Why is $49^{-\frac{1}{2}}=\frac{1}{7}$?

Why is $$49^{-\frac{1}{2}}=\frac{1}{7}$$?

I know that $$\sqrt[n]{m^p}=m^{\frac{p}{n}}$$ so I figured I can state the above is $$-\sqrt{49}=-7$$, but that is incorrect. I can't put the negative inside the root because the solution doesn't contain the imaginary number, $$i$$. So I'm stuck here.

I also know I can just set the equation equal to $$x$$ and easily get the solution as follows $$x=49^{-\frac{1}{2}}$$ $$x^2=49^{-1}$$ $$x=\pm\frac{1}{7}$$

But can anybody give me an intuitive explanation why this is the answer? I'd like to be able to just look at the expression and think "ah, that's obviously $$\frac{1}{7}$$"

• $49^{-1/2}=\frac{1}{49^{1/2}}$
– Surb
Apr 22, 2019 at 7:17
• $a^{-b}=\dfrac1{a^b}$.
– user65203
Apr 22, 2019 at 8:08

It is $$49^{-1/2}=(7^2)^{-1/2}=7^{-1}$$

$$a^{\frac{1}{2}}\cdot a^{\frac{1}{2}}= a^{\frac{1}{2}+\frac{1}{2}}= a^1=a.$$ It follows logically from the above that $$a^{\frac{1}{2}}$$ is that number which, when multiplied by itself, gives you $$a$$. This in turn means that $$a^{\frac{1}{2}}$$ must be $$\sqrt{a}$$ because $$\sqrt{a}\cdot\sqrt{a}=a$$ (also note that $$a\ge0$$ because the square root function is only defined for nonnegative numbers).

$$b^{-1}\cdot b=b^{-1+1}=b^0=1\\ b^{-1}\cdot b=1\implies\\ b^{-1}=\frac{1}{b},\ b\ne0.$$

Therefore:

$$49^{-\frac{1}{2}}= \left(49^{\frac{1}{2}}\right)^{-1}= \frac{1}{\sqrt{49}}=\frac{1}{7}.$$

$$x^2=\frac{1}{49}\implies x=\pm\frac{1}{7}.$$ This is indeed true (just plug in $$\frac{1}{7}$$ and $$-\frac{1}{7}$$ back into $$x$$). But that is different from $$49^{-\frac{1}{2}}$$. $$x^2=\frac{1}{49}$$ is an equation with two solutions and $$49^{-\frac{1}{2}}$$ is a single number. The statement $$x=49^{-\frac{1}{2}}$$ received an extra solution precisely at that moment when you squared both sides.

• Yes, I knew that given the constraint $x \ge 0$ for $\sqrt{x^2}$ then $-\frac{1}{7}$ could be ruled out. I was just too lazy to write that out Apr 22, 2019 at 8:19
• Aright, but did I answer your question or not? Apr 22, 2019 at 9:28
• Well yes, I just already immediately understood the problem once @Surb commented, but I do appreciate your answer Apr 22, 2019 at 18:54

I see that this answer has been down-voted. I would like to know what is wrong with it. I have already made a revision and I assume that this was the only error. Thanks for your help.

$$49^{-\frac{1}{2}}=$$

$$\frac{1}{\sqrt{49}}$$

You know that: $$\sqrt{49}= 7$$ (by definition of square root, we only consider the positive value not the $$-7$$) so,

$$\frac{1}{\sqrt{49}}=+\frac{1}{7}$$ so,

$$49^{-\frac{1}{2}}=\frac{1}{\sqrt{49}}=+\frac{1}{7}$$

Note/Edit: I have removed the value: $$-\frac{1}{7}$$ from the above answer as I have reviewed the comments below and revisited that convention for $$\sqrt{}$$, which denotes only positive value. Also, raising a number to a positive value requires that the number be positive otherwise, the result would be a complex number, which is also not the case here.

• $x \mapsto x^{-\frac 1 2}$ is a function, hence it cannot have two images. Apr 22, 2019 at 7:47
• @nicomezi, I am not sure I get your point since $(\frac{-1}{7})^{2}=(\frac{1}{7})^{2}$. The equation $x^2=1/\sqrt{49}$ has the above 2 solutions not only one. The positive value is the principal root. Apr 22, 2019 at 8:56
• I guess the reason why your answer received downvotes (not from me) has to do with the fact that in it you stated that $49^{−\frac{1}{2}}=±\frac{1}{7}$. How can a single number be equal to two different numbers at the same time? Moreover, from your calculations it follows that $\frac{1}{7}=-\frac{1}{7}$. Is that really true? That's pure nonsense. $49^{−\frac{1}{2}}$ is not an equation to have two or more solutions. It's a single number. Apr 22, 2019 at 9:36
• @MichaelRybkin, thank you for sharing this thought. I see the square root as a "relation" not a function but will dig more to learn - That is why I joined the group after all. Apr 22, 2019 at 9:38
• I personally didn't understand the downvotes. You're always supposed to start with finding the domain of a root, then you have to include both positive and negative solutions to avoid extraneous solutions. Then you can exclude one of the solutions if they're not within the domain of the root. Apr 22, 2019 at 18:56