Math Is Fun Forum

Are a, b, x and y rational numbers? If they are, we can use the comparison method.

a is the sum of the terms involving even powers of y√2001 and b√2001 is the sum of the terms involving odd powers of y√2001 in the binomial expansion of (x+y√2001)[sup]2000[/sup], i.e.

Note that a is the sum of products of even powers and positive coefficients; ∴ a ≥ 0.

CASE 1: b ≤ 0

Then a ≥ 0 ≥ 44b.

CASE 2: b > 0

Note also that

Je ne suis pas une française mais je crois écrire le français assez bien.

You mean

Then

Well, I’m only guessing anyway. I’m guessing that “face” cards refer to jacks, queens and kings (though I’ve never heard them described that way).

Actually, soha said “red face cards”. I thought that meant red cards with faces on them – i.e. the jacks, kings and queens of hearts and diamonds?

The probabilities would then be

(a) 2⁄46 = 1⁄23
(b) 20⁄46 = 10⁄23
(c) 13⁄46

Erm, there seems to be a slight mistake with

I think you mean

Yeah?

I just saw this thread for the first time while browsing around. So let me add my comments.

The way to go about proving this is through the axioms for the real numbers. True, the real numbers can be built up from the rational numbers, which can be built up from the integers, which in turn can be built up from the natural numbers by the Peano axioms. (And you can go back even further by building the Peano axioms from set-theoretic axioms, particularly the axiom of infinity.) However, in an elementary textbook on analysis, one could bypass the complications and just start by axiomatizing the real numbers as a complete ordered field. (In other words, one merely takes the set of real numbers as “given”, defined by set of axioms which together uniquely characterize it.) The way to prove that a > 0 ⇒ a[sup]−1[/sup] > 0 is then through the axioms for an ordered field.

First, note that (−x)y = x(−y) = −(xy). These can be easily proved using the distributive axiom. We then use the following axioms for an ordered field: (i) If 0 ≤ x and 0 ≤ y, then 0 ≤ xy. (ii) If x ≤ y, then x+z ≤ y+z.

Assume to the contrary that 0 < a but a[sup]−1[/sup] ≤ 0, i.e. 0 ≤ −a[sup]−1[/sup]). Then 0 ≤ a·(−a[sup]−1[/sup]) = −1. But 0 ≤ −1 and 0 < a ⇒ 0 ≤ (−1)a = −a ⇒ a ≤ 0. This contradicts the fact that 0 < a. Hence it can’t be true that a[sup]−1[/sup] ≤ 0; ∴ a[sup]−1[/sup] > 0. (Note: An ordered field is a totally ordered set and so the law of trichotomy for totally ordered sets applies: Given any x. y, exactly one of the following holds: x < y, x = y, x > y.)

I think I see where the flaw in the second part of my proof lies. It was in going from

to

I had assumed that y[sub]1[/sub]⁄y[sub]2[/sub] was continuous. This might not be true if y[sub]2[/sub] = 0 in some parts of [a,b]. (Note: y[sub]2[/sub] was assumed to be not identically zero in [a,b] but it could be zero for some values in [a,b].)

For example, suppose y[sub]1[/sub] = x and y[sub]2[/sub] = |x|, and [a,b] = [−1,1]. Then y[sub]1[/sub]⁄y[sub]2[/sub] = 1 if x > 0 and y[sub]1[/sub]⁄y[sub]2[/sub] = −1 if x < 0, so

but

This should be where the fact that y[sub]1[/sub]⁄y[sub]2[/sub] are solutions to the second-order homogeneous linear differential equation comes in. Can anybody show that if y[sub]1[/sub]⁄y[sub]2[/sub] are solutions to such a differential equation, then y[sub]1[/sub]⁄y[sub]2[/sub] would be continuous? I’m afraid I don’t have a clue myself.

I think (b) is false:

Nondecreasing is not the same as increasing.

And (c) would be true if f is continuous.

I don’t hate you – just
Your spam, which you spread like dust!
Now stop this, you must!

More of lightning’s spam!
He just won’t stop, will he? Da­mn!
I’m so mad, I am!