How do you prove square root of 2 exists?

Question

This is probably the most inane question I've ever seen. Not only is it in the textbook though, but the teacher actually assigned it as work.

The question reads:

$\sqrt{2}$ exists.

OK, there's a preface for the group of questions:

Prove using the IVT.

That is to say, the Intermediate Value Theorem.

I mean... intermediate value theorem only works for functions... This is just a number. I went ahead and just said The function $f(x)=\sqrt{x}$ is continuous for all positive, real numbers. $2$ is a positive, real number. $\sqrt{2}$ exists.

But that's not using the "IVT". How the heck does IVT relate to this?

Answer 1

You are close to the proof but to define the function $f(x)=\sqrt{x}$ you essentially take for granted that $\sqrt2$ exists.

I think what your teacher had in mind would be to use the function $g(x)=x^2$.Since $g(0)=0$ and $g(2)=4$ by the IVT there exists a $p$ such stat $g(p)=2\Rightarrow p^2=2 \Rightarrow \sqrt2=p$.

Answer 2

Hint: Use IVT on $f(x)=x^2 - 2$.

Answer 3

Hint:

Consider $f(x)=x^2$. We have $f(1)=1$ and $f(2)=4$. Can you finish?

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