Sanj
Shared on Mon, 03/06/2006 - 13:55Sanjs blog, stardate 63... wait a minute. Thought I may as well put my geekiness out there right from the start, no point hidng from it. This is the first entry of my blog and I thought Id start out with some maths. Seems like a bit of a strange choice but let me explain a little bit.
I recently read this book by Simon Singh called Fermats Last Theorem which is basically a true story about this British mathematician who solved one of the oldest problems in mathematics. Its a great read (honestly) and I can highly recommend it if youre into that kind of thing. The thing I liked most about it though was that it went through the history of maths from the Greeks right through to the modern day. He talked about a lot of mathematicians and what their contribution to this problem, and their contribution to maths in general, was. As an engineering student a lot of these guys names were familiar to me (their equations however have long since been forgotten). There is some great stories in this book and some of these mathematicians are truly inspirational.
However, there was one part of the book which stayed with me and I want to share it. It is the proof of Pythagoras Theorem. Now Im sure I learnt this in high school but I didnt remember it. I found it so incredibly simple, yet so powerful at the same time. Maybe its because al ot of the maths I see these days is quite complicated that I had forgotten how simple it can be sometimes, but I thought it was really cool, so here it is.
For those of you that dont remember what Pythagoras Theorem is; it says that for a right angled triangle, the square of the long side (hypoteneuse) is the sum of the square of the other two sides.
A^2+B^2=C^2
Now heres how to prove it. If we take a square and draw four equal right angled triangles inside it (see diagram). Then we can use the basic rules about the area of a triangle to calculate the area of the larger equation.
So the area of the square can be calculated by (a+b)*(a+b) denoted (a+b)^2. We can also calculate it as the smaller square (C^2) plus the area of the four triangles (4*(1/2*a*b) = 2ab
We than get the equation (a+b)^2 = C^2 + 2ab. When we multiply out (a+b)^2 we get a^2 +b^2 +2ab = C^2 +2ab
2ab cancels out on each side of the equation and we are left with A^2 + B^2 = C^2. Thats it, simple yet impossible to refute. Thanks for reading.
Sanj
I recently read this book by Simon Singh called Fermats Last Theorem which is basically a true story about this British mathematician who solved one of the oldest problems in mathematics. Its a great read (honestly) and I can highly recommend it if youre into that kind of thing. The thing I liked most about it though was that it went through the history of maths from the Greeks right through to the modern day. He talked about a lot of mathematicians and what their contribution to this problem, and their contribution to maths in general, was. As an engineering student a lot of these guys names were familiar to me (their equations however have long since been forgotten). There is some great stories in this book and some of these mathematicians are truly inspirational.
However, there was one part of the book which stayed with me and I want to share it. It is the proof of Pythagoras Theorem. Now Im sure I learnt this in high school but I didnt remember it. I found it so incredibly simple, yet so powerful at the same time. Maybe its because al ot of the maths I see these days is quite complicated that I had forgotten how simple it can be sometimes, but I thought it was really cool, so here it is.
For those of you that dont remember what Pythagoras Theorem is; it says that for a right angled triangle, the square of the long side (hypoteneuse) is the sum of the square of the other two sides.
A^2+B^2=C^2
Now heres how to prove it. If we take a square and draw four equal right angled triangles inside it (see diagram). Then we can use the basic rules about the area of a triangle to calculate the area of the larger equation.
So the area of the square can be calculated by (a+b)*(a+b) denoted (a+b)^2. We can also calculate it as the smaller square (C^2) plus the area of the four triangles (4*(1/2*a*b) = 2ab
We than get the equation (a+b)^2 = C^2 + 2ab. When we multiply out (a+b)^2 we get a^2 +b^2 +2ab = C^2 +2ab
2ab cancels out on each side of the equation and we are left with A^2 + B^2 = C^2. Thats it, simple yet impossible to refute. Thanks for reading.
Sanj
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Comments
Submitted by BATMANKM on Tue, 03/07/2006 - 12:06
Submitted by LadyisRed on Thu, 03/16/2006 - 20:19