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Given arbitrary real number $x>0,$ and an arbitrary natural number n, prove that there exists a positive real number y, such that $y^{n} = x.$

Could anyone give me a hint?

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If you can prove the function $y^n-x$ is continuous in $y$ for fixed $x$, you can note it runs from $-x$ at $y=0$ to $+\infty$ at $y=+\infty$. Then you can use the intermediate value theorem.

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  • $\begingroup$ but we know that any power function is continuous, right? $\endgroup$ – user426277 Apr 6 '17 at 13:15
  • $\begingroup$ and the intermediate value theorem said that f is continuous on a closed interval $\endgroup$ – user426277 Apr 6 '17 at 13:20
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    $\begingroup$ @Smart There are a couple of hairy details that need to be carefully managed for a proof, but you asked for a hint. If you want a more substantive hint, use the definition of limits at $\pm\infty$ to prove you can construct an interval whose endpoints have opposite signs for $y^n-x$. $\endgroup$ – J.G. Apr 6 '17 at 13:29
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Hint only: Given $x>0,$ define $A=\{u>0 | u^n<x \}.$ Then consider the least upper bound of set $A$ as the candidate for $y.$ [You'll need to show $A$ is bounded above first, in order to apply l.u.b. principle.]

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  • $\begingroup$ why shall I consider the least upper bound of set A as the candidate for y? $\endgroup$ – user426277 Apr 6 '17 at 13:25
  • $\begingroup$ Anything less than lub A would be less than the desired $y,$ and anything more than lub A would be greater than the desired $y.$ $\endgroup$ – coffeemath Apr 6 '17 at 13:37
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Let $f: y \mapsto y^{n}-x$ on the open interval $]0, +\infty[$. Note that by Archemidean Axiom (or as a theorem, dependent on the choice of axioms for the real system) there is some such $y_{1}$ that $y_{1} \in ]0, +\infty[$ and that $y_{1}^{n} > x$; i.e. $f(y) > 0$ for all $y \geq y_{1}$. Note that if $0 < y < x^{1/n}$ then $y^{n} < x$; i.e. $f(y) < 0$ for all $0 < y < x^{1/n}$. Note that $f$ is continuous on $]0,+\infty[$ (try to prove this). So by the intermediate value theorem $f(y) = 0$ for some such $y$ that $y \in ]0,+\infty[$.

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The cases $n=1$ or $x=1$ are trivial.

For $n>1$ and $x\ne 1:$ If $x$ is the $n$th power of a member of $\mathbb Q^+$, we are done.

If not, then

(1). If $0<x<1$ and there exists $z$ with $z^n=1/x$ then let $y=1/z$. So it suffices to consider the case $x>1$.

(2). If $x>1$ then for each $b\in \mathbb N$ let $a_b$ be the largest $a\in \mathbb N$ such that $a^n<xb^n$.... So $$(a_b/b)^n<x< ((a_b+1)/b)^n.$$ And we have $$(\bullet)\quad 0<x-(a_b/b)^n<((a_b+1)/b)^n-(a_b)^n=(a_b/b)^n(1+1/a_b)^n<x(1+1/a_b)^n .$$ Now $(a_b+1)/b)^n>x>1$ so $a_b\geq b$ so $a_b\to \infty$ as $b\to \infty$. Applying this to $(\bullet)$ we have $$\lim _{b\to \infty} (a_b/b)^n=x.$$

Note that $(a_b/b)_{b\in \mathbb N}$ is a bounded sequence because (recalling $x>1$) if $a_b/b>x$ then $(a_b/b)^n>x^n> x >(a_b/b)^n$, which is absurd.

So let $y=\lim_{c\to \infty}\sup_{b>c}a_b/b.$

And take a strictly increasing sequence $(b(i))_{i\in \mathbb N}$ such that $\lim_{i\to \infty}a_{b(i)}/b(i)=y.$

For brevity let $y_i=a_{b(i)}/b_i.$ We have $x=\lim_{i\to \infty}y_i^n$ and $y=\lim_{i\to \infty}y_i$, with all terms being positive. Show that this implies $y^n=x$ by showing that for any $r>0$, either $y^n/x>1+r$ or $y^n/x<1/(1+r)$ leads to a paradox.

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  • $\begingroup$ And the short version is: Take $a>0$ with $a^n>x.$ The function $f(z)=z^n $is continuous on $\mathbb R,$ so it maps the connected space $[0,a]$ to the connected space $B=\{f(z):z\in [0,a]\}.$ We have $0\in B$ and $a^n\in B$, and since a connected subspace of $\mathbb R$ is an interval, we have $[0,a^n]\subset B.$ So $x\in B$ .So there exists $y\in [0,a]$ with $f(y)=x.$ $\endgroup$ – DanielWainfleet Mar 30 '17 at 21:23

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