Obviously if $\frac{n}{3}$ is an integer then $\dfrac{2n}{3}$ is an integer. But what if $\frac{n}{3}$ is not an integer? Can it be proven that $\dfrac{2n}{3}$ is not an integer (or, more specifically, an even multiple of 3) if $\frac{n}{3}$ is not an integer?
I feel that intuitively this must be true, but I am not a mathematician and my idle algebraic musings aren't helping me. Is there a quick, simple proof one way or the other?
[Sorry if this is mistagged.]
Simply notice that $\rm\displaystyle\ \frac{n}3 + \frac{2n}3 = n\in \mathbb Z\ \ $ thus $\rm\displaystyle\ \frac{n}3\in\mathbb Z \iff \frac{2n}3\in \mathbb Z.\,$ This works precisely because $\rm\mathbb Z$ is an additive subgroup of $\rm\:\mathbb Q,\:$ i.e. a subset closed under subtraction. For if $\rm\:S\:$ is a subgroup of a group and $\rm\ a+b\ = s \in S\ $ then $\rm\ a = b-s \in S\!\iff\! a+s = b\in S,\, $ so your property holds. Conversely if the property holds and $\rm\:a,b\in S\ $ then since $\rm\, (a-b)+b = a \in S\, $ the property implies that $\rm\: a-b\in S\:,\: $ so (nonempty) $\rm\:S\:$ is closed under subtraction, so $\rm\:S\:$ is a subgroup, by the subgroup test.
See also the below complementary form of the subgroup property from my prior post.
Theorem $\ $ A nonempty subset $\rm\:S\:$ of abelian group $\rm\:G\:$ comprises a subgroup $\rm\iff\ S\ + \ \bar S\ =\ \bar S\ $ where $\rm\: \bar S\:$ is the complement of $\rm\:S\:$ in $\rm\:G$
Instances of this are ubiquitous in concrete number systems, e.g.
transcendental algebraic * nonalgebraic = nonalgebraic if nonzero rational * irrrational = irrational if nonzero real * nonreal = nonreal if nonzeroeven + odd = odd additive example integer + noninteger = noninteger
if $2n/3$ were an integer, then $n/3 = 4n/3 - n = 2*(2n/3) - n$ would also be an integer.
Contradiction.
Assume by contradiction that 2n/3 is an integer m and deduce from here that 3 divides n.
Hint: think about the prime factorisation of n.
Every integer $n$ can be written in the form $n = 3a + b$ where $a$ is an integer and $b$ is one of $0,1,2$. Now $n/3$ is an integer exactly when $b = 0$. Consider the representation of $2n$. There are three cases.
If $b = 0$ then $2n = 3(2a)$.
If $b = 1$ then $2n = 3(2a) + 2$.
If $b = 2$ then $2n = 3(2a+1) + 1$ since $4 = 3 + 1$.
So $2n/3$ is an integer only if $b = 0$, i.e. only if $n/3$ is an integer. And vice versa.
Another way, is $\frac{2n}{3}$ is an integer iff $3\mid 2n$, but $3$ being prime implies that $3\mid 2$ or $3\mid n$, as the first cannot be, then $3\mid n$, and $3\mid n$ iff $\frac{n}{3}$ in an integer. See Euclid's lemma, also note that there´s no use of the prime factorization.
If (n/3) is not an integer, then n is not divisible by 3. If n is not divisible by 3, then (2*n) is not divisible by 3. Therefore, if (n/3) is not an integer, neither is ((2*n)/3).
If $2n/3$ is an integer it is either odd or even. If it is even then $n/3$ is also an integer. If it is odd then so is $3\cdot (2n/3)=2n$, but this is clearly even.
