In this note, we will shed some light on the nth roots and fractional exponents of numbers. We give a clear and rigorous definition of these root numbers.

### About nth roots numbers

Although square roots (2nd roots) are standard, they are restrictive. Recall that a number $a$ is the square root of a number $b$ if $b$ is positive ($b\ge 0$) and $a^2=b$. In this case, we write $a=\sqrt{b}$ (or sometimes we write $a=b^{\frac{1}{2}}$). From the equation $a^2=b,$ we see that square roots are positive numbers.

Let's start with this question: Is there a number $x$ such that $x^3=-9$. Notice that $$-9= (-3)(-3)(-3)=(-3)^3.$$ Then $x=-3$ is the solution. In this case, we say that $-3$ is the cubic root (3th root) of $-9$ and we write $\sqrt[3]{(-9)}=-3$. So unlike square roots, cubic roots can be negative.

More generally, let $k$ be a non-zero integer and $b$ a number (not necessarily integer) and ask the question: is there numbers $x$ such that $x^{2k}=b$. Nice that $x^{2k}=(x^k)^2$. Thus necessarily $b$ is positive and in this case $x$ is called $2k$-th root of $b,$ and we write $x=\sqrt[2k]{b}$ or $x=b^{\frac{1}{2k}}$. If $b$ is negative, then there is no solution.

Thus for even numbers, $n,$ (numbers of the form $n=2k$), the nth root must be positive.

Now assume that we have an odd number $n=2k+1,$ and look for numbers $x$ satisfying the algebraic equation $x^n=b$. This is, equivalent to $x^{2k} x =b$. As $x^{2k}$ is positive, it follows that the numbers $x$ and $b$ have the same sign (either both positive or both negative). In this case, the solution exists. Now if $x$ and $b$ have opposite signs, then there is no solution to the above equation.

In order to be more precise on the nth roots, let the following definition of the components of a radical expression

If we set $b=\sqrt[n]{x}$. Then we have the following cases

nice

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