Obviously if $\frac{n}{3}$ is an integer then $2\dfrac{n}{3}$ is an integer. But what if $\frac{n}{3}$ is not an integer? Can it be proven that $2\dfrac{n}{3}$ 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.]

  • 2
    $\begingroup$ To be precise, you might want to include what $n$ is (probably an integer). $\endgroup$ – wildildildlife Apr 11 '11 at 19:09
  • $\begingroup$ @wildildildlife: Yes, thanks. I mean n to be an integer. $\endgroup$ – Robusto Apr 11 '11 at 19:11
  • $\begingroup$ If $n = \frac 32$, then $\frac{2n}{3}$ is an integer but $\frac n3$ isn't. So $n$ kinda has to be an integer for this problem to make any sense. $\endgroup$ – Joe Z. Feb 15 '13 at 13:47

Simply notice that $\rm\displaystyle\ \frac{n}3\ +\ \frac{2\:n}3\ =\ n\in \mathbb Z\ \ $ therefore $\rm\displaystyle\ \frac{n}3\in\mathbb Z\ \iff\ \frac{2\:n}3\in \mathbb Z$

This is true 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 your 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 $\rm\:S\:$ is closed under subtraction, so $\rm\:S\:$ is a subgroup (or empty).

See also this 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.

     algebraic * nonalgebraic  =  nonalgebraic  if  nonzero 
      rational * irrrational   =   irrational   if  nonzero 
          real *   nonreal     =    nonreal     if  nonzero 

         even  +     odd       =      odd          additive example
       integer + noninteger    =   noninteger
  • $\begingroup$ Thanks for this. Quite the beautiful (i.e. simple) answer I was looking for. Am I correct in interpreting the material after "hence" as meaning "n/3 is a member of the set of integers if and only if 2n/3 is a member of the set of integers"? $\endgroup$ – Robusto Apr 11 '11 at 20:00
  • $\begingroup$ @Rob Yes, that's correct. As I mentioned later, it boils down to the property that the integers are a subset of the rationals closed under subtraction. $\endgroup$ – Bill Dubuque Apr 11 '11 at 20:26

if $2n/3$ were an integer, then $n/3 = 4n/3 - n = 2*(2n/3) - n$ would also be an integer.



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.


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).


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.


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