**Proposition** For any $q \in \mathbb{N}$ and $y >0$, $\exists! x \in \mathbb{R}$ such that $x^{q} =y$. See [[Comparison of square root of 2 and nth root of y proofs]]. **Proof** WLOG suppose $y>1$ since if $y<1$ then $\frac{1}{y} >1$. - Consider the set $S=\{ x \in \mathbb{R}: x^{q}<y \}$. $S$ is non-empty since $1 \in S$. $S$ is bounded above since if $x \in S$ we have $x^{q}<y<y^{q}$ (since $y>1$ and $q \in \mathbb{N}$) so $x<y$. Hence $y$ is an upper bound for $S$. So by [[Real numbers]], $\exists r \;(r=\sup S)$. Since $1 \in S$ we have $r\geq 1$. - Consider $(x+\epsilon)^{q}= \sum_{k=0}^{q} {q \choose k} x^{k} \epsilon^{q-k}$If $x\geq 1$ and $\epsilon<1$ then $x^{k}\leq x^{q}$ for every $k=0,\dots q$ and $\epsilon^{k} <\epsilon$ for $k=1,\dots,q$. so $(x+\epsilon)^{q} \leq x^{q} \left( \sum_{k=0}^{q} {q \choose k} \epsilon \right) \leq x^{q} \left( 1+ \epsilon \sum_{k=1}^{q} {q \choose k} \right)$Note that $\sum_{k=1}^{q} {q \choose k} < \sum_{k=0}^{q} {q \choose k} 1^{k} 1^{q-k} = (1+1)^{q} = 2^{q}$so $(x+\epsilon)^{q} \leq x^{q}(1+ \epsilon 2^{q})$. - Suppose that $r^{q} <y$. We want to show that $r+\epsilon \in S$. Choose $\epsilon<(yr^{-q} - 1)/2^{q}$ then since $r \geq 1$ $(r+\epsilon)^{q} \leq r^{q}(1+ \epsilon 2^{q}) < y$so $r+\epsilon \in S$. But $r+\epsilon>r$, so $r$ cannot be an upper bound for $S.$ - Suppose $r^{q}>y$. For $\frac{\epsilon}{r} < 1$, using [[Bernoulli's Inequality]], $(r-\epsilon)^{q}=r^{q}\left( 1-\frac{\epsilon}{r} \right)^{q} \geq r^{q}\left( 1- \frac{\epsilon q}{r} \right).$ Choose $\epsilon< \min(1, rq^{-1}(1-yr^{-q}) )$ then $(r-\epsilon)^{q}>y$ which contradicts the assumption that $r$ is the least upper bound of $S$. So we must have that $r^{q}=y$ by trichotomy axiom. We can prove uniqueness similar to how we did here [[There is a unique real number that is the square root of 2]]. **Remark** If $x>0$ and $y=x^{q}$ then we write $y= x^{\frac{1}{q}}$. Now we can now define the [[Rational Power of Real Number]].