if an integer is divisible by a and by b, is it always divisible by ab?
If an integer is divisible by a and by b, is it always divisible by ab?
Alexander Perez
Owen Davis
>if an integer is divisible by a and by b, is it always divisible by ab?
No.
Aaron Sullivan
Hey, brainlet
2|4
4|4
8|4 ?!
Nathan Martin
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.
Robert Turner
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.
Aaron Williams
[math]\sqrt{2}[/math]*
Mason Reyes
>Why is it that 2|a2⇒2|a?
en.wikipedia.org
Anthony Phillips
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
Jonathan Martin
the integer is divisible by gcd(a,b) and ab divides lcm(a,b)
Angel Martin
>But given that there is no prime whose square is 2
Prove it.
Jose Miller
en.wikipedia.org
a divides c if and only if for all p prime integer, Vp(a)
Elijah Gray
as long as ab doesn't equal 0, yes.
Some people in this thread find it difficult to not breathe manually.
Ayden Nguyen
>the frequency
What are you talking about?
Your shit isn't periodic
Henry Anderson
>No
>4÷8 = NaN
>12÷9 = NaN
Kayden Butler
>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
Justin Rogers
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).
Jaxson Martinez
b8
one divided into 2 things is 2 things
1÷2 = 2
Nolan Hill
>4 is not divisible by 8
Jayden Jackson
a * 8 = 4
What integer a makes this equality true?
Noah Wood
2 is divisible by 2 and 2, but not by 4...
Charles Morales
Nothing implied A and B had to be integers
Nolan Lee
Back to poorly trolling leddit.
reddit.com
Charles Jenkins
Let p be a prime with p^2 = 2. Since 1^2 = 1 and 2^2 = 4, then in particular 1
Aiden Lewis
suffices to show one example
24 is divisible by 3, 6, 2 and probs more but not by 3*6=18
Jose James
>browsing reddit
Benjamin Gray
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.
Samuel Rivera
2 of those lines (about s^2) are redundant but whatever.
Juan Robinson
2,3,6,12
Jose Thompson
dubs
Juan Turner
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
Christopher Russell
>220
>a:5
>b:10
>ab:50
Nope
Logan Scott
OP's statement isn't only true for coprime a & b. For example, 8 is divisible by a = 2, b = 4, ab=8.
Matthew Young
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