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[ToonyMan]

Let's see what happens tomorrow, shall we?

[Vector]

You disappoint me, Professor Poland.

"Prove that every group of order 91 is cyclic?"

Who the hell do you think I am, anyway?  That isn't very hard at all...

[ein]

You disappoint me, Professor Poland.

"Prove that every group of order 91 is cyclic?"

Who the hell do you think I am, anyway?  That isn't very hard at all...

First thing that came to mind after reading this: "Yours is the TI-84 that will pierce the tangent!"

[Acanthus117]

I have no idea what you just said, but I lol'd.

[JoshuaFH]

What exactly does a proof entail anyway?

[penguinofhonor]

All my dwarves dying of infection and/or underground critter attacks is making me sad.

[ein]

What exactly does a proof entail anyway?

Imagine writing some lines of code using lots of shortcuts.
Now imagine doing it the super long way.
Now add comments.
On every single line.

[Vector]

What exactly does a proof entail anyway?

Imagine writing some lines of code using lots of shortcuts.
Now imagine doing it the super long way.
Now add comments.
On every single line.

No, that's a bad proof.  A proof takes a short statement: "If A, then B" and fills in all the mental trickery with a maximum of clarity and elegance.  There should be no extraneous references or points at which one must make running mental leaps.

Such as the below:

We now show that the equation

(1)

p²=2

is not satisfied by any rational p.  If there were such a p, we could write p = m/n where m and n are integers that are not both even.  Let us assume that this is done.  Then (1) implies

(2)

m² = 2n².

This shows that m² is even.  Hence m is even (if m were odd, m² would be odd), so m² is divisible by 4.  It follows that the right side of (2) is divisible by 4, so that n² is even, which implies that n is even.

The assumption that (1) holds thus leads to the conclusion that both m and n are even, contrary to our choice of m and n.  Hence (1) is impossible for rational p.

This is the extremely spelled-out version of the above proof.  If I were to prove the same thing, I'd say:

Suppose p = m/n, where m, n ∈ N are not both even.  Then m² = 2n², so 2 | m².  Thus 2 | m, so 4 | m².  But then 4 | 2n², so 2 | n.  Contradiction.

[JoshuaFH]

Let me give you another proof Vector:

Mind = Blown

[ein]

So I don't need the overly complicated thing that is harder to read than the Silmarillion?
That will save my life.

[Tack]

Ahem. I shall take the teachings of Vector, and attempt this.


'Every group of order 91 is cyclic. Don't go trying to dispute it, because it's true, so there. You were the one who taught me it in the first place, Poland, so you'd want to have gotten your facts right. Point. Proven.'


I changed my mind. I'm GREAT at maths!

[ein]

Ahem. I shall take the teachings of Vector, and attempt this.


'The guys I support are super awesome and everybody else is wrong. Don't go trying to dispute it, because it's true, so there. You were the one who taught me it in the first place, Poland, so you'd want to have gotten your facts right. Point. Proven.'


I changed my mind. I'm GREAT at politics/religion/any debate topic ever, but not math!

Fixed that for you.

[Tack]

I support YOU, Ein. But professor Poland didn't teach me that.

... You did.



On an actual sad note:

Playing DF (Who woulda guessed?) and found out that, although you CAN set it so you can exit out of things with the space bar now, you can't make an order for a 'rock table', because pressing space makes you get out of the managment screen! 'Oh, Damn. I'll have to change it back to esc'. BUT NO. Seems that you can't bind esc to a key, because if you try, it takes you to the menu.

[Akigagak]

Isn't there a 'reset to default' option?

[Duke 2.0]

Isn't there a 'reset to default' option?

Download it again, place in different folder, copy the relevant text file over. Although I'm sure other keybinding schenanegans can be had.