/sci/ - Science & Math

>>9421200
>if an integer is divisible by a and by b, is it always divisible by ab?
No.

>>9421200
Hey, brainlet
2|4
4|4
8|4 ?!

>>9421200

No. ab could be too large.

12 is divisible by 3 & 3, but it is not divisible by 9.

12 is divisible by 6 and 3, but it is not divisible by 18.

You can define ab in such a way that it could be made true. Something along the lines of prime factorization theorem.

Why is it that [math] 2|a^2 \Rightarrow 2|a [/math]? It's something to do with 2 being prime.
It's the one thing about the proof of the irrationality of [math]sqrt{2}[/math] that I don't understand.

>>9421213
[math]\sqrt{2}[/math]*

>>9421213
>Why is it that 2|a2⇒2|a?
https://en.wikipedia.org/wiki/Fundamental_theorem_of_arithmetic

>>9421213
if 2 is prime and 2 divides a^2, then in the prime decomposition of a^2 there has to be at least one 2. But given that there is no prime whose square is 2, then there's at least two 2s in the prime decomposition of a^2 (in fact, an even amount). Then in the prime decomposition of a, where you divide all the prime powers in the decomposition of a^2 by 2, there is at least one 2.
qed

>>9421200
the integer is divisible by gcd(a,b) and ab divides lcm(a,b)

>>9421245
>But given that there is no prime whose square is 2
Prove it.

>>9421200
https://en.wikipedia.org/wiki/P-adic_number

a divides c if and only if for all p prime integer, Vp(a) <= Vp(c).

Vp(ab) = Vp(a) + Vp(b)

Therefore no.

as long as ab doesn't equal 0, yes.

Some people in this thread find it difficult to not breathe manually.

What function describes the frequency of [math]Cos(x^2x!)[/math]? I am stuck. I have never tried finding the frequency of a function that wasn't first-order before.

>>9421347
>the frequency
What are you talking about?
Your shit isn't periodic

>>9421203
>No
>4÷8 = NaN
>12÷9 = NaN

>>9421389
>hurr i dont know how to divide
Okay imagine a thing. You cut the thing in two with 1 cut. You now have two things.
1 ÷ 1 = 2

>>9421320
4 is divisible by 4 and by 2, but not by 8.

Think before you insult everyone else. OP's statement is only true if a and b are coprime (a proof of this is left as an exercise to the poster).

>>9421399
b8

one divided into 2 things is 2 things
1÷2 = 2

>>9421403
>4 is not divisible by 8

>>9421406
a * 8 = 4

What integer a makes this equality true?

>>9421200
2 is divisible by 2 and 2, but not by 4...

>>9421411
Nothing implied A and B had to be integers

>>9421418
Back to poorly trolling leddit.
https://www.reddit.com/r/xkcd/comments/7mvjyl/xkcd_1935_2018/drx2z9d/

>>9421267
Let p be a prime with p^2 = 2. Since 1^2 = 1 and 2^2 = 4, then in particular 1<p<2. But there are not integers between 1 and 2. Contradiction

>>9421200
suffices to show one example

24 is divisible by 3, 6, 2 and probs more but not by 3*6=18

>>9421427
>browsing reddit

>>9421200
WLOG, let [math] s,\ a,\ b \in \mathbb{Z_{+}} [/math] with s divisible by a and b, i.e. [math] \exists s_a,\ s_b \in \mathbb{Z} [/math] such that:
[math] s = as_a [/math]
[math] s = bs_b [/math]
[math] s^2 = abs_as_b [/math]
then
[math] \frac{s^2}{s_as_b} = ab [/math]
suppose now that ab divides s, i.e.
[math] \frac{s}{ab} = \frac{s_as_b}{s} \in \mathbb{Z} [/math]
then [math] \exists x,y \in \mathbb{Z} [/math] such that [math] s_b = x\frac{s}{s_a} =
xa [/math], and similarly [math] s_a = yb [/math].
but [math] aby = as_a = s = bs_b = abx [/math] so [math] x = y [/math].

it is easy to see that this is true if and only if x = 1, so s is divisible by the product of two of its divisors if and only if it is the product of those divisors.

>>9421642
2 of those lines (about s^2) are redundant but whatever.

>>9421642
2,3,6,12

>>9421411
dubs

yall are trying way too hard

if x is divisible by a, x = ai
if x is divisible by b, x = bj
when x is divisible by both
x/(a*b) = x/(x/i * x/j) = x/(x^2/ij) = ij/x

so OPs statement is only true if x divides the product of the multiplier things

28 is divisible by 7 and by 14 but 28 doesnt divide 4*2

28 is divisible by 7 and 2 and divides 4*14

>>9421200
>220
>a:5
>b:10
>ab:50
Nope

>>9421403
OP's statement isn't only true for coprime a & b. For example, 8 is divisible by a = 2, b = 4, ab=8.

>>9423173
\thread