I would like to prove the following result: "Let $ x, y geq 0 $ don't be negative real and let $ n, may 1 $ be positive integers. If $ y = x ^ { frac {1} {n}} $, then $ y ^ {n} = x $"This is Lemma 5.6.6 (a) from the Analysis 1 book by Terence Tao.

The nth root is defined as follows. $ x ^ { frac {1} {n}}: = $sup$ {y in mathbb {R}: y geq 0 $ and $ y ^ {n} leq x } $.

The following lemma was previously proven. ""$ textbf {Lemma 5.6.5:} $ "To let $ x geq 0 $ be a non-negative real and let $ n geq 1 $ be a positive integer. Then the set $ E: = {y in mathbb {R}: y geq 0 $ and $ y ^ {n} leq x } $ is not empty and is also limited at the top. Certain, $ x ^ { frac {1} {n}} $ is a real number. "

Given Lemma 5.6.5, we just have to show that $ y ^ {n} <x $ and $ y ^ {n}> x $ lead to contradictions. For example in the case where $ n = 2 $ and $ y ^ {2} <x $ we can find one $ varepsilon> 0 $ so that $ (y + varepsilon) in E $ only by expanding $ (y + varepsilon) ^ {2} $ and choose $ varepsilon $ reasonable, which contradicts the assumption that $ y = sup E $.

I am aware of how this result can be verified based on identity $ b ^ {n} – a ^ {n} = (b-a) (b ^ {n-1} + b ^ {n-2} a + … + a ^ {n-1}) $, which is used, for example, in Rudin's real analysis book or in the binomial theorem. However, I try to prove the result with just a few pointers in the textbook. The instructions are as follows:

1) Check the evidence that $ sqrt2 $ is a real number (the proof follows the exact outline above).

2) Evidence by contradiction.

3) The trichotomy of order.

4) Theorem 5.4.12

$ textbf {Proposition 5.4.12:} $ "To let $ x $ be a positive real number. Then there is a positive rational number $ q $ so that $ q leq x $and there is a positive integer $ N $ so that $ x leq N $. "

I tried to prove the result using only the four clues given above, but was unable to achieve anything. The four clues are given for the entire lemma, which consists of more than the above statement, so it is not clear that all clues should be used for this particular statement. Exponentiation properties for real numbers and integer exponents have been proven so far, so that these can be used in the proof.

There is a similar question here. Help with a lemma of the nth root (without the binomial formula), but my question is not answered there (and was not answered in any other similar post that I read).

My experiments focused on the following idea: Suppose $ y ^ {n} <x $. Then $ x-y ^ {n}> 0 $what implies the existence of $ q in mathbb {Q} ^ {+} $ so that $ q leq x -y ^ {n} $. We could also assume that $ 0 <q <1 $ to get $ q ^ {n} leq x-y ^ {n} $, although it is not clear to me that this helps. If we accept that $ (y + varepsilon) ^ {n} geq q ^ {n} + y ^ {n} $ for all $ varepsilon> 0 $, then we could get a contradiction by taking the limit as $ varepsilon $ tends to zero. However, limits will only be developed in the next chapter. Instead I tried to find it $ varepsilon $ directly, especially by trying to use hint number four without luck (I think all the messy attempts here would make a long post unreadable).

Any help would be appreciated. Please excuse the long post. Thank you to everyone who takes the time to read this post.