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I was wondering what $x^{1/2}$ is. You know, when I say $x^2$ it's $x \cdot x$ or $x^3 = x \cdot x \cdot x$ etc.

But what is $x ^{1/2}$?

I know it's $$\sqrt x$$ but I mean when you want to explain it like with $$x^2 = x \cdot x$$ How do you explain it?

N. F. Taussig
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Hudhud
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5 Answers5

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You probably know the rule $x^n\cdot x^m=x^{n+m}$ which can simply be motivated for natural numbers $n$ and $m$ from the intuitive undertstanding via

$$x^n\cdot x^m=\underbrace{x\cdots x}_{\text{$n$ times}}\cdot \underbrace{x\cdots x}_{\text{$m$ times}}=\underbrace{x\cdots x}_{\text{$n+m$ times}}=x^{n+m}.$$

Now you introduce $x^{1/2}$ for what reason whatsoever and ask yourself what value this should give. One sensible way could be to do it in such a way so that $x^{1/2}\cdot x^{1/2}=x^{1/2+1/2}=x^1=x$. This means we try to keep the rule from above. In other words: we want

$$(x^{1/2})^2=x,\qquad\text{and $x^{1/2}=\sqrt{x}$ is one reasonable way to do it}.$$

I cannot think of another sensible way to define the meaning of $x^{1/2}$.

M. Winter
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  • No. $y^2 = 1$ does not imply $y = \sqrt{1}$... – user21820 Aug 24 '17 at 05:49
  • @user21820 It was obviously meant as a motivation for the notation and not a rule to solve for $x$. – M. Winter Aug 24 '17 at 08:15
  • If that was not clear to me, I fail to see how it would be clear to students many of whom do not have a proper grasp of basic logic. Note that if you want to express that in logical symbols then the implication needs to be backwards! So, if you don't mind just changing that line "$(x^{1/2})^2=x \quad ⇒ \quad x^{1/2} = \sqrt{x}$" to "If we want $(x^{1/2})^2=x$ then we could get it by defining $x^{1/2} = \sqrt{x}$", then I think your answer would be great. – user21820 Aug 24 '17 at 08:20
  • @user21820 I know it its a long time ;) but better now? – M. Winter Oct 09 '17 at 20:53
  • It's great, thank you! =) – user21820 Oct 10 '17 at 13:02
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We define $x^{\frac12}$ to be equal to $\sqrt x$ because we like the property $$\forall x\in\mathbb R:\forall a,b\in\mathbb N: (x^a)^b = x^{a\cdot b}$$ and we want this equality to hold even if $a$ and $b$ are not integers.

So, whaterver $y=x^\frac12$ is equal to, we want $y^2$ to be equal to $x$, and that means that $y$ must be $\pm \sqrt x$. The fact that there are two possibilities already presents a problem, and in fact the rule cannot be extended as-is, but rather in the slightly changed form

$$\forall x>0:\forall a,b\in\mathbb Q: (x^a)^b=x^{a\cdot b}$$

(notice that it's only true for $x>0$ from now on).

We choose the positive one because if we choose the negative, we wouldn't be able to apply the "$^\frac12$" twice in a row.

George N. Missailidis
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5xum
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    Don't we want the property $x^ax^b = x^{a + b}$ to hold here since want $x^{\frac{1}{2}}x^{\frac{1}{2}} = x^{\frac{1}{2} + \frac{1}{2}} = x^1$? – N. F. Taussig Mar 10 '16 at 13:54
  • @N.F.Taussig: It amounts to the same thing here because we want $(x^a)^b = x^{a·b}$ where $b$ is a natural number. Naturally, the $b$-th power of $(x^a)$ is simply the product of $b$ copies of $x^a$, and $a·b$ is simply the sum of $b$ copies of $a$. – user21820 Aug 24 '17 at 06:18
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I was once asked this question by a Chinese high school student who was quite bright, but couldn't speak English worth beans.

He wrote the following on a piece of paper and looked at me questioningly:

$$2^{1/2}=?$$

Here is how I answered him.

(I wrote each of the following on a sheet of paper one after another and pushed the paper at him, and he wrote the answer to the question mark.) (And actually I used dots for multiplication rather than $\times$, but whatever.)

$$5\times5\times5=5^?$$ $$7\times7\times7\times7\times7=7^?$$ $$(5\times5)\times(5\times5\times5)=5^?$$ $$5^2\times5^3=5^?$$ $$7^4\times7^5=7^?$$ $$37^4\times37^{13}=37^?$$ $$17^5\times17^5\times17^5=17^?$$ $$(17^5)^3=17^?$$ $$(17^3)^5=17^?$$ $$(1357^{17})^{10}=1357^?$$ $$(x^2)^{1/2}=x^?$$

At this point he grabbed the paper from me and wrote:

$$2^{1/2}=\sqrt 2$$

(Note: it was implied by our written dialogue that $x$ represented some counting number. Not a negative number! When you allow negative numbers to get involved with fractional exponents, things get more interesting—and also very beautiful—but that's a different lesson. :) )

Wildcard
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In the same way you explain $\frac{1}{2}$ formally: it's a solution to $y + y = 1$. (This formal stipulation works even when you don't have a sensible way to cut your representations of whole numbers in half.)

In your case, $x^{\frac{1}{2}}$ describes the solutions to $y \times y = x$.

George N. Missailidis
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mad_algebraist
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It's a positive number (remember that square roots are "always" positive, and are defined for positive numbers if you want real solution) y such that y*y = x ;

ab123
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