# Its about square root

The reason is not difficult if one knows about the definition of square root of real numbers.

**Definition:** Let $x$ be a non-negative number. Then a non-negative number $r$ is called square root of $x$ iff $r^2=x$.

Square root of $x$ is denoted by $\sqrt{x}$.

**From this definition:**

- 2 is square root of 4 as $2^2=4$.
- 3 is square root of 9 as $3^2=9$.

This definition can be extended to include the case, when square root can be negative.

**Extended definition:** Let $x$ be a non-negative number. Then a number $r$ is called square root of $x$ iff $r^2=x$.

**From the extended definition:**

- 2 is square root of 4 as $2^2=4$.
- -2 is square root of 4 as $(-2)^2=4$.

From above examples we see 2 and -2 are square root of 4 but it doesn't mean they must be equal. If we use the extended definition then it means we cannot consider to take square root on both side of equation. As square root of left side of equation and right side of equation may have different answers.

But if we consider the first definition then no doubt $\sqrt{4}=\sqrt{2^2}=\sqrt{(-2)^2}=2$.

If we use the second definition then $\sqrt{4}=\sqrt{2^2}=\sqrt{(-2)^2}=\pm 2$. We cannot consider one of 2 or -2 at some place, we have to consider both the answers.

Conventionally in mathematics $\sqrt{4}$ only denotes positive square root, that is, 2. But it should be clear that squae root of non-negative number only exists in real number, e.g if we write $\sqrt{x}$, then it is clear $x\geq0$. Also this is not a good thing to write $\sqrt{x}=x^{\frac{1}{2}}$.

- dyk/3
- Last modified: 2 years ago
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