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Function with multiple possible arguments that will never return the same value given any sequence of arguments

Let's say you have a function that takes both x and y, real numbers that are integers, as arguments.
What would you put inside that function, using only mathematical operators, so that no two given sequences of arguments could ever return the same value, be it any kind of value?
Example of a function that fails at doing this:
function myfunction(x,y){
return x * y;
}
// myfunction(2,6) and myfunction(3,4) will both return 12
// myfunction(2,6) and myfunction(6,2) also both return 12.

As already noted in comments, at the level of JavaScript numbers such a function can't exist, simply because assuming that we're working with integer-valued IEEE 754 binary64 floats there are more possible input pairs than possible output values.
But to the mathematical question of whether there is a simple, injective function from pairs of integers to a single integer, the answer is yes. Here's one such function that uses only addition and multiplication, so should fit the questioner's "using only mathematical operators" constraint.
First we map each of the inputs from the domain of integers to the domain of nonnegative integers. The polynomial map x ↦ 2*x*x + x will do that for us, and maps distinct values to distinct values. (Sketch of proof: if 2*x*x + x == 2*y*y + y for some integers x and y, then rearranging and factoring gives (x - y) * (2*x + 2*y + 1) == 0; the second factor can never be zero for integers x and y, so the first factor must be zero and x == y.)
Second, given a pair of nonnegative integers (a, b), we map that pair to a single (nonnegative) integer using (a, b) ↦ (a + b)*(a + b) + a. It's easy to see that this, too, is injective: given the value of (a + b)*(a + b) + a, I can recover the value of a + b by taking the integer square root, and from there recover a and b.
Here's some Python code demonstrating the above:
def encode_pair(x, y):
""" Encode a pair of integers as a single (nonnegative) integer. """
a = 2*x*x + x
b = 2*y*y + y
return (a + b)*(a + b) + a
We can easily check that there are no repetitions for small x and y: here we take all pairs (x, y) with -500 <= x < 500 and -500 <= y < 500, and find the set containing encode_pair(x, y) for each combination. If all goes well, we should end up with a set with exactly 1 million entries, one per input combination.
>>> all_outputs = {encode_pair(x, y) for x in range(-500, 500) for y in range(-500, 500)}
>>> len(all_outputs)
1000000
>>> min(all_outputs)
0
But perhaps a more convincing way to establish the injectivity is to give an explicit inverse, showing that the original (x, y) can be recovered from the output. Here's that inverse function. It makes use of Python's integer square root operation math.isqrt, which is available only for Python >= 3.8, but is easy to implement yourself if you need it.
from math import isqrt
def decode_pair(n):
""" Decode an integer produced by encode_pair. """
a_plus_b = isqrt(n)
a = n - a_plus_b*a_plus_b
b = a_plus_b - a
c = isqrt(8*a + 1)
d = isqrt(8*b + 1)
return ((2 - c%4) * c - 1) // 4, ((2 - d%4) * d - 1) // 4
Example usage:
>>> encode_pair(3, 7)
15897
>>> decode_pair(15897)
(3, 7)
Depending on what you allow as a "mathematical operator" (which isn't really a particularly well-defined term), there are tighter functions possible. Here's a variant of the above that provides not just an injection but a bijection: every integer appears as the encoding of some pair of integers. It extends the set of mathematical operators used to include subtraction, division and absolute value. (Note that all divisions appearing in encode_pair are exact integer divisions, without any remainder.)
def encode_pair(x, y):
""" Encode a pair of integers as a single integer.
This gives a bijective map Z x Z -> Z.
"""
ax = (abs(2 * x + 1) - 1) // 2 # x if x >= 0, -1-x if x < 0
sx = (ax - x) // (2 * ax + 1) # 0 if x >= 0, 1 if x < 0
ay = (abs(2 * y + 1) - 1) // 2 # y if y >= 0, -1-y if y < 0
sy = (ay - y) // (2 * ay + 1) # 0 if y >= 0, 1 if y < 0
xy = (ax + ay + 1) * (ax + ay) // 2 + ax # encode ax and ay as xy
an = 2 * xy + sx # encode xy and sx as an
n = an - (2 * an + 1) * sy # encode an and sy as n
return n
def decode_pair(n):
""" Inverse of encode_pair. """
# decode an and sy from n
an = (abs(2 * n + 1) - 1) // 2
sy = (an - n) // (2 * an + 1)
# decode xy and sx from an
sx = an % 2
xy = an // 2
# decode ax and ay from xy
ax_plus_ay = (isqrt(8 * xy + 1) - 1) // 2
ax = xy - ax_plus_ay * (ax_plus_ay + 1) // 2
ay = ax_plus_ay - ax
# recover x from ax and sx, and y from ay and sy
x = ax - (1 + 2 * ax) * sx
y = ay - (1 + 2 * ay) * sy
return x, y
And now every integer appears as the encoding of exactly one pair, so we can start with an arbitrary integer, decode it to a pair, and re-encode to recover the same integer:
>>> n = -12345
>>> decode_pair(n)
(67, -44)
>>> encode_pair(67, -44)
-12345
The encode_pair function above is deliberately quite verbose, in order to explain all the steps involved. But the code and the algebra can be simplified: here's exactly the same computation expressed more compactly.
def encode_pair_cryptic(x, y):
""" Encode a pair of integers as a single integer.
This gives a bijective map Z x Z -> Z.
"""
c = abs(2 * x + 1)
d = abs(2 * y + 1)
e = (2 * y + 1) * ((c + d)**2 * c + 2 * (c - d) * c - 4 * x - 2)
return (e - 2 * c * d) // (4 * c * d)
encode_pair_cryptic gives exactly the same results as encode_pair. I'll give one example, and leave the reader to figure out the equivalence.
>>> encode_pair(47, -53)
-9995
>>> encode_pair_cryptic(47, -53)
-9995

I'm no math wiz but found this question kinda fun so I gave it a shot. This is by no means scalable to large number since I'm using prime numbers as exponents and gets out of control really quick. But tested up to 90,000 combinations and found no duplicates.
The code below has a couple extra functions generateValues() and hasDuplicates() that is just there to run and test multiple values coming from the output of myFunction()
BigNumber.config({ EXPONENTIAL_AT: 10 })
// This function is just to generate the array of prime numbers
function getPrimeArray(num) {
const array = [];
let isPrime;
let i = 2;
while (array.length < num + 1) {
for (let j = 2; (isPrime = i === j || i % j !== 0) && j <= i / 2; j++) {}
isPrime && array.push(i);
i++;
}
return array;
}
function myFunction(a, b) {
const primes = getPrimeArray(Math.max(a, b));
// Using the prime array, primes[a]^primes[b]
return BigNumber(primes[a]).pow(primes[b]).toString();
}
function generateValues(upTo) {
const results = [];
for (let i = 1; i < upTo + 1; i++) {
for (let j = 1; j < upTo + 1; j++) {
console.log(`${i},${j}`)
results.push(myFunction(i,j));
}
}
return results.sort();
}
function hasDuplicates(arr) {
return new Set(arr).size !== arr.length;
}
const values = generateValues(50)
console.log(`Checked ${values.length} values; duplicates: ${hasDuplicates(values)}`)
<script src="https://cdnjs.cloudflare.com/ajax/libs/bignumber.js/8.0.2/bignumber.min.js"></script>
Explanation of what's going on:
Using the example of myFunction(1,3)
And the array of primes [2, 3, 5, 7]
This would take the 2nd and 4th items, 3 and 7 which would result in 3^7=2187
Using 300 as the max generated 90,000 combinations with no duplicates (However it took quite some time.) I tried using a max of 500 but the fan on my laptop sounded like a jet engine taking off so gave up on it.

If x and y are some fixed size integers (eg 8 bits) then what you want is possible if the return of f has at least as many bits as the sum of the number of bits of x an y (ie 16 in the example) and not otherwise.
In the 8 bit example f(x,y) = (x<<8)+y would do. This is because if g(z) = ((z>>8), z&255) then g(f(x,y)) = (x,y). The impossibility comes from the pigeon hole principle: if we want (in the example) to map the pairs (x,y) (of which there 2^16) 1-1 to some integer type, then we must have at least 2^16 values of this type.

function myfunction(x,y){
x = 1/x;
y = 1/y;
let yLength = ("" + y).length
for(let i = 0; i < yLength; i++){
x*=10;
}
return (x + y)
}
console.log(myfunction(2,12))
console.log(myfunction(21,2))

Based on your question and you comments, I understood the following:
You want to pass 2 real numbers into a function. The function should use mathematical operators to generate a new result.
Your question is, if there is any kind of mathematical equation/function you could use, that would ALWAYS deliver a unique result.
If that's so, then the answer is no. You can make your function as complicated as possible and get a result(c) using the two numbers (a & b).
In this case I would look for another combination which could give me the result(c) using the same equation/function. Therefore I would use the system of linear equation to solve this mathematical issue.
In general, a system with fewer equations than unknowns has infinitely many solutions, but it may have no solution. Such a system is known as an underdetermined system.
In our case we would have one equation which gives us one result and two unknowns, therefore it would have infinitely many solutions because we already have a solution, so there is no way for the system to have no solutions at all.
More about this topic.
Edit:
I just recognized that some of us understood the domain of the function in a different way. I was thinking about real numbers (R) but it seems many assumed you talk about integers (Z) only.
Well I guess
real integers
wasnt clear enough, at least for me.
So if we would use integers only, I have no idea if that is possible to always have different results. Some users suggested a topic about that here I am also interested to take a look into that too.

Algorithm to round about n * 10

I am trying to find an algorithm to result:
45 => 100
458 => 1000
8456 => 10000
I would like to "round" this way, i looked into Math.round() but it doesn't think to be enough.

For a single line of code you can do what Rup said. A base 10 log will give you the closest power of 10, then do ten to the power of the ceiling of the result of the log
var x = 45;
Math.pow(10, Math.ceil(Math.log(x) / Math.LN10));

easiest thing to come up with without thinking:
var n = 455;
var rounded;
for(var i = 1000000000 i>= 1; i/=10)
{
if(i < n)
{
break;
}
else
{
rounded = i;
}
}

Why isn't my attempted solution working?

When I run the code in the console, the browser just stops working (am assuming stack overflow).
I've come up with several different algorithms for solving this problem, but I thought this one would not cause any SOs.
The problem:
The sequence of triangle numbers is generated by adding the natural numbers. So the 7th triangle number would be 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28. The first ten terms would be:
1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...
Let us list the factors of the first seven triangle numbers:
1: 1
3: 1 3
6: 1,2,3,6
10: 1,2,5,10
15: 1,3,5,15
21: 1,3,7,21
28: 1,2,4,7,14,28
We can see that 28 is the first triangle number to have over five divisors.
What is the value of the first triangle number to have over five hundred divisors?
Failing solution:
function divisors(n){
var counter = 0;
var triangle = 3;
var triangle_add = 2;
while (counter < n){
for (var i = 1; i = triangle; i++){
if (triangle % i === 0){
counter++;
}
};
if (counter < n){
triangle_add++;
triangle = triangle + triangle_add;
counter = 0;
};
};
return triangle;
};
console.log(divisors(501));

Your solution is not working because, most probably, it is very slow. This problem can be solved much faster by the following method:
Find all the prime numbers smaller than some N (put, for example, N=100'000) using Sieve of Eratosthenes. It is quite fast.
As we know from elementary math each number can be written in the form X=p1^i1*p2^i2*...*pn^in where pj is prime number and ij is the power of corresponding prime number. The number of divisors of X is equal to (i1+1)*(i2+1)*...*(in+1) since that many different ways we can form a number which will be divisor of X. Having an array of prime numbers the number of divisors for X can be calculated quite fast (the code still has place to be optimized):
int divisorCount(long long X)
{
int c = 1;
for (int i = 0; PRIMES[i] <= X; ++i)
{
int pr = PRIMES[i];
if (X % pr == 0)
{
int p = 1;
long long r = X;
while (r % pr == 0)
{
r = r / pr;
++p;
}
c *= p;
}
}
return c;
}
Iterate through all triangle numbers and count divisor numbers for them using the above function. The i-th triangle number is i * (i + 1) / 2, so no need to keep a variable, increment it and add it each time.

breaking an integer into parts

I'm trying to add all the digits in an integer value until i get a value below 9 using Javascript.
for an example, if i have 198, I want to add these together like 1 + 9 + 8 = 18, and since 18 is higher than 9 add 1 +8 again = 9.

Rather than giving you full code I would just explain how to do it.
You can do modulo math by number%10 and subsequent int divide by 10 (number/10) until you get 0 to get all the digits of a number. Sum the individual digits and until sum > 9 repeat the above process in a loop.
Edit: okay here is the code for you:
<script>
var num=198;
n = num;
var sum;
do {
sum = 0;
while (n>0) {
rem = (n % 10);
sum += rem;
n = (n - rem)/10;
}
n = sum;
} while(sum>9);
alert("sum is: " + sum);
</script>

function foo(var x)
{
while(x > 9)
{
var y = 0;
while(x!=0)
{
y += x%10;
x = Math.floor(x/10);
}
x = y;
}
return x;
}

Here are two hints: (i % 10) gives the least significant decimal digit of i, while i /= 10 removes the least significant digit from i. The rest is left as an exercise for the reader.

Subtracting long numbers in javascript

Why is q == 0 in the following script?
<script>
var start = 1234567890123456789;
var end = 1234567890123456799;
var q = end - start;
alert(q);
</script>
I would think the result should be 10. What is the correct way to subtract these two numbers?

Because numbers in JavaScript are floating-point. They have limited precision.
When JavaScript sees a very long number, it rounds it to the nearest number it can represent as a 64-bit float. In your script, start and end get rounded to the same value.
alert(1234567890123456789); // says: 1234567890123456800
alert(1234567890123456799); // says: 1234567890123456800
There's no built-in way to do precise arithmetic on large integers, but you can use a BigInteger library such as this one.

As of January 2020, BigInt datatype is going to be added to Javascript. The proposal is currently in Stage 4. It will enable precise calculation for number which are more than 2^53-1 (Number.MAX_SAFE_INTEGER).
BigInt has been shipped in Chrome, Node, Firefox, and is underway in Safari. Read more here.
var start = BigInt('1234567890123456789');
var end = BigInt('1234567890123456799');
var q = end - start;
alert(q)
A BigInt is created by appending n to the end of an integer literal — 10n — or by calling the function BigInt(). It is also different from Number so 1 + 1n will fail.
You can read more about it here from MDN pages

Jason already posted the why. For a solution, you can get a Javascript BigInt library at http://www-cs-students.stanford.edu/~tjw/jsbn/

const subtract = (a, b) => [a, b].map(n => [...n].reverse()).reduce((a, b) => a.reduce((r, d, i) => {
let s = d - (b[i] || 0)
if (s < 0) {
s += 10
a[i + 1]--
}
return '' + s + r
}, '').replace(/^0+/, ''))
Better use big-integer library for these things so as to handle all different test cases.This is just for the a general case you can use....

It is explained in the JavaScript documentation:
According to the ECMAScript standard, there is only one number type: the double-precision 64-bit binary format IEEE 754 value (numbers between -(253-1) and 253-1). There is no specific type for integers.
Wikipedia page about double precision floating point format explains:
Between 252= 4,503,599,627,370,496 and 253= 9,007,199,254,740,992 the representable numbers are exactly the integers. For the next range, from 253 to 254, everything is multiplied by 2, so the representable numbers are the even ones, etc.
(All integer numbers smaller than 252 are represented exactly.)
1234567890123456789 and 1234567890123456799 are larger than 260= 1152921504606846976. At this magnitude only about 1% of the integer numbers are stored exactly using the double-precision floating point format.
These two cannot be stored exactly. They both are rounded to 1234567890123456800.
The JavaScript documentation also explains how to tell if a an integer number is stored exactly:
[...] and starting with ECMAScript 6, you are also able to check if a number is in the double-precision floating-point number range using Number.isSafeInteger() as well as Number.MAX_SAFE_INTEGER and Number.MIN_SAFE_INTEGER. Beyond this range, integers in JavaScript are not safe anymore and will be a double-precision floating point approximation of the value.

function add(x, y) {
//*********************************************************************//
// This function adds or subtracts two extremely large decimal numbers //
// Inputs x and y should be numbers, i.e. commas are removed already //
// Use this function to remove commas and convert to number: //
// x = parseFloat(strNumber.replaceAll(",","").trim()); //
// Inputs x and y can be both positive, or both negative, //
// or a combination (i.e. one positive and one negative in any //
// position whether as x or as y) which means subtraction //
//*********************************************************************//
var temp, borrow=false, bothNeg=false, oneNeg=false, neg=false;
if (x < 0 && y < 0) { bothNeg = true; x = -x; y = -y; }
else if (x < 0 || y < 0) {
oneNeg = true;
if (Math.abs(x) == Math.abs(y)) { x = 0; y = 0; }
else if (x < 0 && Math.abs(x) > Math.abs(y)) { neg = true; x = -x; y = -y; }
else if (x < 0 && Math.abs(x) < Math.abs(y)) { temp = y; y = x; x = temp; }
else if (y < 0 && Math.abs(x) < Math.abs(y)) { neg = true; temp = y; y = -x; x = -temp; }
}
x = parseInt(x*1000000000/10).toString();
y = parseInt(y*1000000000/10).toString();
var lenx=x.length, leny=y.length, len=(lenx>leny)?lenx:leny, sum="", div=0, x1, y1, rem;
for (var i = 0; i < len; i++) {
x1 = (i >= lenx) ? 0 : parseInt(x[lenx-i-1]);
y1 = (i >= leny) ? 0 : parseInt(y[leny-i-1]);
y1 = (isNaN(y1)) ? 0 : y1;
if (oneNeg) y1 = -y1;
if (borrow) x1 = x1 - 1;
if (y < 0 && x1 > 0 && Math.abs(x1) >= Math.abs(y1)) { borrow=false; div=0; }
if (y < 0 && y1 <= 0 && (x1 < 0 || Math.abs(x1) < Math.abs(y1))) { borrow=true; rem=(x1+y1+div+10)%10; div=10; }
else { rem=(x1+y1+div)%10; div=Math.floor((x1+y1+div)/10); }
sum = Math.abs(rem).toString() + sum;
}
if (div > 0) sum = div.toString() + sum;
sum = parseFloat(sum*10/1000000000);
if (bothNeg || neg) sum = -sum;
return sum;
}