is this float or integer number? If it's float how can I round it to two decimals (to get 2.29)? and if it's integer, I suppose it's very large number, and I have another problem than.
Thanks

Technically, as said in comments, this is a Number.
What you can do if you want the number (not its string representation):
var x = 2.2932600144518896e+160;
var magnitude = Math.floor(Math.log10(x)) + 1;
console.log(Math.round(x / Math.pow(10, magnitude - 3)) * Math.pow(10, magnitude - 3));
What's the problem with that? Floating point operation may not be precise, so some "number" different than 0 should appear.
To have this number really "rounded", you can only achieve it through string (than you can't make any operation).

JavaScript only has one Number type so is technically neither a float or an integer.
However this isn't really relevant as the value (or rather representation of it) is not specific to JavaScript and uses E-Notation which is a standard way to write very large/small numbers.
Taking this in to account 2.2932600144518896e+160 is equivalent to 2.2932600144518896 * Math.pow(10,160) and approximately 229 followed by 158 zeroes i.e. very flippin' big.

Why do these binary representations yield the same number?

According to the documentation, one can convert a binary representation of a number to the number itself using parseInt(string, base).
For instance,
var number = parseInt("10100", 2);
// number is 20
However, take a look at the next example:
var number = parseInt("1000110011000011101000010100000110011110010111111100011010000", 2);
// number is 1267891011121314000
var number = parseInt("1000110011000011101000010100000110011110010111111100100000000", 2);
// number is 1267891011121314000
How is this possible?
Note that the binary numbers are almost the same, except for the last 9 bits.

1267891011121314000 is way over Number.MAX_SAFE_INTEGER (9007199254740991). It can't safely represent it in memory.
Take a look at this classic example:
1267891011121314000 == 1267891011121314001 // true

Because that's a 61-bit number and the highest-precision type that Javascript will store is a 64-bit floating point values (using IEEE-754 doubles).
A 64-bit float does not have 61 bits of integral precision, since the 64 bits used are split into the fraction and exponent. As you can see from the graphic of the bit layout, only 52 bits are available to the mantissa (or fraction), which is used when storing an integer.
Many bignum libraries exist to solve this sort of thing, as it's a very common problem in scientific applications. They tend not to be as fast as math with hardware support, but allow much more precision. Pulling from Github results, bignumber.js may be what you need to support these values.

There are upper and lower limits on numbers in JavaScript. JavaScript represents number as 64 bit floating point. IEEE754
Look into a BigDecimal implementation in JavaScript if you need larger support.

Math.sin() Different Precision between Node.js and C#

I have a problem in precision in the last digit after the comma.The javascript code generates one less Digit in compare with the C# code.
Here is the simple Node.js code
var seed = 45;
var x = Math.sin(seed) * 0.5;
console.log(x);//0.4254517622670592
Here is the simple C# code
public String pseudorandom()
{
int seed = 45;
double num = Math.Sin(seed) * (0.5);
return num.ToString("G15");//0.42545176226705922
}
How to achieve the same precision?

The JavaScript Number type is quite complex. It looks like floating point number will probably be like IEEE 754-2008 but some aspects are left to the implementation. See http://www.ecma-international.org/ecma-262/6.0/#sec-number-objects sec 12.7.
There is a note
The output of toFixed may be more precise than toString for some
values because toString only prints enough significant digits to
distinguish the number from adjacent number values. For example,
(1000000000000000128).toString() returns "1000000000000000100", while
(1000000000000000128).toFixed(0) returns "1000000000000000128".
Hence to get full digit accuracy you need something like
seed = 45;
x = Math.sin(seed) * 0.5;
x.toFixed(17);
// on my platform its "0.42545176226705922"
Also, note the specification for how the implementation of sin and cos allow for some variety in the actual algorithm. It's only guaranteed to within +/- 1 ULP.
Using java the printing algorithm is different. Even forcing 17 digits gives the result as 0.42545176226705920.
You can check you are getting the same bit patterns using x.toString(2) and Double.doubleToLongBits(x) in Java.

return num.ToString("G15");//0.42545176226705922
actually returns "0.425451762267059" (no significant digit + 15 decimal places in this example), and not the precision shown in the comment after.
So you would use:
return num.ToString("G16");
to get "0.4254517622670592"
(for your example - where the significant digit is always 0) G16 will be 16 decimal places.