3
$\begingroup$

Basically, the square root operation was discovered and proved rigorously from the Pythagorean theorem, it was denoted by square root of a rational number say $n$ as $\sqrt{n}$, but at a later stage, it used to be denoted by fractional power equals to $1/2$, also for multi-square root operation for say a prime number $p$ as $\sqrt[2^{n}]{p}$ or $p^{2^{- n}}$, for some positive integer $n$, and then this was so generalized to any fraction number

But, the power or exponent operation was basically defined for positive integer numbers, where the number of using the multiplication operation is basically a natural number such that it makes sense, and this was simply generalised to a ricipricical or the inverse of the integer number of power operation to be extended to negative integers

To illustrate that simply in examples for say $n^3 = n*n*n$, so we used the multiplication operation twice which is an integer number of times

And generally, we use the multiplication operation as ($k - 1$) times, to express a power integer $n^k$ for $k$ positive integer

But what does it mean when the exponent provided is a fraction number in accordance with the sensible basic definition of power integer?

Does that mean that an integer is multiplied by itself a fractional number of times? which seems meaningless

Improve this question
$\endgroup$
3
  • $\begingroup$ What does it mean a fractional exponent ? $\dfrac 1 2$ is the multiplicative inverse of $2$: $x= x^{\frac 1 2}x^{\frac 1 2}=(x^{\frac 1 2})^2=x^{2(\frac 1 2)}=x^1=x$. $\endgroup$
    – Mauro ALLEGRANZA
    Commented Jul 9, 2017 at 18:23
  • $\begingroup$ Of course, this is the known way among mathematicians, and also seems valid basically for very important reason that the square root operation is the only proved root operation in mathematics, but the other fractional of odd prime root operations were simply concluded and never proven, also note that $x$ can be expressed from its original defined operation as $x = (\sqrt{x})* (\sqrt{x}) = (\sqrt{x})^2 = x$, without using the adopted fraction as $\frac{1}{2}$, $\endgroup$
    – Bassam Karzeddin
    Commented Jul 9, 2017 at 18:37
  • $\begingroup$ But when it comes to another fraction exponent as $\frac{1}{3}$, it becomes more doubtful, since the cube root operation was never having any rigorous proof of existence but only a mere conclusion by comparison and APPROXIMATION, not at all a similar for the case of square root operation which was proven rigorously from the Pythagorean theorem, so does that mean really the $\sqrt[3]{2}$ for example that the number of multiplication operation we use should be (-2/3), which isn't any natural number to make sense the same way that exponent power number was basically and originally was defined! $\endgroup$
    – Bassam Karzeddin
    Commented Jul 9, 2017 at 18:44

1 Answer 1

Reset to default
4
$\begingroup$

Nicole d'Oresme was the first to use fractional exponents. https://en.wikipedia.org/wiki/Nicole_Oresme

In Algorismus proportionum and De proportionibus proportionum, Oresme developed the first calculation-method of powers with fractional irrational exponents. http://www.nicole-oresme.com/seiten/oresme-biography.html

The exponent 1/2 means that the square root of $x$ has to be multiplied by itself in order to give $x$.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Also, it seems valid for the valid square root proved operation, but the explanation on how many times we use the multiplication operation in accordance with the original power number as $n^k$ seems unconvincing, and the notations of square root only seems more suitable, however, this might become more complex when using other fractional not of the form as power of two, since none of the other root operation was proved in geometry rigorously as the case of square root operation, but only concluded $\endgroup$
    – Bassam Karzeddin
    Commented Jul 9, 2017 at 12:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.