# Fun little thing I'm curious about.

• Jul 16th 2010, 01:42 PM
Mazerakham
Fun little thing I'm curious about.
From reading books and stuff, everyone says that "ccompletenesss is the important thing about the real numbers that gives them the properties necessary fro calculus.

My fun little question is this: do the real numbers need completeness to have the intermediate value theorem?

To put my question more concisely, suppose we have a function which maps the rational numbers into the rational numbers. Also, suppose it is continuous. That is, for any positive RATIONAL epsilon, there is a positive RATIONAL delta such that... you get the picture. My question is this: does the intermediate value theorem hold? Counter-example, anyone? I've been trying to come up with one.

The point I'm trying to make here is that the epsilon-delta definition of continuity seems kind of...independent of the completeness axiom. So, yea. Would the intermediate value theorem hold for a rational-to-rational function, with continuity defined as above?
• Jul 16th 2010, 03:33 PM
Ackbeet
Counter-example: Let $\displaystyle f(x) =(x-\sqrt{2})(x+\sqrt{2})$ over the interval from $\displaystyle [0,3].$ Now, $\displaystyle f(0)<0,$ and $\displaystyle f(3)>0.$ By the Intermediate Value Theorem, you'd expect there to be a rational root in the interval. But there isn't.

In addition, the delta-epsilon definition of continuity is entirely equivalent to the "calculus" definition, where the limit of a function at a point is equal to the function's value at that point. All you have to do is look at an irrational function value, such as $\displaystyle \sqrt{2},$ to see that if the irrationals are not included, you can't have continuity.
• Jul 16th 2010, 04:57 PM
Mazerakham
Doy!
Thank you so much. I feel rather silly for not coming up with a counterexample so simple. I'm just trying to solidify my understanding of the reals, and their relationship to the integers and rationals. Your post was educational. (Happy)
• Jul 16th 2010, 05:27 PM
Ackbeet
You're very welcome. Have a good one!
• Jul 18th 2010, 05:58 AM
As Ackbeet said. it all goes back to the definition of irrationals. Due to the completeness axiom, we were able to discover the number $\displaystyle \sqrt{2}$ by taking the set of rationals whose square is less than 2 and then use the completeness axiom to show that there is a least upper bound for this set and is actually the irrational $\displaystyle \sqrt{2}$ as we define it. based on this, all real analysis is built and hence, you can't even talk about real functions without giving credit to the completeness axiom.