Is it possible to form a square using sheets of $A4$ sized paper (without them overlapping)? Any sheet of A series paper has an irrational number as its aspect ratio; $\sqrt2$. My intuition tells me that there is no way to combine these sheets of paper into a square without them overlapping — but I can't find any way on how I would go about proving this. Anybody got any idea?
 A: Here's a more powerful theorem that solves this problem. First, we have to show that the component rectangles have sides parallel to the sides of the entire rectangle.
Lemma: If there is a polygon whose sides are axis-aligned and that can be subdivided into rectangles, then the component rectangles are axis-aligned.
Proof: By induction on the number of component rectangles. The case with one component rectangle is obvious. Now assume there is more than one. There must be at least one 90° angle (in fact, there must be at least four). The only way to cover this angle is with the corner of one of the rectangles, and this rectangle must also be axis-aligned. If you remove that rectangle, you'll be left with one or more smaller polygons, each with only 90° and 270° angles. 
Theorem: Suppose a rectangle can be divided into multiple rectangles, where each rectangle has at least one side of integral length. Then the original rectangle has at least one side of integral length.
Proof: Consider this function from the plane to the complex numbers: $$f(x, y) = e^{2\pi i(x + y)}$$
If you integrate $f$ over an axis-aligned rectangle $[x_1,x_2]\times[y_1,y_2]$, you get $${(e^{2\pi i x_2} - e^{2\pi i x_1})(e^{2\pi i y_2} - e^{2\pi i y_1})}\over{-4\pi^2}$$ This vanishes if and only if at least one of the dimensions is an integer. The integral over the entire rectangle is the sum of the integrals over the components, so the integral over the entire rectangle is zero, so at least one of the sides is an integer. □
Therefore the square has integral side length, and the area is also integral. But the area is also equal to $n\sqrt{2}$, where $n$ is the number of sheets. But this is irrational, a contradiction.
A: Suppose this was possible.
We have used $n$ number of sheets (say, A4) to do this. The area of each sheet is $\sqrt2$ (let us consider each sheet of dimensions $1 \times \sqrt 2 $).
Total area of sheets $= n\sqrt 2$.
Now let $a$ be the number papers whose short side makes a boundary on any particular side, and $b$ be the number of papers whose long side is involved in that particular boundary.
Side length of square $l = a \times 1+b \times \sqrt 2$.
Now equate the area of square to $n\sqrt 2$.
We've $$n\sqrt 2= l^2 =(a+b\sqrt 2)^2$$
Thus, $$\sqrt 2 = \frac{a^2+2b^2}{n-2ab}$$
Which isn't possible, obviously, since $\sqrt 2$ is irrational.
Therefore leading to a contradiction. Hence, our assumption was false. 
Thus, there cannot exist a square consisting of vertically and horizontally placed A4 papers.
A: For completeness' sake, as with much mathematics, what we might assume in theory doesn't always hold in reality; a model can only go so far in telling us what can or cannot happen.
A4 paper has dimensions defined as 210mm by 297mm. These are both divisible by 3, so that the ratio is 70:99.
70 lengths are equivalent to 99 widths. If a grid of sheets are aligned edge to edge and corner to corner, they can be used to make a square without overlaps or gaps.
A: Here is a sketch.
If units are chosen so the sides of the paper rectangle are $1$ and $\sqrt 2$, the side of the square is $a+b\sqrt 2$ and considering a corner we must have $a,b \gt 0$.
Now the area of the square is $(a+b\sqrt 2)^2=a^2+2b^2+2ab\sqrt 2$
The area of an individual rectangle is $\sqrt 2$ and $\sqrt 2$ is irrational.
A: Another generalization. You can't tile a square with a (finite) mix of (virtual) $An$-dimensional sheets. To prove that, use the smallest sheet in the mix to tile the others and invoke any of the other proofs here.
This argument is for the virtual paper (where $A(n+1)$ is what you get by halving $An$) since (as another answer points out) real $An$ paper dimensions are integral numbers of millimeters.
