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?
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6$\begingroup$ @DaveCousineau: The $\sqrt{2}$ factor that appears for A0-A8 paper has to do with the silver ratio, preserving aspect ratios, and for manufacturing simplicity related to the mathematics. See here and here. The term "ratio" does not mean rational. In your example, $W$ and $H$ need not be integers; $W/H$ will still be their ratio. $\endgroup$– Eric NaslundNov 26, 2017 at 2:01
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6$\begingroup$ @DaveCousineau : Is your position that no one can cut a square piece of paper along its diagonal? $\endgroup$– Eric TowersNov 26, 2017 at 4:03
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2$\begingroup$ @DaveCousineau, While you cant even measure to see if your physical sheet is $\sqrt{2}$ in length, you can still try (within the tolerances your machinery allows you). To see where the $\sqrt{2}$ comes from read up more about its history... Also, for fun, fold a corner of an A sized sheet at 45deg so that it meets it's adjacent edge, now hold up the resultant hypotenuse to another sheet's longer edge. Look like they're about the same size?? $\endgroup$– Lamar LatrellNov 26, 2017 at 4:08
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2$\begingroup$ (continued) Your argument rest on faulty assumptions about physical nature of molecules and, more importantly, how rational/irrational number ought to behave. For some reason you seems to think that by default if things cannot be proven to be irrational it must be rational. That is absurd and not logical in anyway. Mathematically speaking, the set of rational has measure zero on real line so if we are to take a random length of something (whatever that means), you're more likely (100% to be exact) to find out that the length is of irrational quantity (assuming some fixed unit length). $\endgroup$– BigbearZzzNov 27, 2017 at 20:23
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2$\begingroup$ Putting all that aside and back to the main problem. I could care less about if it's meaningful or not to be able to "measure" the actual length of a paper physically, my main concern is that you defaulted to "since it's absurd for things to be irrational, it must be rational". You could say that it's either rational or irrational and I'd probably agree. You could say that the length can't be determined physically and I'd agree even more. However, for some reason you think that irrational requires infinite precision while rational number do not, this I cannot agree to. $\endgroup$– BigbearZzzNov 27, 2017 at 22:51
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5 Answers
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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.
answered Nov 25, 2017 at 19:27
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5$\begingroup$ Like Mark's answer, you answer doesn't include the proof that there cannot exist a square consisting of vertically and horizontally placed A4 papers. $\endgroup$– user499203Nov 25, 2017 at 19:29
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2$\begingroup$ @ThePirateBay Both solutions are perfectly fine. $\endgroup$ Nov 25, 2017 at 19:42
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3$\begingroup$ This answer is interesting, but it does not appear to be complete. In general, 'a' and 'b' are not necessarily the same for all sides. (For a trivial case, imagine making a square out of 1:2 rectangles. By rotating pairs of rectangles, you could make cases where no two sides of the square have the same number of long/short sides...) $\endgroup$– TLWNov 25, 2017 at 21:38
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10$\begingroup$ @TLW It suffices to consider any one side (or in fact any cross section parallel to any side) $\endgroup$ Nov 25, 2017 at 22:03
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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.
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2$\begingroup$ Just to emphasize the difference between theory and practice. "A4 paper has dimensions defined as 210mm by 297mm. These are both divisible by 3", AS PER STANDARD ISO 216, en.wikipedia.org/wiki/ISO_216#A_series. Otherwise we have $ 2^{1/4}/4 $ and $ (1/(2^{1/4}))/4 $ giving 297.302.... and 210,224...., so ratio is $sqrt2$ $\endgroup$ Nov 25, 2017 at 23:05
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1$\begingroup$ @cgiovanardi What are your “otherwise” numbers based on? What standard exists in the world where A4 is defined as having sides like $2^{1/4}/4$? $\endgroup$ Nov 26, 2017 at 0:55
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2$\begingroup$ I mean that in theory we have an aspect ratio of SQRT(2). But in practice the aspect ratio is 99/70, as @Nij stated in his answer. $\endgroup$ Nov 26, 2017 at 0:59
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4$\begingroup$ For those wondering why ISO 216 uses the ratio $\frac{99}{70}$ to specify A4 paper, note that it is the sixth convergent in the continued fraction expansion of $\sqrt{2}$. $\endgroup$ Nov 26, 2017 at 2:17
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2$\begingroup$ I have reverted an edit by user StephenG, that had edit summary “Expand on explanation of why sqrt(2) is not the correct ratio for A4 paper”. The reason is that the edit was longer than this answer itself; so it's up to the author of this answer to incorporate it if they choose. $\endgroup$ Nov 26, 2017 at 3:42
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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.
answered Nov 25, 2017 at 19:24
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1$\begingroup$ You didn't prove that there cannot exist a square consisting of vertically and horizontally placed A4 papers. $\endgroup$– user499203Nov 25, 2017 at 19:27
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1$\begingroup$ @ThePirateBay I assumed the side of the square has $a$ vertical and $b$ horizontal. I am implicitly using that if $a+b\sqrt 2 = c+d\sqrt 2$ then $a=c$ and $b=d$ which follows from irrationality. Alternatively if there is a square with all orientated the same, I can rotate it and make a square with $a,b\gt 0$ which is all in need, because $a^2+2b^2$ is not an integer multiple of $\sqrt 2$. $\endgroup$ Nov 25, 2017 at 19:31
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$\begingroup$ What about if the sides do not have the same 'a' and 'b'? $\endgroup$– TLWNov 25, 2017 at 21:39
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2$\begingroup$ @TLW It is a square - I can either use what was in my previous comment, or simply that the sides are equal because it is a square. $\endgroup$ Nov 25, 2017 at 21:40
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8$\begingroup$ @TLW But If one side is $a+b\sqrt 2$ so is the other, however it is composed, because it is a square. $\endgroup$ Nov 25, 2017 at 22:01
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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.
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$\begingroup$ I can‘t quite see how you can directly exclude cancellation of nonzero terms in your last step? $\endgroup$– AlexRNov 26, 2017 at 9:52
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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.
answered Nov 26, 2017 at 13:24