Math Is Fun Forum

Wow! I thought of Steve Davis and it correctly guessed my character. I’d say that was just a fluke.

Some friends of mine have come up with some of their own haiku here: clickety click.

Are you familiar with Lagrange’s theorem? One consequence of Lagrange is that the order of an element of a finite group must divide the order of group. The group you’re dealing with is of order 10; hence the possible orders for 2 are 1, 2, 5 and 10. You therefore only need to check 2[sup]1[/sup], 2[sup]2[/sup], 2[sup]5[/sup] and 2[sup]10[/sup].

Okay, more ideas. Let

(1) If D varies and the other coefficients are fixed, then [1] is obviously the general solution to the differential equation

(2) If C varies and the other coefficients are fixed, then

and so [1] is the general solution to the differential equation

(3) If B varies and the other coefficients are fixed, then using the same trick in (2) shows that [1] is the general solution to the differential equation

(3) And if A varies and the other coefficients are fixed, then [1] is the general solution to the differential equation

And, for more fun, you can try varying some of the coefficients and keeping the others fixed. For instance, varying two of the coefficients and keeping the other two fixed will result in a second-order differential equation.

Obviously varying D just shifts the curve up and down, but varying A can also be fairly easily visualized. If A is positive and you make it more positive, or if it’s negative and you make it more negative, you squash the curve sideways, making it narrower. This would be because the curve tends to ±∞ more and more quickly as you increase the absolute value of the leading coefficient.

For B and C, things may be more complicated. I experimented with some cubic curves but my results are not very conclusive – what actually happens may very well depend on the sign of A. For B this could be what might happen:

Increasing the absolute value of B appears to increase the vertical distance between the local maximum and minimum points. In addition: (a) If A > 0 then (i) if B > 0 and you make it more positive, the curve tends to be shifted in a NW direction, and (ii) if B < 0 and you make it more, the curve tends to be shifted in a SE direction; (b) If A < 0 then (i) if B > 0 and you make it more positive, the curve tends to be shifted in a NE direction, and (ii) if B < 0 and you make it more, the curve tends to be shifted in a SW direction.

It contains the subset ℝ which is well known to be uncountable. Is that proof enough?

Given

Then

So yes,

I’m sure the other five relations all work out the same way.

Oops. I was thinking only of “countably infinite” there. Why I was doing that I have no idea.

But yes, cxc001’s problem clearly takes the term countable to mean “finite or countably infinite”.

It depends on which how you define “finite set”. The two most commonly used definitions are:

(1) A set is said to be infinite iff there is a bijection from itself to a proper subset of itself. It is said to be finite iff it is not infinite (i.e. there is no bijection from itself to a proper subset of itself).

From the first definition, we see that ℕ is infinite (e.g. there is a bijection from itself to the set of all even natural numbers). From the second definition, the fact that ℕ is not finite can be proved by induction on n.

Let’s do the proof of the problem using both definitions.

(1) Using this definition, it is more straightforward to prove the contrapositive statement: If |ℕ| ≤ |A| then A is infinite. |ℕ| ≤ |A| means there is an injective function f from ℕ into A. Since ℕ is infinite, so is f(ℕ). And since A has an infinite subset, it must itself be infinite.

(2) For this, we use contradiction and the axiom of choice. It may be that you can do it without the axiom of choice (ask Ricky ) but since the axiom of choice simplifies life a great deal, I don’t see why I shouldn’t be justified in using it.

The fact that there is an injective function from A into ℕ is a simple consequence of the definition of finiteness. Suppose there is a surjective function f from A onto ℕ. By the axiom of choice, there is a bijection f* from a subset A* of A to ℕ (i.e. define the equivalence relation ~ on A by x ~ y iff f(x) = f(y), then select a unique element from each equivalence class to form A* and take f* to be the restriction of f to A*). As A is finite, so is A*. Let g be a bijection from A* to {1,2,…,n} for some natural number n. Then g∘f*[sup]−1[/sup] is a bijection from ℕ to {1,2,…,n}, implying that ℕ is finite – a contradiction.

No, no, it is (√2)r.