Well-defined function that doesn't satisfy IVT

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In summary, the conversation discusses the function $f:\mathbb{Q}\rightarrow \mathbb{Q}$, defined as $f(x)=x^2$. It is shown that $f$ is well-defined, with each input having a unique output in the rational numbers. The function also satisfies $f(1)=1$ and $f(3)=9$, and it is proven that $f$ does not satisfy the intermediate value theorem on the interval $[1,3]\cap \mathbb{Q}$ by showing that there is no rational number $c$ such that $f(c)=2$.
  • #1
mathmari
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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:
 
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  • #2
Hey mathmari!

It looks all correct to me. (Nod)
 
  • #3
For 1 you should explicitly show that for any rational number, x, x2 is a RATIONAL number.
 
  • #4
Country Boy said:
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.
 
  • #5
Thank you! (Smile)
 
  • #6
Klaas van Aarsen said:
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.
 

FAQ: Well-defined function that doesn't satisfy IVT

1. What is a well-defined function?

A well-defined function is a mathematical rule that assigns a unique output to every input. This means that for every input value, there is only one possible output value, and the function is not ambiguous or undefined.

2. What is the Intermediate Value Theorem (IVT)?

The Intermediate Value Theorem (IVT) is a fundamental theorem in calculus that states if a continuous function has a value less than one point and a value greater than another point, then it must also have a value between those two points. In other words, the function must take on all intermediate values between two given points.

3. Can a well-defined function violate the Intermediate Value Theorem?

Yes, it is possible for a well-defined function to violate the Intermediate Value Theorem. This can occur when the function is not continuous or when there are "jumps" in the graph that prevent the function from taking on all intermediate values between two given points.

4. Why is it important for a function to satisfy the Intermediate Value Theorem?

The Intermediate Value Theorem is important because it guarantees the existence of solutions to certain problems and allows us to make conclusions about the behavior of a function. It is also a necessary condition for certain mathematical techniques and proofs.

5. How can we determine if a function satisfies the Intermediate Value Theorem?

To determine if a function satisfies the Intermediate Value Theorem, we can graph the function and check if it is continuous and if it takes on all intermediate values between two given points. We can also use mathematical techniques such as the Mean Value Theorem to prove that a function satisfies the IVT.

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