What is wrong with the claim that an 8x8 square can be broken into pieces that can be reassembled to form a 5x13 rectangle?

Question

it says

 

"hint: where is the extra square unit"

 

am i supposed to use a proof here? if so is this Fibonacci, generalized Fibonacci, Lucas number?

Answer 1

I am not familiar with number theory as a whole but perhaps this observation will help.

 

The Fibonacci numbers as you know are   1 1 2 3 5 8 13.

 

Notice 8 lies between 5 and 13 which are the dimensions of the proposed rectangle.

 

Basically we start with an 8 x 8 square

 

Cut it into 2 pieces, 5 x 8 and 3 x 8

 

Cut the 3 x 8 piece into 2 pieces, 3 x 5 and 3 x 3

 

Cut the 3 x 3 piece into 2 pieces 2 x 3 and 1 x 3

 

Cut the 1 x 3 piece into 2 pieces 1 x 2 and 1 x 1

 

If we take the larger pieces from each stage, starting from the bottom, we get

 

1 x 2, 2 x 3, 3 x 5, 5 x 8

 

notice the pattern?

 

You assemble the above rectangles to make a total rectangle with dimensions 5 x 13.

 

You get the 5 inch width and the 13 inch length but there is still a piece in the corner missing.

 

But we do have the 1 x 1 piece left over.

 

But despite cutting and cutting and getting closer and closer to the full rectangle, we come up short.

 

But given the Fibonacci sequence in the dimensions of the cut pieces, I suspect you need to use that in some way in your proof.

Answer 2

The area of a rectangle (and a square) is = width times height. So the original square has 8*8 = 64 units of area.

 

A 5 by 13 rectangle requires 5*13=65 units of area, so there is not enough.