Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.]

share|improve this question
To be precise, you might want to include what $n$ is (probably an integer). –  wildildildlife Apr 11 '11 at 19:09
@wildildildlife: Yes, thanks. I mean n to be an integer. –  Robusto Apr 11 '11 at 19:11
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. –  Joe Z. Feb 15 '13 at 13:47

6 Answers 6

up vote 20 down vote accepted

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
share|improve this answer
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"? –  Robusto Apr 11 '11 at 20:00
@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. –  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.

share|improve this answer


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.

share|improve this answer

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.

share|improve this answer

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).

share|improve this answer

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.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.