MHB Well-defined function that doesn't satisfy IVT

[mathmari]

Hey! :giggle:

Let $f:\mathbb{Q}\rightarrow \mathbb{Q}$, $f(x):=x^2$.
Show that :
(i) $f$ is well-defined.
(ii) $f(1)=1$, $f(3)=9$
(iii) $f$ does not satisfy the intermediate value theorem (e.g. not on $[1,3]\cap \mathbb{Q}$) For (i) do we just say that $f$ is well-defined from $\mathbb{Q}$ to $\mathbb{Q}$, since each input $x$ has a unique output $x^2$ ?

For (ii) we have that $f(1)=1^2=1$ and $f(3)=3^2=9$. That's it?

For (iii) we have that $1=f(1)<2<f(3)=9$. So there must be $c\in (1,3)$ such that $f(c)=2 \Rightarrow c^2=2 \Rightarrow c=\pm \sqrt{2}\notin \mathbb{Q}$. Is that correct?

:unsure:

 

[I like Serena]

Hey mathmari!

It looks all correct to me. (Nod)

 

[HOI]

For 1 you should explicitly show that for any rational number, x, x² is a RATIONAL number.

 

[I like Serena]

For 1 you should explicitly show that for any rational number, x, x² is a RATIONAL number.

The rational numbers are closed for multiplication, so just saying that if $x$ is rational, then $x^2$ is also rational should be enough.

 

[mathmari]

Thank you! (Smile)

 

[HOI]

The rational numbers are closed for multiplication, so just saying that if $x$ is rational, then $x^2$ is also rational should be enough.

Yes, it is enough. My point was that mathman, in the original post, said only "since each input x has a unique output x²", not that that output was a rational number.