How can we prove that the square root of any number is equal to the statment given below

Question

Is there any theorem that can explain this $\sqrt[n]{x}= x^{1/n} $, is there any practical example of $ x^{1/n} $, 1/n times of a number.

Answer 1

Your equality is the definition of the root of power $n$.

To compute such values for $x > 0$, use $x^{1/n} = \exp\left(\frac{\ln x}{n}\right)$ - this is how computers do it, it works for $\forall x > 0, \forall n \in (\mathbb{R}\setminus{0})$

A practical example is the interest on a saving account. Say, a bank pays $2\%$ per year interest rate. How would you compute an interest that accrues in a month? If you compute the interest as $\frac{2\%}{12}$ (there are 12 months in a year, assume that we have to pay the same interest each month), then the yearly interest will be $\left(1+\frac{0.02}{12}\right)^{12} - 1 > 0.02$.
Correct interest computation will be $\left(1+0.02\right)^{\frac{1}{12}} - 1$.

Answer 2

Look at this answer which shows that $$\sqrt[b]{x^a}=x^ {a/b}$$ and so your question asks about the particular case of $a=1$

(In fact the answer referenced to begins by defining $x^{1/b}$ then building up to $x^{a/b}$)

Answer 3

Here is a very practical example of a use of $x^{1/n}$: tuning a piano.

What we generally perceive as the "interval" between two notes is actually the ratio of their frequencies. The ratio of frequencies between middle C and high C (one octave up) on a piano is exactly $2.$ There are $13$ keys in this interval, producing $12$ smaller intervals (called half-steps) if you start at middle C and then strike every key in succession until you reach high C. In order to make these half-steps all sound uniform, each key is tuned to a frequency $2^{1/12}$ times the frequency of the key before it.