There is an impossible triangle.
a+a=2a, so you may say that it is possible. However, it is impossible.
a^2+b^2=a^2, so b^2=0. The height must be zero. It is impossible.
a+b=c, and b must be a. As I mentioned, it is ridiculous. However, if a+a=c=2a were possible, you would need to put fractal triangles in it.
a*=b*=a/2. c*=c/2=a
These are fractal triangles.
a* would be a/2,a/4,a/8,・・・a/2^x.
c* would be a,a/2,a/4,・・・a/2^x-1.
2a*=(a/2^x)×2=a/2^x-1=c*=a*+a*.
a*=a/2,so a/2+a/2=a.
∴a+a=2a=c
Therefore, an impossible triangle is possible. When x is ∞, a*=a/2^∞ would be close to zero. If you saw the zero point, you could see the line which includes infinite triangles. It must be fractal.
2012年5月28日月曜日
2012年5月13日日曜日
Square
Prime numbers are also expressed as squares like ancient Babylonian. I found the pattern of prime numbers which were based on fractal. At first, any prime numbers are squared to find none prime numbers.
3^2=9
Then you pile a rectangle. 3*2=6. 9+6=15
You continue to pile rectangles. 15+6=21,21+6=27,27+6=33
You can expand prime numbers.
Then I put numbers in each squares and rectangles.
Any prime numbers which are squared are the sum of numbers in rectangles and squares.
For example, 7^2=9+6+6+10+4+14=49.
11^2=7^2+6+6+4+4+4+4+44=121.
Moreover, you can put the Pythagorean Theorem in it.
7^2=5^2+(√24)^2=25+24=49
11^2=7^2+(√72)^2=49+72=121
3^2=9
Then you pile a rectangle. 3*2=6. 9+6=15
You continue to pile rectangles. 15+6=21,21+6=27,27+6=33
You can expand prime numbers.
Then I put numbers in each squares and rectangles.
Any prime numbers which are squared are the sum of numbers in rectangles and squares.
For example, 7^2=9+6+6+10+4+14=49.
11^2=7^2+6+6+4+4+4+4+44=121.
Moreover, you can put the Pythagorean Theorem in it.
7^2=5^2+(√24)^2=25+24=49
11^2=7^2+(√72)^2=49+72=121
2012年5月8日火曜日
Ancient Babylonian mathematics
I have read Hidden Harmonies: The Lives and Times of the Pythagorean Theorem , and I'm interested in Ancient Babylonian mathematics. Ancient Babylonians seem not to have the concept of algebra, but they know how to use geometry. Now we have studied the Pythagorean Theorem, and we use it as it is. However, ancient Babylonians seem to have the almost same stem of it. These are their geometries.
They look like piling squares. You see two of them which are white and blue.
Then, you put dots in each squares.
There are two squares which are purple and blue. You see a dot in the purple square. You also see four dots in blue. However, you distinguish the colors, so there are three dots in blue. It looks like L shape.
Then you see three squares. Therefore, you can add 5 dots in green.
This is the infinite pattern.You would add odd numbers in infinite shape of L. Ancient Babylonians have counted dots.
1+3+5=9=3^2
1+3+5+7=16=4^2
1+3+5+7+9=25=5^2
1+3+5+7+9+11=36=6^2
1+3+5+7+9+11+13=49=7^2
1+3+5+7+9+11+13+15=64=8^2
1+3+5+7+9+11+13+15+17=81=9^2
∞
Now we know the Pythagorean Theorem which is a^2+b^2=c^2. See Ancient Babylonian mathematics.
3^2+4^2=5^2
You can find the same. Of course, you have to search all numbers in Ancient Babylonian mathematics to fit the Pythagorean Theorem because there is no concept about square root in it.
Now we can put algebra into Ancient Babylonian mathematics.
There are three squares which are white, blue, and green. You see also two rectangles.
Therefore, (a+1)^2=a^2+2a+1.
when you put square root in it, you would see the Pythagorean Theorem.
(a+1)^2=a^2+(√2a+1)^2
a must be dots, and 2a+1 is like L shape. You count dots.
In this case, a+1=3, a=2, 2a+1=5.
Therefore, 3^2=2^2+(√5)^2
9=4+5
They look like piling squares. You see two of them which are white and blue.
Then, you put dots in each squares.
There are two squares which are purple and blue. You see a dot in the purple square. You also see four dots in blue. However, you distinguish the colors, so there are three dots in blue. It looks like L shape.
Then you see three squares. Therefore, you can add 5 dots in green.
This is the infinite pattern.You would add odd numbers in infinite shape of L. Ancient Babylonians have counted dots.
1+3+5=9=3^2
1+3+5+7=16=4^2
1+3+5+7+9=25=5^2
1+3+5+7+9+11=36=6^2
1+3+5+7+9+11+13=49=7^2
1+3+5+7+9+11+13+15=64=8^2
1+3+5+7+9+11+13+15+17=81=9^2
∞
Now we know the Pythagorean Theorem which is a^2+b^2=c^2. See Ancient Babylonian mathematics.
3^2+4^2=5^2
You can find the same. Of course, you have to search all numbers in Ancient Babylonian mathematics to fit the Pythagorean Theorem because there is no concept about square root in it.
Now we can put algebra into Ancient Babylonian mathematics.
There are three squares which are white, blue, and green. You see also two rectangles.
Therefore, (a+1)^2=a^2+2a+1.
when you put square root in it, you would see the Pythagorean Theorem.
(a+1)^2=a^2+(√2a+1)^2
a must be dots, and 2a+1 is like L shape. You count dots.
In this case, a+1=3, a=2, 2a+1=5.
Therefore, 3^2=2^2+(√5)^2
9=4+5
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