# Why there aren't any squares of 2 divisible by 3?

A friend of mine recently told me that it is not possible to perfectly divide a cake in three pieces because 1/3 is an repeating decimal. Now, this is clearly a silly statement as 0.33333... is an repeating decimal but it is a real number nonetheless. My friend is right in the sense that is not possible to divide a cake in any "perfect" way, e.g. in two halves; more precisely, it is not possible to measure anything "perfectly" so that you will never know if you division was actually perfectly accurate.

I wanted to prove my friend wrong using the following argument: let's imagine you are right, you cannot divide a cake in three pieces but you can divide it in two pieces. Then you can divide those two pieces in two pieces and so on. At one point, you are going to have x pieces of cakes where x is divisible by three. Thus, your original statement must be false. While I was thinking this, however, I noticed that there is no number $2^n$ that can be divided by three without a remainder.

Has this been proved (I guess so)? Is there any intuitive explanation?

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1/3 is not irrational at all! – Ferra Feb 22 '14 at 0:12
This explains why $2^n$ is not divisible by $3$. – John Habert Feb 22 '14 at 0:13
Sorry, oversights. I corrected everything. – Secret Parrot Feb 22 '14 at 0:16
Clarification: The set of rational numbers is $\mathbb{Q}=\left\{\frac{a}{b}: a,b\in\mathbb{Z}, b\neq 0\right\}$. – Zilliput Feb 22 '14 at 0:16
Thank you for the clarification. I corrected the post, as I intended "repeating decimal" and not "irrational". Now it should make sense. – Secret Parrot Feb 22 '14 at 0:17

To divide a cake equally among three persons, using only divisions in half:

1. Divide the cake in half twice, producing 4 equal pieces.
2. Give one equal piece to each person.
3. Using the same method, divide the remaining piece equally among the three persons.
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Truly a parsimonious solution. – Kevin Carlson Feb 22 '14 at 0:28
I am afraid I do not get this. You divide your cake in four pieces, fine. You give one pieces to each person, then you divide the remaining piece into four. Now, if you give one of these sub-pieces to each person, you remain with one sub-piece. Am I missing something? – Secret Parrot Feb 22 '14 at 9:20
The method has three steps. When you divide the leftover piece, using the same method, you have to use all three steps. Your description above stopped after step 2. – MJD Feb 24 '14 at 14:07

$2^n/3=2\cdot2\cdot\ldots\cdot2/3$. Proof by induction:

Let $n=1$. $2/3$ is just $0$ with remainder $2$. Hence, there exists a remainder for $2/3$.

How about $2^{n+1}/3$? Using the previous proof that $2/3$ has a remainder, and expanding $2^{n+1}/3$ as $2^n\left(2/3\right)$ shows that $2^{n+1}/3$ has a remainder. Therefore, $\forall n, 2^n/3$ has a remainder.

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According to the Fundamental theorem of arithmetic, every integer greater than $1$ has a unique prime factorization. Since $2$ and $3$ are prime, $2^n$ ($n>0$) is already the prime factorization, and does not have $3$ as a factor. Hence it is not divisible by $3$.

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$1.$ If people would only be able to split things into tenths, then it would be indeed impossible for them to cut a pie into three pieces by a finite number of steps. But obviously this is not the case.

$2.$ Halving the whole means doubling the number of parts. Obviously, since $3$ is an odd number, it cannot be obtained by doubling any integer.

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Six in an even number; if you double it, you get twelve. Both are divisible by three. – Secret Parrot Feb 22 '14 at 9:15
The doubling does not cause that property. (It keeps it, but it does not create it). – Lucian Feb 22 '14 at 11:14