The h Bar

@BernardMeurer, you can't call me (sorry), but you can vent via the computer to me if you want. As an aside, you should bake cookies and brownies and such - my mom's brownies are literally to die for.

Bernard Meurer

@heather Ask @ACuriousMind for his predictions on what will happen if I try to bake

heather

1:11 AM

@BernardMeurer, just follow the recipe, it's not that hard.

Bernard Meurer

And thank you, but I would traumatize you for life after a rant, ask me again when you're over 16 and I'll rant for you any time :D

@heather I have set my kitchen on fire 3 times

heather

@BernardMeurer, gotcha, I'll wait until I'm 16.

Wait, how does that work?! 3 times? You have to be joking.

Bernard Meurer

I wish I was

heather

How did you do that? I am impressed! =P

user218912

how do I prove that if a number is divisible by 3 and 5 it's also divisible by 15? @0celo7

Bernard Meurer

1:14 AM

It works like this, you put oil in the pan and you go heat it up and you fall asleep, or you try and hack your toaster but you break the timer on it so it sets the bread on fire since you don't remember to go check on it. Things like that

@obe Let's try

ACuriousMind

@BernardMeurer You need to lean how to cook

heather

@obe, hmm. My first thought would be that if you have a number $x$ and $x \mod 3 = 0$ then $x \mod 3n = 0$. Then, you can say that $n = 5$ in this case, so therefore $x \mod 15 = 0$. (sorry, by the way, I don't know the congruence symbol.)

Bernard Meurer

So, $$a,\alpha,\beta\in \mathbb R; \frac{a}{\alpha}\in\mathbb Q, \frac{a}{\beta}\in\mathbb Q \implies a/(\alpha\beta)\in\mathbb Q$$

user218912

@BernardMeurer not $\mathbb R$

heather

@BernardMeurer, oh, that's the approach? Huh, that's interesting.

Bernard Meurer

1:16 AM

@obe yeah, I was on my way to fix it

@heather I dunno, I'm just writing the problem down

heather

@BernardMeurer, oh

user218912

okay but that's not generally true though

Bernard Meurer

@ACuriousMind I know, I know, I think finding myself a girlfriend that loves cooking is easier though

heather

@obe, and then my next thought would be that $x \mod n = x \mod a*b$ where $a$ and $b$ are two factors of $n$ (I think).

Bernard Meurer

@obe Why?

user218912

1:17 AM

is it?

Bernard Meurer

@obe That's what we're trying to prove

heather

@obe, does that make sense, or am I being going the completely wrong direction?

Bernard Meurer

@heather It makes sense, sadly that's not a proof though :P

Well, it's obvious, it's the result of a division it will always be rational

$$\mathbb Q = \{\frac{c}{d}\forall c,d \in \mathbb R\}$$

user218912

what?

user218912

doesn't $c, d$ have to be integers?

Bernard Meurer

1:22 AM

Do they?

user218912

i could be wrong, idk math.

Bernard Meurer

$$\mathbb Q = \left\{{\dfrac p q: p \in \mathbb Z, q \in \mathbb Z_{\ne 0}}\right\}$$

There

That

user218912

wait no they can be rational

user218912

change it

Bernard Meurer

Let's not dwell on this, that definition seems good

$a\alpha = a\alpha^{-1}$

$a\beta=a\beta^{-1}$

user218912

1:27 AM

huh?

Bernard Meurer

Hmm $\mathbb Q$ seems funky

It's not a closed set, definitely not

but it doesn't seem to be open either

user218912

can we just do the specific case for $3, 5$

Bernard Meurer

@obe I'm just thinking out loud, I don't know the answer :)

user218912

well

user218912

@0celo7 help!!!

user218912

1:29 AM

:(

heather

@obe, given: that $x \mod 3 = 0$ and $x \mod 5 = 0$
set two equations equal: $x \mod 3 = x\mod 5$
and then somehow get that equation to $x \mod 15 = 0$. but I don't know how to do that, I'm thinking about it.

what would be the inverse of $\mod 5$?

Bernard Meurer

@heather I'm not sure that will yield anything :/

heather

@BernardMeurer, yeah, but I feel like trying =)

Bernard Meurer

@heather I know, I know, I'm just trying thinking with you

Okay

heather

darn, why is this so hard to prove?

user218912

1:39 AM