Proof verification: Proving that any positive real number has a square root using the Intermediate Value Theorem

Question

I am tasked with using the intermediate value theorem to prove that every positive real number $y\in\mathbb{R}$ has a square root, using the intermediate value theorem (as the title suggests). For this purpose, consider the function $f(x)=x^2$. We note that $f(0)=0$, and therefore $f(0)\leq y$, since $y$ is a positive value. We note also that $f(x)\to+\infty$ as $x\to+\infty$. That is, $\forall m\in\mathbb{R}$, $\exists x\in\mathbb{R}$ such that $f(x)>m$. For this reason, pick $m>y$, then we have that $\exists x_0\in\mathbb{R}$ such that $f(x_0)>m$. Therefore $f(0)\leq y<f(x_0)$, and, since $f(x)$ is continuous everywhere on $\mathbb{R}$, we have therefore that, by the intermediate value theorem, $\exists c\in[0,x_0]$ such that $f(c)=y$, or, more appropriately, $\exists c\in[0,x_0]$ such that $\sqrt{y}=c$. This completes the proof.

Of course I am biased, since this is my own proof, and therefore believe the proof to be correct. However, the solution bank which I am using to verify this proof has used a different technique (which I will disclose if I am so prompted), so I was hoping someone could verify the accuracy of this proof. Any responses are appreciated, thank you.

Answer 1

The proof looks just fine.

I guess I would say that you don't actually need to apply the fact that $x^2 \to +\infty$ as $x \to +\infty$ in order to get the value of $x_0$ that you need. You can just take $x_0 = \max\{y,1\}$ and it follows that $f(x_0) \ge y$.

Answer 2

My old professors would probably accept this proof, but at the same time accuse you of killing a fly with a sledgehammer. You're not only making an appeal to IVT but also to the fact that $x\mapsto x^2$ is continuous.

By contrast, a proof without IVT would have fewer moving parts. You just need to note that given $x\in\mathbb R^+$ that the set $\{y\in\mathbb R^+\mid y^2>x\}$ is non-empty and bounded below. Therefore that set has an infimum by the construction of the real numbers, which is in fact what we define as $\sqrt x$.

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Edit: it was pointed out in comments that you were actually ordered to kill a fly with a sledgehammer. In that case, you get bonus points from me for using $f(x)=x^2$ as your function instead of $f(x)=\sqrt x$. Now, if you'll excuse me, I'm off to lament the state of real analysis education....

Answer 3

Basically, you want to prove that $x^2 = l$. Consider $f(x) = x^2$ , There exists a k > 0 such that $f(k) > l$. Don't believe me? if $l$ > 0, choose $k = l$. If $l<0$, choose $k = 1$.

So, we now have $ f(0) < l < f(k) $ because $l$ is a positive number but lower than some number $k$. By IVT, there is some number x in $[0,k]$ such that $f(x) = l$. Or $x^2 = l$ which is exactly what we need.