I am thinking of getting the smashbox thing, but I can't decide on the color. I have dark brown hair and fair (if a bit yellow) skin. My hair is dyed auburn or red.

I can't decide between Auburn or Brunette

HIVEMIND ALERT!

We have an "admin" section of our company's little newsletter and since my boss left, my headquarters counterpart and I are in charge of it. It's where we announce stuff like new hires, benefits info, etc. Can anyone think of cool names for it? I was hoping for a song title or something slightly snarky and fun. Right now "Paperwork Rules" is the lead contender.

Can anyone here outline the proof for why the square root of 8 is irrational? I've written one, I'm just not sure it's right.

Damn. It's been ages since I've seen a proof like that. IIRC, you start by assuming that the square root of 8 is rational, and then eventually you show that this leads to a contradiction.

It'd be fun to see if I could remember / come up with the proof myself, but I'm a bit busy at the moment.

eta: Assume the square root is rational. So you can express the root as a fraction a/b. a/b can be reduced to lowest terms, so assume that it is.

Then a miracle occurs. I forget what you do next....

Finally, you can show that a/b is not in lowest terms, thus the contradiction completes the proof.

Sophia, pick what goes with your brow color, not your hair color. In my case, my brows match my natural hair color, which is a light brown. So taupe works perfectly.

Thanks Kristen!

Can anyone here outline the proof for why the square root of 8 is irrational? I've written one, I'm just not sure it's right.

Emily, is there a reason it's 8, of all numbers? I've only done it with 2.

With 2, you assume that its root is rational, m/n, and that both m and n have no common factors. Therefore, its square is m²/n²=2, so m²=2*n².

Now there are several possible cases for m and n: If both m and n are odd, their squares, both odd as well, can't have a factor of 2 between them, and we get a contradiction. If m is odd and n is even we get the same contradiction.

If m is even then its square is a multiplication of 4. n can't also be even (because if it were, 2 would have been a common factor between m and n and we ruled that out already), so n can only be odd. But a square of an odd number is an odd number and multiplying it by 2 won't give a number that is a multiplication of 4, so we get a contradiction again.

All possibilities led to a contradiction, when we assumed that the square root of 2, m/n, is rational, so it has to be irrational. QED.

With 8 I think there should only be slight correction in the last part, the "m is even" part, right?

[Edit: geekily enough, I think I wrote the whole thing down just so that I could write "QED" at the bottom. We only ever got to write the Hebrew initials of "what was to proove", so I never got to write "QED" on anything. Until now. Thanks, Emily!]

Ah yes, Nilly knows it.

Good times... good times....

Can you just prove that the square root of 2 is irrational, and since the square root of 8 = 2 * square root of 2 that means that the square root of 8 is also irrational?

OK, I bitch about my cats when they get noodgy, obnoxious or pissy, but at least the are under 20 lbs. Pissy Pet

I'm happy with my proof that sqrt(5) and sqrt(6) are irrational. With 8, I've gone with the following:
a/b=sqrt(8)
a=sqrt(8)b
a*a=8b*b=2*4b*b
Now the 4 in there means a*a is divisible by 4, so a is divisible by 2. But the other 2 means that a*a/4 is also divisible by 2, and since it's not a square, a must be divisible by another 2, so a must be divisible by 2*2, so a=4c
4c=sqrt(8)b
16c*c=8b*b
2c*c=b*b so b must be divisible by 2 as well, so a and b weren't respectively prime. I'm just not sure if the part where I explain why a must be divisible by 4 is clear enough.

Then again, she may just mark me down for not doing it in the Greek style (oh god, she can't, can she? Because that's a totally different proof, involving lengths of lines, and... and she didn't teach us that, so no. Whew).