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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# [Math] Prove that there exists a positive real number y, such that $y^{n} = x.$

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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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## Best Answer

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.