Me: If any one religion was true, there wouldn't be any other religions.
Me: Do you back up your computer data regularly?
He: I did it an year back.
Me: Same here!
I don't want Congress to come to power. More so because RaGa may become the PM. AAP can't take majority seats by itself. They are most likely here to prove a point by defeating some highly corrupt politicians. If they get a significant chuck, they may have to make a coalition Government with Congress. Fear of another Congress rule. Coalition Governments are never good. They don't allow development work to happen and more chances of re-election, which is an economical burden to the country. BJP is any day better than Congress and NaMo is a great PM candidate because of his administrative style. So, we should vote for BJP this time. I don't mean to say BJP is corrupt-free. I also fear some Hindu religious enthusiasts may hijack the Government's agenda and affect the development work.
PS: It's just my opinion as of today. I couldn't follow much about politics this time. So, my opinion could be too uninformed!
If you add up the first n numbers and square the result, this produces the same answer as adding the first n cubes. The picture gives a wordless proof of this result.
The picture shows the case n=5. In this case, the sum of the first 5 natural numbers is 1+2+3+4+5=15, which squares to 15x15=225. On the other hand, the sum of the first n cubes is (1x1x1)+(2x2x2)+(3x3x3)+(4x4x4)+(5x5x5)=1+8+27+64+125, which also adds up to 225. There is nothing special about the number 5 here: an analogous identity holds for any other positive integer, and it can be illustrated by a similar picture.
So how exactly does the picture illustrate the equation? Well, the big square is a grid whose area is 15x15=225, where our unit of length is the side length of the smallest box appearing in the picture. The vertical strip to the right of the square and the horizontal strip below the square are each 1+2+3+4+5=15 boxes long. It follows that the total number of boxes in the big square is (1+2+3+4+5)^2.
On the other hand, consider the area of the red part of the big square. There are 5 red squares, each of which measures 5x5. The total red area is thus 5x5x5=125, or 5 cubed. Something similar happens for the dark blue squares, provided that the two half-squares on the left edge and top edge are combined to form one big 4x4 square. We then have 4 dark blue squares, each of which measures 4x4, giving a total dark blue area of 4x4x4=64, or 4 cubed. Adding up the areas of each colour gives a total area of 1+8+27+64+125, which is the sum of the first five cubes.
We have counted the integer-valued area of the big square in two ways. Both methods must produce the same answer, from which it follows that the equation must hold. This technique of counting a quantity in two different ways and equating the two results is what mathematicians call a combinatorial proof. The “sum of cubes” identity can be proved in a fairly mechanical way using the standard technique of mathematical induction, but that proof does not give any intuition for why the result should be true. A combinatorial proof, such as the one in the picture, gives a better idea of why the result should be true, and is therefore more satisfying to a mathematician than the other proof.
One might expect, given that there are cubes of integers appearing on one side of the equation, that a combinatorial proof of the result would involve summing volumes, but remarkably, it reduces to summing areas.
Combinatorial proofs are also known as counting arguments. Here's another post by me in which I explain, using a counting argument, why it is not possible for everybody on Google+ to have more than 5,000 followers (https://plus.google.com/101584889282878921052/posts/YV7j9LRqKsX). That post provoked some surprisingly negative reactions.
The picture is based on a picture from Brian R. Sears' website (http://users.tru.eastlink.ca/~brsears/math/oldprob.htm#p32). I found the picture in a recent paper by Johann Cigler (http://arxiv.org/abs/1403.6609) which examines some generalizations of the sum of cubes result.
- Pragathi InfoTechCo-Founder & Engineering Manager, 2013 - present
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