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.