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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 number size

i just wanted to know javascript number size because i want to send lot of them via network per frame and i must know a measure of how many im gonna send per second.
As i readed:
According to the ECMAScript standard, there is only one number type: the double-precision 64-bit binary format IEEE 754 value (number between -(2^53 -1) and 2^53 -1).
So if im gonna send lot of diferent numbers(example later) if all numbers between -(2^53 -1) and (2^53 -1) use same memory i may just combinate them like 567832332423556 and then locally split them locally when received instead of sending a lot of diferent numbers, because anyway that unique number "567832332423556" sends same information as a separated 5,6,7,8... but in one so its supossed to waste many less if it haves same size as a single 5.
Is this true or just im so confused? pls explain me :(.
var data = Array2d(obj.size); //Size can be between 125 and 200;`
Array2d: function (rows) { //The number of rows and files are same
var arr = [];
for (var i=0;i<rows;i++) arr[i] = [];
return arr;
},
...
if (this.calculate()) {
data[x][y] = 1;
} else {
data[x][y] = 0;
}
and somewhere in the code i change those 1 to any number from 2 to 5 so numbers may be from 0 to 5 depends of the situation.
Example:
[
[0,0,2,1,3,4,5,0,2,3,4,5,4(200 numbers)],
[0,5,2,1,5,1,0,2,3,0,0,0,0(200 numbers)]
...(200 times)
]
*And i really need All numbers, i cant miss even one.
If in therms of size 5 is shame as 34234 so i could just do something like:
[
[0021345023454...(20 numbers 10 times)],
[0021345023454...(20 numbers 10 times)]
...(200 times)
]
and it may use 20 times less because if 5 size is the same as 2^53 i just stack numbers 20 by 20 and they should waste lot less (ofc, 20 numbers less by stacking 20, at least in the network, maybe the local split is a little big but locally i do few things so i can handle that).

Precise limits on numbers are covered in What is JavaScript's highest integer value that a Number can go to without losing precision? - 9007199254740991 for regualr arithmetic operations, 2^32 for bit operations.
But it sounds like you are more interested in network representation than memory usage at run-time. Below is list of options from less to more compact. Make sure to understand your performance goals before moving away from basic JSON solution -as cost and complexity of constructing data rises more compatc representation you pick.
Most basic solution - JSON representation of existing array gives pretty decent ~2 characters per value representation:
[[0,1,5,0,0],[1,1,1,1,1],[0,0,0,0,0]]
Representing all numbers in a row as one big string gives ~1 character number:
["01500","11111","00000"]
Representing same values as concatenated numbers does not bring much savings - "11111" as string is about as long as the same 11111 as number - you add pair of quotes per row for string but one coma pare 16 values when packing as numbers.
You can indeed pack values to number in more compact form since the range is 0-5 using standard 6-ary value you get ~6^20 per on JavaScript number which is not significant savings over 16 values per number which you get with just representing as digits concatenation.
Better packing would be to represent 2 or 3 values as one character - 2 values give 36 combinations (v1 * 6 + v2) which can be written with just [A-Z0-9], 3 - 216 value which mostly fits into regular characters range.
you can go strictly binary representation (3-4 bit per value) and send via WebSockets to avoid cost of converting to text with regular requests.
Whether you go with binary or text representation one more option is compression - basic RLE compression may be fine if your data have long sequences of same value, other compression algorithms may work better on more random data. There are libraries to perform compression in JavaScript too.

Convert Excel Formula to Javascript

I have this formula on excel
=(1130000000000*F11^1.85)/(F19^1.85*(F13*(1-2/F15))^4.8655)
when
F11 = q
F19 = 150 (Constant)
F13 = d
F15 = sdr
I convert it to this
Math.round((1130000000000*Math.pow(q,1.85))/(Math.pow(150,1.85)*Math.pow(d*(1-2/sdr)),4.8655))
but the results are wrong
when
q = 120
d = 200
sdr = 17
the result should be 8.76
but I am getting long numbers
any help ?
Thanks

From YUI blog
JavaScript has a single number type: IEEE 754 Double Precision floating point. Having a single number type is one of JavaScript’s best features. Multiple number types can be a source of complexity, confusion, and error. A single type is simplifying and stabilizing.
Unfortunately, a binary floating point type has some significant disadvantages. The worst is that it cannot accurately represent decimal fractions, which is a big problem because humanity has been doing commerce in decimals for a long, long time. There would be advantages to switching to a binary-based number system, but that is not going to happen. As a consequence, 0.1 + 0.2 === 0.3 is false, which is the source of a lot of confusion.
http://www.yuiblog.com/blog/2009/03/10/when-you-cant-count-on-your-numbers/

Completely ignore my previous answer, although it is a problem which you will definitely encounter should you continue with these large numbers, your actual issue is a misplaced bracket in your code. If you use this (simply replaced variable names with values):
Math.round((1130000000000*Math.pow(120,1.85))/(Math.pow(150,1.85)*Math.pow((200*(1-2/17)),4.8655)))
It returns 9 (a rounded 8.76), the bracket you misplaced is around the 1-2 mark.
For future reference, the largest number Javascript can comprehend without fault is +/- 9007199254740992.
References:
What is JavaScript's highest integer value that a Number can go to without losing precision?
ECMA-262 - The Number Type

Math.pow() not giving precise answer

I've used Math.pow() to calculate the exponential value in my project.
Now, For specific values like Math.pow(3,40), it returns 12157665459056929000.
But when i tried the same value using a scientific Calculator, it returns 12157665459056928801.
Then i tried to traverse the loop till the exponential value :
function calculateExpo(base,power){
base = parseInt(base);
power = parseInt(power);
var output = 1;
gameObj.OutPutString = ''; //base + '^' + power + ' = ';
for(var i=0;i<power;i++){
output *= base;
gameObj.OutPutString += base + ' x ';
}
// to remove the last comma
gameObj.OutPutString = gameObj.OutPutString.substring(0,gameObj.OutPutString.lastIndexOf('x'));
gameObj.OutPutString += ' = ' + output;
return output;
}
This also returns 12157665459056929000.
Is there any restriction to Int type in JS ?

This behavior is highly dependent on the platform you are running this code at. Interestingly even the browser matters even on the same very machine.
<script>
document.write(Math.pow(3,40));
</script>
On my 64-bit machine Here are the results:
IE11: 12157665459056928000
FF25: 12157665459056929000
CH31: 12157665459056929000
SAFARI: 12157665459056929000

52 bits of JavaScript's 64-bit double-precision number values are used to store the "fraction" part of a number (the main part of the calculations performed), while 11 bits are used to store the "exponent" (basically, the position of the decimal point), and the 64th bit is used for the sign. (Update: see this illustration: http://en.wikipedia.org/wiki/File:IEEE_754_Double_Floating_Point_Format.svg)
There are slightly more than 63 bits worth of significant figures in the base-two expansion of 3^40 (63.3985... in a continuous sense, and 64 in a discrete sense), so hence it cannot be accurately computed using Math.pow(3, 40) in JavaScript. Only numbers with 52 or fewer significant figures in their base-two expansion (and a similar restriction on their order of magnitude fitting within 11 bits) have a chance to be represented accurately by a double-precision floating point value.
Take note that how large the number is does not matter as much as how many significant figures are used to represent it in base two. There are many numbers as large or larger than 3^40 which can be represented accurately by JavaScript's 64-bit double-precision number values.
Note:
3^40 = 1010100010111000101101000101001000101001000111111110100000100001 (base two)
(The length of the largest substring beginning and ending with a 1 is the number of base-two significant figures, which in this case is the entire string of 64 digits.)

Haskell (ghci) gives
Prelude> 3^40
12157665459056928801
Erlang gives
1> io:format("~f~n", [math:pow(3,40)]).
12157665459056929000.000000
2> io:format("~p~n", [crypto:mod_exp(3,40,trunc(math:pow(10,21)))]).
12157665459056928801
JavaScript
> Math.pow(3,40)
12157665459056929000
You get 12157665459056929000 because it uses IEEE floating point for computation. You get 12157665459056928801 because it uses arbitrary precision (bignum) for computation.

JavaScript can only represent distinct integers to 253 (or ~16 significant digits). This is because all JavaScript numbers have an internal representation of IEEE-754 base-2 doubles.
As a consequence, the result from Math.pow (even if was accurate internally) is brutally "rounded" such that the result is still a JavaScript integer (as it is defined to return an integer per the specification) - and the resulting number is thus not the correct value, but the closest integer approximation of it JavaScript can handle.
I have put underscores above the digits that don't [entirely] make the "significant digit" cutoff so it can be see how this would affect the results.
................____
12157665459056928801 - correct value
12157665459056929000 - closest JavaScript integer
Another way to see this is to run the following (which results in true):
12157665459056928801 == 12157665459056929000
From the The Number Type section in the specification:
Note that all the positive and negative integers whose magnitude is no greater than 253 are representable in the Number type ..
.. but not all integers with large magnitudes are representable.
The only way to handle this situation in JavaScript (such that information is not lost) is to use an external number encoding and pow function. There are a few different options mentioned in https://stackoverflow.com/questions/287744/good-open-source-javascript-math-library-for-floating-point-operations and Is there a decimal math library for JavaScript?
For instance, with big.js, the code might look like this fiddle:
var z = new Big(3)
var r = z.pow(40)
var str = r.toString()
// str === "12157665459056928801"

Can't say I know for sure, but this does look like a range problem.
I believe it is common for mathematics libraries to implement exponentiation using logarithms. This requires that both values are turned into floats and thus the result is also technically a float. This is most telling when I ask MySQL to do the same calculation:
> select pow(3, 40);
+-----------------------+
| pow(3, 40) |
+-----------------------+
| 1.2157665459056929e19 |
+-----------------------+
It might be a courtesy that you are actually getting back a large integer.

Compute the double value nearest preferred decimal result

Let N(x) be the value of the decimal numeral with the fewest significant digits
such that x is the double value nearest the value of the numeral.
Given double values a and b, how can we compute the double value nearest N(b)-N(a)?
E.g.:
If a and b are the double values nearest .2 and .3,
the desired result is the double value nearest .1,
0.1000000000000000055511151231257827021181583404541015625,
rather than than the result of directly subtracting a and b,
0.09999999999999997779553950749686919152736663818359375.

As a baseline: In Java, the Double.toString() provides the N(x) function described in the question, returning its value as a numeral. One could take the strings for a and b, subtract them with the elementary-school method, and convert the resulting string to double.
This demonstrates solving the problem is quite feasible using existing library routines. This leaves the task of improving the solution. I suggest exploring:
Is there a function D(x) that returns the number of significant digits after the decimal place for the numeral described in N(x)? If so, can we multiply a and b by a power of ten determined by D(a) and D(b), round as necessary to produce the correct integer results (for situations where they are representable as double values), subtract them, and divide by the power of ten?
Can we establish criteria for which b-a or some simple expression can be quickly rounded to something near a decimal numeral, bypassing the code that would be necessary for harder cases? E.g., could we prove that for numbers within a certain range, (round(10000*b)-round(10000*a))/10000 always produces the desired result?

You can convert to 'integers' by multiplying then dividing by a power of ten:
(10*.3 - 10*.2)/10 == 0.1000000000000000055511151231257827021181583404541015625
It may be possible to work out the appropriate power of ten from the string representation of the number. #PatriciaShanahan suggests looking for repeated 0's or 9's.
Consider using a BigDecimal library such as javascript-bignum instead.

You could also inquire in Smalltalk Pharo 2.0 where your request translates:
^(b asMinimalDecimalFraction - a asMinimalDecimalFraction) asFloat
Code could be found as attachment to issue 4957 at code.google.com/p/pharo/issues - alas, dead link, and the new bugtracker requires a login...
https://pharo.fogbugz.com/f/cases/5000/Let-asScaledDecimal-use-the-right-number-of-decimals
source code is also on github, currently:
https://github.com/pharo-project/pharo-core/blob/6.0/Kernel.package/Float.class/instance/printing/asMinimalDecimalFraction.st
The algorithm is based on:
Robert G. Burger and R. Kent Dybvig
Printing Floating Point Numbers Quickly and Accurately
ACM SIGPLAN 1996 Conference on Programming Language Design and Implementation
June 1996.
http://www.cs.indiana.edu/~dyb/pubs/FP-Printing-PLDI96.pdf