Why is $\sqrt[4]{4} = \sqrt{2}$

Question

Why is $\sqrt[4]{4} = \sqrt{2}$ ?

Sorry for the basic question, but having a bit of a hard time trying to understand this one.

Answer 1

$\sqrt[4]{4}=4^{1/4}=(2^2)^{1/4}=2^{1/2}=\sqrt{2}$

Answer 2

Let's give $\sqrt2$ a name: call it $x$, so $x^2=2$. Then what is $x^4$?

$$x^4=(x^2)^2=2^2=4$$

Since the $4$th power of $x$ is $4$, we can say $x=\sqrt[4]{4}$.

Answer 3

The equation $x^4=4$ can be written $x^4-4=0$ and this factorises as $(x^2-2)(x^2+2)=0$

There are four roots of the quartic equation over the complex numbers, two of which are real numbers, one positive and one negative. The convention for roots of real numbers is that the principal $n^{th}$ root of a positive number is the positive real root.

So the fourth root of $4$ which we pick out using the surd notation is by convention the positive real root (belonging to the factor $x^2-2$) which is $\sqrt 2$.

Answer 4

Let $\sqrt[4]4=a$.

Thus, $$a^4=4$$ and $a>0$.

Hence, $a^2=2$ and $a=\sqrt2$.

Answer 5

The reason is that the function $\sqrt[n]{\cdot}$ for $n\in\Bbb N_{>0}$ is defined as

$$\sqrt[n]{\cdot}:[0,\infty)\to[0,\infty),\quad x\mapsto r\ge0: r^n=x$$