Why You Can’t Fold a Piece of Paper 8 Times
Using maths to prove a well-known property of an A4 piece of paper
Many of you may have heard of the phrase “you can’t fold a piece of paper 8 times” and I bet many of you have tried at least once only to be faced with the harsh reality after your 7th fold that you can’t achieve the 8th fold. I found myself thinking of a way to describe why using mathematics.
To do this we first need the dimensions of A4 paper. On average A4 dimensions are roughly 297mm x 210mm x 0.075mm. I will assume each time we fold the paper the length of the longest side of the paper will be affected, so the first fold is along the 297mm side.
Fold 1
Let us look at what happens for the first fold. The thickness doubles to 0.1mm as there are now two layers of paper and the 297mm side is halved. However, the fold itself has a small amount of length too so we must consider this when calculating the new dimensions. For each fold, I will assume the crease makes a semicircle with negligible distance between each layer. Let's see the first fold looking at the crease where we can see each layer of paper.
Let's find the length of the circumference of the semicircle radius 0.05mm which is 0.075π. So now we can find the new dimensions of the paper which are (297–0.075π)/2mm x 210mm x 0.15mm.
Fold 2
Let's look at one more fold then we will see a pattern for the dimensions of each fold.
Now the circumference of the crease is 0.15π and the thickness is 0.3mm, so the new dimensions are (297–0.075π)/2mm x (210–0.15π)/2mm x 0.3mm.
Let's table our dimensions after each fold and we will see the pattern for each dimension.
The thickness is doubling each time which you can understand from the fact you are putting the amount of layers on top of itself. The length and width are becoming smaller. The length of every odd number fold changes by taking away the new crease circumference and halving and the width follows a similar process changing every even fold.
So now let's look at the 6th, 7th and 8th fold.
After 7 folds, the value of the length is 9.95mm (3 sig fig) and the value of the width is 21.9mm (3 sig fig). Now we will perform one more ‘fold’ by continuing the sequence for the length and width.
The length will stay constant and the width is (21.9–9.6π)/2, however, this value is negative as 21.9 < 9.6π, therefore, another fold is impossible to achieve.
Thank you for reading. If you have any thoughts on the proof let me know in the comments
