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I came across a puzzle where we need to determine the error in the following proof.

False Theorem. $420 > 422$

Proof. We will demonstrate this fact geometrically. We begin with a $20 \times 21$ rectangle which has area $420$. Now we cut along the diagonal and slide the upper piece parallel to the cut until it has moved exactly $2$ units leftward. This leaves a couple stray corners which are $2$ units wide and just over $2$ units high. Finally, we snip off the two corners and place them together to form an additional small rectangle. Now we have two rectangles, a large one with area just over $(20 + 2) \times 19 = 418$ and a small one with area just over $2 \times 2 = 4$. Thus, the total area of the resulting figure is a bit over $418 + 4 = 422$. By conservation of area, $420$ is equal to just a little bit more than $422$.

Can someone please explain what the error is in the above argument ?

Thank You!

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    See also the famous Fibonacci-based dissection paradoxes, e.g. [see here.](http://www.mathematik.uni-bielefeld.de/~sillke/PUZZLES/jigsaw-paradox.html)2012-04-26

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From your description, you’ve started with the rectangle oriented so that the $21$ unit side is vertical. You’ve slid the upper triangular half $2$ units to the left, so that it rises a little more than $2$ units, and you’ve snipped off the triangular projections at upper left and lower right. Your large rectangle loses two units of width, so its area is a bit more than $(21+2)\cdot 18=414$. The area of the small rectangle is a bit over $4$, so altogether you’ve a bit more than $418$, which is no contradiction.

In fact the small rectangle is $2\times 2.1$ units, so its area is $4.2$, and the large rectangle is $23.1\times 18$ units, so its area is $415.8$, and the total is still $420$, as it should be.

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    I was wondering why my comment earlier didn't inspire the "ahah moment"... until I looked back and saw that I forgot to say *which side* is less than$2$inches!2012-04-26