Show that if $y = x^{1/n}$, then $y^n =x$. Lemma 5.6.6. Let $x, y \ge 0$ be non-negative reals, and let $n,m \ge 1$ be positive integers. Show that if $y = x^{1/n}$, then $y^n =x$. 
In Tao's Analysis 1, he defines $x^{1/n} = \sup\{z \in \mathbb{R}: z\ge 0, z^n\le x\}$. I am trying to prove the statement by contradiction. Suppose that $y^n \not=x$. Then, $y^n >x$ or $y^n <x$ (both cases should lead to contradiction). First, consider $y^n >x$. Then, there exists $0 \le \epsilon \le 1$ such that $(y-\epsilon)^n > x$ (I don't know how to justify this sentence). This implies that $y$ is not the least upper bound (contradiction). Another case can be shown similarly. 
If I expand $(y-\epsilon)^n$ out, I get $y^n + \epsilon^n + f(y, \epsilon)$. How can I choose $\epsilon$ to make $(y-\epsilon)^n > x$?   Is there easier way to solve the question? 
 A: Consider that $(y\pm \epsilon)^n = y^n + \sum_{k=1}^n(\pm 1)^k \epsilon^ky^{n-k}{n\choose k}$.
Trick is to choose $\epsilon$ so that $\begin{cases}\sum_{k=1}^n \epsilon^ky^{n-k}{n\choose k} < x-y^n&\text{if we assume }y^n < x\\0<-\sum_{k=1}^n(- 1)^k \epsilon^ky^{n-k}{n\choose k}<y^n-x&\text{if we assume }y^n > x\end{cases}$
(So if we assume $y^n < x$ then $(y+\epsilon)^n =y^n +\sum_{k=1}^n \epsilon^ky^{n-k}{n\choose k} < y^n + x-y^n=x$ so $y$ is not an upper bound.)
(And if we assume $y^n > x$ then $(y-\epsilon)^n = y^n -(-\sum_{k=1}^n(- 1)^k \epsilon^ky^{n-k}{n\choose k})>y^n-(y^n-x)=x$ so $y$ is not the least upper bound.)
But how to find such $\epsilon$s?
Note that if $\epsilon < 1$ then we have $e^k < e < 1$, $\min(y,y^n)\le y^{n-k} \le \max(y,y^n)$ and $\min({n\choose j} \le {n\choose j}\le \max({n \choose j}$.
So 
$\begin{cases}\sum_{k=1}^n \epsilon^ky^{n-k}{n\choose k} <\\n\cdot \epsilon n\max(y,y^n)\max{n\choose j}<n\max(y,y^n)\max{n\choose j}&\text{if we assume }y^n < x\\\\-\sum_{k=1}^n(- 1)^k \epsilon^ky^{n-k}{n\choose k}<\\\lceil \frac n2\rceil\cdot\epsilon\max(y,y^n)\max{n\choose j}-\lfloor \frac n2\rfloor\cdot\epsilon^k\min(y,y^n)\min{n\choose j}<\\n\cdot \epsilon\max(y,y^n)\max{n\choose j}<\\n\max(y,y^n)\max{n\choose j}&\text{if we assume }y^n > x\end{cases}$
So let $\epsilon=\begin{cases}\frac {x-y^n}{n\max(y,y^n)\max{n\choose j}}&\text{if we assume }y^n < x\\\frac {y^n-x}{n\max(y,y^n)\max{n\choose j}}&\text{if we assume }y^n > x\end{cases}$
That's it.
.....
But Rudin (Th. 1.21) has an easier (????) solution if you are clever enough to notice
For $0 \le a < b$ then $b^n - a^n < (b-a)n$ so $(y+\epsilon)^n - y^n > \epsilon*n$ and $y^n - (y-\epsilon)^n < \epsilon*n$.
To wit:
(This is stolen from Baby Rudin Th 1.21)
The trick is to realize that for $0\le a < b$ we have $b^n - a^n = (b-a)(b^{n-1} + b^{n-1}a + ... + a^{n-1})< (b-a)(n\cdot b^{n-1})$.
So you want to choose $\epsilon_1$ and $\epsilon_2$ so that:
If $y < x$ then $(y+\epsilon_1)^n - y^n < (y+\epsilon_1-y)n(y+\epsilon_1)^{n-1}=\epsilon_1 (y+\epsilon_1)^{n-1}< x-y^n$
That's doable if $\epsilon_1 < \frac {x-y^n}{(y+\epsilon_1)^{n-1}}< \frac {x-y^n}{(y+1)^{n-1}}$ (assuming $\epsilon_1 < 1$).
We can do the same for assuming $y^n > x$ and choosing an $\epsilon_2$ so that:
$y^n - (y-\epsilon_2)^n < \epsilon\cdot n(y-\epsilon_2)^{n-1} < y^n -x$
Which is doable but selecting $\epsilon_2 < \frac {y^n -x}{  n(y-\epsilon_2)^{n-1}} < \frac {y^n -x}{ n(y)^{n-1}}$
