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## Beginner's mistake:

```javascript
const numbers = [10, 1, 3, 20];
numbers.sort();

console.log(numbers); // ["1", "10", "20", "3"] - string sorting
```

By default `sort()` converts elements to strings
and sorts them **in lexicographic order** (`"10"` comes before `"2"`).

---

## The correct way: with a compare function

```javascript
const numbers = [10, 1, 3, 20];
numbers.sort((a, b) => a - b);

console.log(numbers); // [1, 3, 10, 20]
```

- The function `(a, b) => a - b` tells JavaScript **how to compare numbers**:
  - if `a - b` < 0 -> `a` comes first
  - if `a - b` > 0 -> `b` comes first
  - if `a - b` = 0 -> the order stays unchanged

---

## Sorting in descending order:

```javascript
numbers.sort((a, b) => b - a);
console.log(numbers); // [20, 10, 3, 1]
```

---

## If you don't want to mutate the original array:

```javascript
const arr = [5, 2, 9, 1];
const sorted = [...arr].sort((a, b) => a - b);

console.log(sorted); // [1, 2, 5, 9]
console.log(arr);    // [5, 2, 9, 1] (the original is unchanged)
```

---

## In short:

| Task | Code | Result |
|---|---|---|
| Ascending | `arr.sort((a, b) => a - b)` | `[1, 2, 3, 4]` |
| Descending | `arr.sort((a, b) => b - a)` | `[4, 3, 2, 1]` |
| Without changing the original array | `[...arr].sort(...)` | a new sorted array |

---

> **Summary:**
> To sort numbers in ascending order,
> **always pass a compare function:**
>
> ```javascript
> arr.sort((a, b) => a - b);
> ```

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