- #1

mathmari

Gold Member

MHB

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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: