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I'm self-studying proof theory, and working on the following problem:

Consider $s\in\mathbb{R}$, with $s>0$. Apply the Intermediate Value Theorem to prove the existence of $\sqrt{s}$.

I figure I can use the theorem to prove the existence of a root $r$ for $f(x)=x^2-s$. So, $r^2-s=0 \iff r^2=s \iff r=\sqrt{s}$.

Now, in order to apply the theorem to my function, I need to define a closed interval $[a,b]$ such that $f(a)<0$ and $f(b)>0$. If I let $a=0$, then $f(a)=-s<0$ (since $s>0$), and I'm good. But for $b$ things are not so clear. Here's what I have so far: $$f(b)>0 \iff b^2-s>0 \iff b^2>s$$ I'm not sure I can take the square root of both sides of $b^2>s$, as that would rely on the existence of $\sqrt{b}$, which is essentially what I'm trying to prove in the first place. Thinking about it intuitively, I realize that if $s>1$, then $s^2>s$, and if $s \leq 1$, then $s^2 \leq s$. Thus, $$s>1 \implies f(s)=s^2-s>0,$$ in which case I could simply set $b=s$ and know that $f(b)>0$. But if it turns out that $s \leq 1$, I'm in trouble, since $$s \leq 1 \implies f(s)=s^2-s \leq 0$$ I imagine I should set $b$ equal to something like $s+1$, or $s+\frac{1}{2}$, and it looks like $\frac{1}{2}$ is the minimum I can add to satisfy all conditions, but I'm not sure how to show that explicitely.

ivan
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2 Answers2

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Here's one way:

As above let $f(x) = x^2-s$. Choose $n$ large enough so that $s < n^2$. There is no need to take square roots, and we know that $\lim_n n^2 = \infty$, so such an $n$ always exists.

Then $f(0) <0$ and $f(n) = n^2-s > 0$, so you can take the interval to be $[0,n]$.

copper.hat
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  • Right, I guess what I'm trying to figure out is how to express $n$ in terms of $s$ in a way that ensures $s<n^2$ for all $s>0$ – ivan Feb 23 '13 at 19:10
  • You don't need a formula. You just need to know that such an $n$ exists. If you want you could define it as $n(s) = \min { n \in \mathbb{N} | s < n^2 }$. – copper.hat Feb 23 '13 at 19:14
  • You're probably right. I guess I was making it harder than necessary :) – ivan Feb 23 '13 at 19:17
  • I understand. I prefer to have concrete formula where possible. – copper.hat Feb 23 '13 at 19:18
  • @copper.hat, that is a common mistake. Always remember that theory is easier than practice; in this case, in theory it's enough to argue something must exist, in practice they'll ask you to exhibit it. In this case, though, $\lceil s \rceil^2$ will do. – vonbrand Feb 23 '13 at 20:15
  • @vonbrand: Indeed, $s+1$ will do. My initial solution was to choose $n$ large enough so $\frac{s}{n^2} < 1$, with $n$ an integer so that $\sqrt{s} = n \sqrt{\frac{s}{n^2}}$ (the point being that the square root of $n^2$ is known). Then I realized that it was horribly complicated, but didn't simplify the choice of upper bound after that... – copper.hat Feb 23 '13 at 20:22
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You could also skip the polynomial root method and just use theorem directly.

Since $f(x) = x^2$ is a continuous function, if we can find $s$ in an interval $[f(a), f(b)]$, then there is some $c$ in $[a, b]$ such that $f(c) = s$, then $c^2 = s$. We just need to construct the interval $[f(a), f(b)]$ for any $s$.

We can find $a^2 \leq s$ and $b^2 \geq s$. Set $a=0$ and $b$ sufficiently large such that $b^2 \geq s$.

Bob
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