### 64 = 65??

Guess you had also got a mail with this subject. It partitions a 8X8 square into 4 pieces and re-arranges it in such a way that it forms a rectangle of dimensions 5X13 and try to prove that

8X8 = 5X13

==>64 = 65!!!!!!!!

==>64 = 65!!!!!!!!

This video might help you visualise it. Check out the comments for possible errors or solutions.

This problem was first pointed by one of the mathematician Charles Lutwidge Dodgeson who also penned Alice in the Wonderland with the pen name Lewis Caroll.

Okay! Lemme tell you. That video or the mail you saw assumes that the slopes of all the cut pieces to be the same which in reality is not. They have different slopes across their slant edges. You may try it out on a graph sheet to get confirmed.

Okay I'll post what I did on my graphic editor here.

First of all, we have a 8X8 square. Let us cut them into 4 pieces as seen in the video.

Lets try arranging these four pieces into a 5X13 rectangle. Here is what we get.

You can notice that as the slopes of the slant efges of all the four pieces vary, there is a gap between them. In face it is a parellogram with the area 1 square unit which is exactly the difference between 64 and 65 (8X8 and 5X13).

Let us try another similar problem. Let us take a square with side 5 units and divide it into 4 parts as given before and try it to arrange inside a 3X8 rectangle. If those who didnt knew, they would equate 24 to 25 as done for the previous problem.

Here again, the slopes vary, but now, instead of leaving a gap at the centre, pieces overlap , which exactly explains why we cannot equate 24 to 25. In fact, the area covered by the overlapped pieces equates to 1 square unit.

One can't tweak math so easily, aint it?

If one diggs more into this, they'll find a pattern in the sequence. Lets try to find it.

In the first problem, the lengths considered were these numbers. 3, 5, 8, and 13.

Similarly in the second one, the lengths considered were 2,3,5,8.

didn't you notice that these are fibonacci numbers in order?

Forget the first number in both the examples, we have 8 square = (5 X 13) - 1 ,

Similarly, 5 square = (3 X 8) + 1

We have the fibonacci sequence as

0,1,1,2,3,5,8,13,21,34,55,89..... ==> F0, F1, F2, F3, F4, F5, F6, F7, F8, F9, F10, F11......

Writing the above equations in F form, we have

F6 X F6 = F5 * F7 - 1

F5 X F5 = F4 * F6 + 1

In general, Fibonacci sequences have the property,

Fn X Fn = Fn-1 * Fn+1 + (-1)^n

This problem is just a specific version of the above property of fibonacci numbers.

This is just one small property of the innumerous of fibonacci numbers...

Just came to know about this fact yesterday from a book that I'm reading since a week. Thought of sharing it with you guys.

Interested to know more about such properties, read that book...

Alfred S. Posamentier and Ingmar Lehmann, The (Fabulous) Fibonacci Numbers,

Thank you...

This problem was first pointed by one of the mathematician Charles Lutwidge Dodgeson who also penned Alice in the Wonderland with the pen name Lewis Caroll.

Okay! Lemme tell you. That video or the mail you saw assumes that the slopes of all the cut pieces to be the same which in reality is not. They have different slopes across their slant edges. You may try it out on a graph sheet to get confirmed.

Okay I'll post what I did on my graphic editor here.

First of all, we have a 8X8 square. Let us cut them into 4 pieces as seen in the video.

Lets try arranging these four pieces into a 5X13 rectangle. Here is what we get.

You can notice that as the slopes of the slant efges of all the four pieces vary, there is a gap between them. In face it is a parellogram with the area 1 square unit which is exactly the difference between 64 and 65 (8X8 and 5X13).

Let us try another similar problem. Let us take a square with side 5 units and divide it into 4 parts as given before and try it to arrange inside a 3X8 rectangle. If those who didnt knew, they would equate 24 to 25 as done for the previous problem.

Here again, the slopes vary, but now, instead of leaving a gap at the centre, pieces overlap , which exactly explains why we cannot equate 24 to 25. In fact, the area covered by the overlapped pieces equates to 1 square unit.

One can't tweak math so easily, aint it?

If one diggs more into this, they'll find a pattern in the sequence. Lets try to find it.

In the first problem, the lengths considered were these numbers. 3, 5, 8, and 13.

Similarly in the second one, the lengths considered were 2,3,5,8.

didn't you notice that these are fibonacci numbers in order?

Forget the first number in both the examples, we have 8 square = (5 X 13) - 1 ,

Similarly, 5 square = (3 X 8) + 1

We have the fibonacci sequence as

0,1,1,2,3,5,8,13,21,34,55,89..... ==> F0, F1, F2, F3, F4, F5, F6, F7, F8, F9, F10, F11......

Writing the above equations in F form, we have

F6 X F6 = F5 * F7 - 1

F5 X F5 = F4 * F6 + 1

In general, Fibonacci sequences have the property,

Fn X Fn = Fn-1 * Fn+1 + (-1)^n

This problem is just a specific version of the above property of fibonacci numbers.

This is just one small property of the innumerous of fibonacci numbers...

Just came to know about this fact yesterday from a book that I'm reading since a week. Thought of sharing it with you guys.

Interested to know more about such properties, read that book...

Alfred S. Posamentier and Ingmar Lehmann, The (Fabulous) Fibonacci Numbers,

Thank you...

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