I am in a weird mood, so I feel like making this topic. In this topic, post math questions that are flawed in some way. (In other words, there should be some problem with the question itself.) Feel free to post questions about the properties of the empty set's members, things that are ill defined (in other words, not defined at all), questions that make absolutely no sense (like having fractional numbers of people), and other silliness.

Of course, if you feel like answering (for some definition of "answering") any of the questions posted in this thread, feel free to do so.

Anyway, I'll start:

Let p and q be even primes, with q > p. Prove that q - p is even.

If I have one, and she has one, and I give her one, do I still have one?

depends on whether he is one or not

I know everyone knows it, but I still like the proof that 1 = 2. Start by taking two values which are equal to each other, A and B.

B = A then multiply both sides by A
AB = A^2 then subtract B squared from both sides
AB - B^2 = A^2 - B^2 Reduce the expression
B(A-B) = (A+B) (A-B) Cancel out the like elements
B = A+B Since A equals B, restate
B = 2B Divide by B
1 = 2

A big "wow" moment for me was the time I learned that 2 + 2 does not equal 4, but actually 2 + 2 = 3.99999999999.

Technically, you could solve the problem simply by not using the fact that p and q are prime. Of course, you would be proving something about nothing, as there is only 1 even prime.

(Incidentally, if we replaced "even" with "odd" in the problem, then we end up with a true (and easily provable) statement about numbers that actually *do* exist.)

Well, you could disregard the q>p. Then both of them can only be 2 and in fact can only be 2, with 2 being the oddest prime, because it's the only one that's not odd. That makes p-q automatically 0, which is even (by all common definitions).

Person X owns 207 games on GOG, he has played 82 of them. Every weekly sale X ends up buying another 2 games and every holiday sale ends up buying another 10 games so he has spent enough to get the 3 games they give as a reward to those who spend a certain much in the sale. X completes Point and Clicks in 3 days, action games in 7 days and RPG's in 28 days, strategy games remain unplayed forever as X can't convince his small circle of friends to play with him. Twice a year X goes into a mood, declaring that all new games suck, and plays a rpg he's already beaten 5 times before to get out of said mood. Assuming that X starts a game, gets half way through, switches to a different game, gets half way through THAT game before going back to the first one to finally complete it, how many unplayed games will there be in his backlog in 5 years time?

If hers was different from yours, you now probably have two, same as she.

I happen to like a proof of this fact that uses mathematical induction.

Prove that, for every pair of positive integers x and y, that x = y.

Base case: Suppose x, y <= 1. then:
x = 1
y = 1
Therefore x = y.

Now, suppose we've shown this to be true for x, y <= n -1. If x, y <= n:
Then x - 1, y - 1 <= n - 1
Therefore, x - 1 = y - 1
Hence, x = y
QED.

(From this theorem, it's clear that 1 = 2.) Fun fact: Many mathematical theorems start with "let p be an odd prime".

Ah that's a good one :) I'm going to have to remember that. It's a good exercise for students to find the error in the proof.

Why do we disregard negative integers? 2>-2.

"Billy, if you had 9 apples and Suzie wanted three, how many would you have left?"

"Nine."

"But you gave Suzie three apples."

"I ain't giving Suzie any of my apples!"

Yes.