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Could someone explain how/why the square root of $x$ equals $x$ to the one half power? I know by definition it does, but is there any mathematical process we can go through to get from one to the other?

In addition, I know the $x$ to the 2ed power is equal to $x*x$, but what is $x^{1/2}$ equal to? Is it one half of $x$ times one half of $x$? No!

I know all this stuff by definition, but I want to know why. Thanks

pshmath0
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user1404617
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    $$\sqrt x=x^n\iff\big(\sqrt x\big)^2=x^{2n}\iff x=x^{2n}\iff x^1=x^{2n}\iff2n=1\iff n=\frac12$$ – Lucian Jan 29 '14 at 17:15

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We can define the symbol $\sqrt{x}$ to be the number that which when squared gives $x$ back, e.g. $$(\sqrt{x})^2=x.$$ We would like to write $\sqrt{x}$ in the form $x^a$ so that $$(x^a)^2=x.$$ Using the fact that $(x^a)^b=x^{ab}$ then we have $$x^{2a}=x^1,$$ since $x \equiv x^1$. Hence, $$2a=1,$$ or in other words $a=1/2.$ This gives $$\sqrt{x}\equiv x^{1/2}.$$ This notation is consistent with our original definition because $$(\sqrt{x})^2=(x^{1/2})^2=x^{(1/2)\times 2}=x^1=x.$$ Note also that our definition is consistent with the other rules of exponentiation, e.g. $x^ax^b=x^{a+b}$.

pshmath0
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We have defined $x^{\frac{1}{2}}$ to mean $\sqrt{x}$, because is the only choice that fits the different rules for exponentials you learn, like $x^a \cdot x^b = x^{a + b}$ or $(x^{a})^b = x^{ab}$. These rules are easily verified when you have natural exponents, and only if you assume that $x^{\frac{1}{n}}$ means $\sqrt[n]{x}$ will these rules still apply. So it's the only definition that makes sense.

Arthur
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For simplicity, assume that $x>0$.

$\sqrt{x} \cdot \sqrt{x} = \left(\sqrt{x}\right)^2 =\ ?$

$\sqrt{x} \cdot \sqrt{x} = \sqrt{x^2} =\ ?$

What happens when you evaluate these? Both should give the same result (since $x>0$) which is $x^1$. Let's make up an exponent $a$ to represent the square root. Then

$x^a \cdot x^a = x^1 \Longleftrightarrow x^{2a} = x^1 \Longleftrightarrow 2a = 1 \Longleftrightarrow a = \dfrac{1}{2}$

John Habert
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One way we have $\sqrt(x)$*$\sqrt(x)$ = $x$ When $x$ is ofcourse nonnegative On the other hand by rules of exponents we have $x^.5$*$x^.5$ = $x^1$ So from here it follows.

imranfat
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Law of Exponents $\,\Rightarrow\,x = a^{1/2}\,$ is a positive root of $\, x^2 - a,\ $ assuming $\ a > 0.$

Further, by definition, $\,\sqrt{a}\ $ is also a positive root of $\, x^2 - a.$

If $\ a^{1/2}\neq \sqrt{a}\ $ then the quadratic $\,x^2-a\,$ has $> 2$ roots: $\,a^{1/2},\ {\pm}\sqrt{a},\,$ contradicting the theorem that a nonzero polynomial over a field has no more roots that its degree.

Thus $\,\ a^{1/2} = \sqrt{a}\ \ \ $ QED

Bill Dubuque
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