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I am trying to understand this proof.

Theorem. If $a$ is a nonnegative real number and $n$ is a positive integer, there exists a real number $b\geq 0$ such that $b^n=a$.

Proof. Let $$X=\{x\in\mathbb{R}\mid x\geq 0 \text{ and }x^n\leq a\}.$$ First $X\neq \varnothing$ since $0\in X$. We argue by contradiction to show that $X$ is bounded above by $a+1$. Suppose that there exists $x\in X$ such that $x>a+1$. Then $$a\geq x^n>(a+1)^n=\sum\limits_{k=0}^n \binom{n}{k}a^k\geq na$$ which is impossiblle; so $X$ is bounded above.

By the least-upper-bound axiom, $X$ has a supremum $b\in\mathbb{R}$. We will prove that $b^n=a$. We know that exactly one of these relations holds $$b^n<a,b^n>a,b^n=a$$ We show that the first two possibilities cannot occur.

Suppose $b^n<a$, and let $\delta=a-b^n>0$. Choose positive integers $m_0,...,m_{n-1}$ such that $$\binom{n}{k}b^k \frac{1}{m_{k}^{n-k}}<\frac{\delta}{n},k=0,1,...,n-1$$ Let $m=\max\{m_0,...,m_{n-1}\}$. Then $$(b+\frac{1}{m})^n=\sum\limits_{k=0}^n \binom{n}{k}b^k\frac{1}{m^{n-k}}$$ $$=\sum\limits_{k=0}^{n-1} \binom{n}{k}b^k\frac{1}{m^{n-k}}+b^n$$ $$<\sum\limits_{k=0}^{n-1}\frac{\delta}{n}+b^n$$ $$=\delta+b^n=a.$$ Therefore $b+\frac{1}{m}\in X$ but $b\leq b+\frac{1}{m}$ which is impossible; thus $a\leq b^n$. Similarly one shows that $a<b^n$ is false, and therefore $a=b^n$.

My problem understanding it is in the choice of the $m_k$. Why can we make that choice? I think when the author says

Suppose $b^n<a$, and let $\delta=a-b^n>0$. Choose positive integers $m_0,...,m_{n-1}$ such that $$\binom{n}{k}b^k \frac{1}{m^{n-k}}<\frac{\delta}{n},k=0,1,...,n-1$$

he is using the fact that the order is archimedean, but I don´t see the form... If we have $w,r\in\mathbb{R}^+$ there exists $q\in\mathbb{N}$ such that $w<qr$. But I don't see which is the $w,q,r$...

LeviathanTheEsper
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2 Answers2

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Turn it around. Since everything is positive, this is the same as saying $\binom{n}{k}b^k \frac{n}{\delta} < m_k^{n-k}$, and if we also pick $m_k \ge 1$, then it is sufficient to pick $\binom{n}{k}b^k \frac{n}{\delta} < m_k$.

Paul Sinclair
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The argument is meant to choose a number $x$ larger than $b$ such that $x^n < a$, contradicting the least upper bound property of $b$.

The argument uses a clever trick to find a such a number of the form $b + 1/m = x$, where $m$ is an integer. This is done by choosing integers $m_1, \ldots, m_{n-1}$ such that the $k+1$th term in the expansion of $(b+1/m_k)^n$ is less than $\delta/n$. Since there are fewer than $n$ terms in this expansion, the total contribution of the terms (not including $b^n$) will be less than what room there is between $b^n$ and $a$.

This argument is essentially showing that $x \mapsto x^n$ is a continuous function via the binomial expansion.

Edit addressing the comment: This is indeed the Archimedian property.

Since $n-k > 0$ we know that $1/m_k^{n-k}$ gets arbitrarily small as $m$ gets large. Smaller than $\delta/(n b^k (n \text{ choose } k))$ say, To see this, take reciprocals and use an argument like the proof did to show that $a+1$ is an upper bound for $X$. This implies that such an $m_k$ exists. Then the chosen $m$ will have the desired property, since $1/m^{n-k}$ is even smaller.

Eric Thoma
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