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Can we write $x=y^q$ where x is a positive real number and y is any real number i.e. whether all positive real numbers can be formed by integral powers of any real number, In other words I want to find the range of the function $f(x,y)=x^y$ where y is a natural number and x is any real or you can say that atleast one qth root of a positive real is real. In more simple words for a given x and all integral values of q does there exist a y always?

I think it can be said but I dont know precisely why or cannot prove it.

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    What are your free variables? Are you asking: *given $x$ and $y$ does there exist a value of $q$ such that $x=y^q$*? Or are you asking *given $x$ does there exist values of $y$ and $q$ such that $x=y^q$*? Or are you asking for *given $x$ and $q$ does there exist a value of $y$ such that $x=y^q$*? – 2017-02-18
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    The last one ie for every q and a given x a y must always exist. – 2017-02-18
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    Note that $f(x,1) = x$. – 2017-02-18
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    As for the range of the function $f~:~\Bbb R\times \Bbb N\to \Bbb R$ given by $f(x,y)=x^y$, clearly it will be all of $\Bbb R$ since the function restricted to when $y=1$ is just $f(x,1)=x^1=x$ and $x$ ranges over all reals and the range of the function as a whole will be a superset of the range of the restricted function, the only superset of $\Bbb R$ that is also a subset of the codomain is again all of $\Bbb R$. – 2017-02-18
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    If you are asking for a proof as to why $\sqrt[q]{x}$ is real for all natural $q$ and all positive real $x$... by modifying the [proof for the existence of the square root of 2](http://math.stackexchange.com/questions/1319580/proof-that-square-root-of-2-exists-and-is-a-real-number) for arbitrary $q$ and $x$ one sees that indeed $\sqrt[q]{x}$ is indeed a real number. – 2017-02-18

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What you're asking is: Given an integer $q \geq 1$ and a real number $x > 0$, does there exist $y > 0$ such that $y^q = x$?

The answer is yes. Take $y = \sqrt[q]{x}$.

The fact that a $q$th root exists is shown by proving that the function $f(t) = t^q$ defined for $t > 0$ has range $(0,+\infty)$, by using the intermediate value theorem.

Edit: Let $y > 0$. We have $\lim_{t \to +\infty} f(t) = +\infty$. Therefore there exists $B > y$ that is in the range of $f$. Similarly, $\lim_{t \to 0} f(t) = 0$, so there exists $A < y$ that is in the range of $f$.

The function $f$ is continuous, both $A$ and $B$ are in its range, and $A < y < B$. Therefore, by the intermediate value theorem, $y$ is in the range of $f$.

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    Can you show how ivt can be used ? – 2017-02-18
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    So y cannot be negative only positive ? If not how can we prove it ? – 2017-02-19
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    Well, you need to take into account signs in the usual way. For example, $-64$ has a cube root but no square root, since $3$ is odd and $2$ is even. These facts are not difficult to prove once you have the theorem about roots of positive numbers. – 2017-02-19