Pocket maths: folding halves into thirds
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I have folded a piece of paper in half hundreds of times in my life. And probably so did you. Folding a piece of paper in half is fairly easy: just bend the piece of paper until the corners meet, and then crease. That is it. And with this method one can also fold a piece of paper in $4$, in $8$, etc. We just have to successively divide the sections of the paper in half.
But what if we wanted to fold a piece of paper into thirds, as in the picture above? Some people are good at doing that, but they don't really measure anything: they just do it approximately by looking at the paper and folding where it seems about right. I guess it goes without saying, but mathematicians don't like things to be "about right", they want them right... and even though I wasn't a mathematician, when I was a child I thought that maybe there was a way for me to successively fold different parts of the paper in half, until one of the creases would be the crease at a third of the paper.
I thought of doing that because dividing a piece of paper in half is easy, as everyone knows. So I decided to find a sequence of folds that would allow me to do that! Except that one such sequence does not exist.
I will give you a second to try and prove that for yourself...
The proof is actually quite simple. If such a sequence of folds existed, that would mean that $\frac{1}{3}$ could be written as a finite sum of some fractions of the form $\frac{1}{2^n}$ (can you see why?). We will assume that such a sequence exists, in order to arrive at a proof by contradiction.
Like I already said, if such a sequence exists, then there are some natural numbers $0 < n_1 < n_2 < \cdots < n_k$ such that $$\frac13 = \frac{1}{2^{n_1}} + \frac1{2^{n_2}} + \cdots + \frac1{2^{n_k}}$$ Of course that we can reduce all the fractions on the right hand side to the same denominator, getting $$\frac13 = \frac{2^{n_k-n_1}}{2^{n_k}} + \frac{2^{n_k - n_2}}{2^{n_k}} + \cdots + \frac{1}{2^{n_k}} = \frac{2^{n_k - n_1} + 2^{n_k-n_2} + \cdots + 1}{2^{n_k}}$$ from which one is able to get $$2^{n_k} = 3\times(2^{n_k - n_1} + 2^{n_k-n_2} + \cdots + 1)$$ which is absurd, because $2^{n_k}$ is not divisible by $3$.
And so the dreams of a young boy were shattered...
But what if we wanted to fold a piece of paper into thirds, as in the picture above? Some people are good at doing that, but they don't really measure anything: they just do it approximately by looking at the paper and folding where it seems about right. I guess it goes without saying, but mathematicians don't like things to be "about right", they want them right... and even though I wasn't a mathematician, when I was a child I thought that maybe there was a way for me to successively fold different parts of the paper in half, until one of the creases would be the crease at a third of the paper.
I thought of doing that because dividing a piece of paper in half is easy, as everyone knows. So I decided to find a sequence of folds that would allow me to do that! Except that one such sequence does not exist.
I will give you a second to try and prove that for yourself...
The proof is actually quite simple. If such a sequence of folds existed, that would mean that $\frac{1}{3}$ could be written as a finite sum of some fractions of the form $\frac{1}{2^n}$ (can you see why?). We will assume that such a sequence exists, in order to arrive at a proof by contradiction.
Like I already said, if such a sequence exists, then there are some natural numbers $0 < n_1 < n_2 < \cdots < n_k$ such that $$\frac13 = \frac{1}{2^{n_1}} + \frac1{2^{n_2}} + \cdots + \frac1{2^{n_k}}$$ Of course that we can reduce all the fractions on the right hand side to the same denominator, getting $$\frac13 = \frac{2^{n_k-n_1}}{2^{n_k}} + \frac{2^{n_k - n_2}}{2^{n_k}} + \cdots + \frac{1}{2^{n_k}} = \frac{2^{n_k - n_1} + 2^{n_k-n_2} + \cdots + 1}{2^{n_k}}$$ from which one is able to get $$2^{n_k} = 3\times(2^{n_k - n_1} + 2^{n_k-n_2} + \cdots + 1)$$ which is absurd, because $2^{n_k}$ is not divisible by $3$.
And so the dreams of a young boy were shattered...
- RGS
