Is it possible to get 1/3 without dividing by 3?

So I need to divide a rectangle into 3 equals parts, but without fractions. It's one of those old "You have two jars of two sizes and need to get an exact amount of some other size" type problems, except a real world example and I don't think I can do much in the way of pouring back and forth.

Here is what I can do:

I can place the 3 parts into 3 "slots" of a larger set, so long as the larger set can be evenly divided using a non-repeating decimal (1/4, 1/5, 1/10, etc) and then set the size of the pieces to exceed their slots by a percentage greater than 100 so that they overflow, pushing each other along the way, to fit 100% of the total slots.

So if I have 4 slots, I would have to set the size of the pieces to 133.(3)%, so not an option. 5 slots requires 166.(6)%, so that's out.

I'm wondering if there is either a basic equation/function I'm not thinking of to solve this or if it's provably impossible.

If it's any help, there's no reason (other than futility if it's impossible and sanity if requires a significant depth), why I couldn't put slots within slots with the pieces at varying depths. I just can't think of the math to pull if off (if it can be).

You can get $1/3$ by diving by two, but that will take you an infinite amount of time:

$$\frac{1}{2} - \frac{1}{4} + \frac{1}{8} - \frac{1}{{16}} + \frac{1}{{32}} - \frac{1}{{64}} + - \cdots = \frac{1}{3}$$

We start with the unit square (I marked the $1/3$ line), and add $1/2$: We take $1/4$ and add $1/8$ We take $1/16$ and add $1/32$ One more • Unfortunately, infinity is something I'm lacking. Is this the only known wat to pull it off? – Anthony Mar 6 '12 at 7:11
• @Anthony Give me a few minutes, I'll give you a geometric proof of this that might or might not help. – Pedro Tamaroff Mar 6 '12 at 7:27
• So it's basically a geometric way of showing how the number is always approaching 1/3 but is infinitely repeating 3 instead. Reminds me of a math class where they proved that .(9) = 1. Oddly enough, the only reason I have a real world dilemma is because google chrome always rounds down a pixel rather than to the nearest whole integer. Maybe I should sent them your proof. – Anthony Mar 6 '12 at 7:56
• You should really draw those figures to scale; it looks like you've proved that your sum is $2/3$... By the way, a nice geometric proof that $1/4 + 1/16 + 1/64 + \cdots = 1/3$ (which is equivalent to what you're proving) is the second figure on this page. – Rahul Mar 6 '12 at 8:37
• In the spirit of @Rahul's idea of summing $1/4+1/16+\cdots$ we can also divide the rectangle into four equal size congruent rectangles (halve both sides), then keep doing the same for the quadrant in top right. After countably infinitely many steps you have 3 rectangles of each size, and can combine them to form the three equal area regions. – Jyrki Lahtonen Mar 6 '12 at 10:32

What you seem to be able to do is start with the number $1$, divide by numbers of the form $2^a 5^b$ (with $a$ and $b$ non-negative integers), and add any numbers you produce.

Your restricted ability to divide (combined with addition) effectively allows you to multiply by any positive number which can be written as a terminating decimal expression, but not by other numbers. So the operations restrict you to such terminating decimals and you cannot produce $1/3$.

• Most of that went over my head, but if I understand you correctly, you're saying that 1/3 can only be derived from either terminating decimals or division by numbers not in my set? I guess this makes sense insofar as I know even if I had 15 slots and multiplied each 3rd by 500, this would still require setting the slots to be equal first, which would require setting them to a width of 6.(6). I just keep thinking I can invert somehow, but I suppose not without one of the numbers being non-terminating. – Anthony Mar 6 '12 at 7:50

This is found in Euclid...

Take the line segment $AB$ that you want to trisect. On a parallel line $\ell$ , pick two points $U,V$, then with your compass construct $W$ on $\ell$ so that $UV = VW$, then construct $X$ on $\ell$ so that $VW = WX$. Draw lines $UA$ and $XB$, extend them until they meet, say at $Z$. Draw line $VZ$, and let $C$ be the point where it meets segment $AB$. Then $AC$ is one-third of $AB$.

• I like this one. (+1) – Pedro Tamaroff Mar 14 '12 at 0:09

Hint $\$ Sums and differences of reals with terminating decimal expansion remain terminating, which is clear either from radix arithmetic, or by arithmetic of fractions of form $\rm\:n/10^k.$

You can't get 1/3 starting from any whole number and dividing by 2 or 5 a finite number of times, because no step in the process can cause 3 to appear in the denominator. If you can divide by more than just 2 and 5 you might be able to do it -- for example you could divide 2/6.

But you can get arbitrarily close, if that's good enough for you.