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Statement:

If $n$ is an odd positive integer and $a<0$. prove that there is exactly one negative $b$ such that $b^n = a$

Proof: let $n= 2k+1$ where $k \in \mathbb N$

let $c < 0$ s.t. $c<a<0$

And consider the function $f(x) =x^{2k+1} $ defined on $[c,0]$, the function is continuous on the interval $(c,0)$.

the two endpoint values will be $ f(c)= c^{2k+1} <0 $ and $f(0) = 0$

here $c^{2k+1} <c<a<0$ implies $f(c)=c^{2k+1}<a<0= f(0)$

thus by the intermediate value theorem we have $f(x)=a$ for some $x\in [c,0]$ say $x=b$.

This proves the existence of at least one negative b such that $b^n = a$

there can't be more than one such $b$ because $f$ is strictly increasing on $(c,0)$.

This completes the proof.

I have proven this theorem in this way. Am i doing it right or it needs correction ? can anyone comment on this proof?

Dr. Sundar
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Prashant Anand
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  • This is valid. It could be shortened provided you can use a similar result concerning uniqueness of positive $r$ for which $r^n=c$ in case $c>0.$ [However to prove that requires an argument not much different from the one you gave.] – coffeemath Apr 13 '22 at 06:14
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    For future reference, you can use $\in$ to write the symbol "$\in$", instead of $\epsilon$. $a\in A$ looks much nicer than $a\epsilon A$. – 5xum Apr 13 '22 at 06:22

1 Answers1

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In your proof, you claim that $c^n < c$. But all you know is that $c<0$, and that is not enough to conclude $c^n < c$. For example, if $c=-\frac12$, and $n=3$, then

$$c^n = \left(-\frac12\right)^3 = -\frac18 > c.$$

Having said that, the main idea of your proof is correct. If you only improve the choice of $c$, you should be able to complete the proof!

5xum
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