I am the beginning of eternity
the end of time and space,
I am the beginning of every end,
and the end of every place.

So I am not sure if I can defend everything in the riddle. I just copied it from a book! I guess that "joyous expectancy" is meant to hint at pregnancy, but actually mean that she is looking forward to cooking with the onion.

That must be an incredible tasting- or smelling- onion :)

<Manic>

Still wondering about #1 and #3 from your post.
#1 Leaving Paris aside, I am having trouble with "a false good feeling"... A bolt of lighntning? A firecracker?

#3 I don't really have any ideas...

I am envisioning a really bad outdoor theater troupe having onions hurled at them. The audience may consider them "serviceable" but the performers wouldn't find the experience so poetic.

Prove that (P-S)^2 is greater than/equal to (4)(S) + 8.

I can prove a weaker result, namely, (P-S)^2 ≥ 4S+5, as follows.

From PS = QR it follows that we can write P = ab, S = cd, Q = ac, R = bd, where a, b, c, and d are positive integers.

From P > Q we have b > c, which means b ≥ c+1, since these are integers, and from P > R we have a > d, which means a ≥ d+1.

Hence (P-S)^2 = (ab-cd)^2 ≥ [(c+1)(d+1) - cd]^2 = (c+d+1)^2 = c^2+2cd+d^2+2c+2d+1 ≥ 4cd+2c+2d+1 ≥ 4cd+5 = 4S+5, since c^2-2cd+d^2 = (c-d)^2 ≥ 0 and 2c+2d ≥ 4.

If I could justify c ≠ d (e.g., S is not a perfect square), this argument would yield the desired result (P-S)^2 ≥ 4S+8, because then I could say 2c+2d ≥ 6 and (c-d)^2 ≥ 1, i.e., c^2+d^2 ≥ 2cd+1, but I don't see any way to do it.

Using the notation of my previous post, suppose d = c.

Then the basic inequalities can be written in the form

P = ab > Q = ac > R = bc > S = c^2, where a, b, and c are positive integers.

It follows that a > b > c, or a ≥ b+1 and b ≥ c+1, which in turn implies a ≥ c+2.

Hence (P-S)^2 ≥ [(c+2)(c+1) - c^2]^2 = (3c+2)^2 = 9c^2+12c+4. This is greater than 4S+8 = 4c^2+8 for all c > 0.29666, hence for all positive integers.

That leaves only the case c ≠ d, which was disposed of earlier.

You have 8 identically shaped objects, however, one of them weights a little more (or less) than the other 7, now, with a balance, can you determine the one that is more/less with just 2 weighings?

you are just asked to find it with two tries (or weighing...) Then again, I may not have remember this puzzle in its entirety...

Another one... 4 people needs to cross a rickety/dangerous bridge. And here is the kicker/puzzle...

You have 1 flash light, it will only last 17 minutes on its battery. And because of the danger of crossing this bridge, you <NEED> the flashlight to be working/on...

Person A can cross the bridge in 1 min.
Person B can cross the bridge in 2 mins.
Person C can cross the bridge in 5 mins.
Person D can cross the briege in 10 mins.

The bridge can at most, hold the weight of 2 people, how do all 4 cross safely?

When I first encountered this puzzle, all kinds of 'things' crossed my mind...

1. Does/can this flashlight need to be passed from one person to another?

2. How far 'ahead' does the light shine? And can the knowledge of this 'rickety/dangerous' bridge's info be passed on?

3. Can I leave it 'securely' tied to the bridge?

4. Can I pass it off to the next person like a track relay?