Difference between revisions of "Math"

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==Combinatorics==
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The way to think of it is this. There are 4*3*2*1 permutations of ABCD because there are 4 ways to do the first item, then for each of those 4 ways there are 3 remaining ways to do the second item, and for each of those there are 2 remaining ways to do the third item, which leaves only 1 way to do the last item.
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If you are only picking a permutation of 3 from the 8 letters ABCDEFGH, then there are 8 ways to pick the first letter, and for each of those there are 7 ways to pick the second, and 6 to pick the third. So the number is 8*7*6.  This is easier than to think of 8!/5!.
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If you are picking a combination of 4 letters, there is only 1, despite the 24 permutations. But if you are picking a combination of 3 from the 8 letters, you start by realizing there are 8*7*6 permutations, but the number of combinations will be smaller. Each  combination consists of 3 letters, and those 3 letters have 3*2*1 permutations. Thus, we must divide the number of permutations by 3*2*1 to get the number of combinations. So the number of combinations is 8*7*6/(3*2*1).
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==Equality==
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*[https://twitter.com/SC_Griffith/status/1528140634422394880 Twitter opn what it means to be equal] (2022).
  
 
==Miscellaneous==
 
==Miscellaneous==
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*See also [[Statistics]].
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*[https://twitter.com/wtgowers/status/1456558355225485317 Gower Twitter thread] on thinking you've found a great result, but you made a mistake (2021).
 
*[https://twitter.com/wtgowers/status/1456558355225485317 Gower Twitter thread] on thinking you've found a great result, but you made a mistake (2021).
  
 
*[https://www.rasmusen.org/rasmapedia/index.php?title=Cedars_Math My 7th Grade Cedars Math Page]
 
*[https://www.rasmusen.org/rasmapedia/index.php?title=Cedars_Math My 7th Grade Cedars Math Page]
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*[https://eml.berkeley.edu/~cshannon/e204_21.html Shannon's math camp notes] on analysis, etc.
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----
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==Erdos Numbers==
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*My Erdos Number is 5, from
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:*Rasmusen-Connell-Farb-Lubotzky-Alon-Erdos, 
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:*Rasmusen-Janssen-Sierksma-Doignon-Fishburn-Erdos 
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:*Rasmusen-Ayres-Rowat-Beardon-Lehner-Erdos
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----
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==Geometry==
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[https://mathcs.clarku.edu/~djoyce/java/elements/elements.html Joyce] has an amazingly good website on Euclid's Elements.
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----
  
 
==Math Education==
 
==Math Education==
 
*[https://math.stackexchange.com/questions/4307332/is-synthetic-division-ever-useful-outside-of-a-low-level-algebra-course?noredirect=1#comment8974303_4307332 My StackExchange question on synthetic division] and whether it is every useful for anything.
 
*[https://math.stackexchange.com/questions/4307332/is-synthetic-division-ever-useful-outside-of-a-low-level-algebra-course?noredirect=1#comment8974303_4307332 My StackExchange question on synthetic division] and whether it is every useful for anything.
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==Irrational Numbers==
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*I read that Kronecker opposed the use of irrational numbers because they can't be constructed from integers by a finite number of steps. But you can  construct pi as the ratio not only as an infinite series, but as the ratio of circumference to diameter. What's wrong with that?
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*If you don't use infinity, you can't derive the area of a circle, can you? Or the length of a curve?
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*Not using proof by contradiction is separate from not using infinity in proofs, right?
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*'''Theorem: The square root of 2 is irrational.'''
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:'''Proof:''' Suppose not.  If the square root of 2 is rational, Sqrt(2) = a/b for some integers a and b. First, note that this implies that Sqrt(2) = x/y for some integers x and y that are not both even. That's because if  Sqrt(2) = a/b  and both a and b are even, then it is also true that a and b are divisible by 2, so Sqrt(2) = .5a/.5b with both .5a and .5b being integers.  If .5a and .5b are both even, we can repeat the division by 2, and we can keep doing this till either the numerator or denominator is odd (or both).
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:So suppose Sqrt(2) = x/y, where x and y are not both even. We will show that this implies that both x and y are even, a contradiction.
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:First, square both sides so 2 = x^2/y^2.  Then 2y^2 = x^2.  This implies that x^2 is even. But then x must be even too,  since if x is odd,  x(x-1) is even because it multiplies an odd number (x) times an even one (x-1), and  so x(x-1)+x is odd because it adds an even number (x(x-1)) to an odd number (x).
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:If x is even, we can write it as x = 2m for some number m. But then  2y^2 = (2m)^2 = 4m^2, so y^2 = 2m^2. But then y^2 is even, so by our earlier argument, y is even too. But that means both x and y are even, which contradicts our starting assumption. So it must be that Sqrt(2) != a/b for any integers a and b. QED.
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*The Real Numbers are Uncountable. Cantor proved this. His Diagonalization Argument is one way. The [https://en.wikipedia.org/wiki/Cantor%27s_diagonal_argument#:~:text=In%20set%20theory%2C%20Cantor's%20diagonal,cannot%20be%20put%20into%20one%2D Wikipedia article on this] is good. The basic idea is that you can convert all irrational number into  a list of  decimals if they're countable, and then contradict this by constructing a new decimal that is different from the first one on your list in the first decimal place, the second one in the second decimal place, the third one in the third decimal place, etc.
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----
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==Line Integrals==
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I should set up a latex file in a rasmapedia directory to explain line integrals.
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They are really "curve integrals".
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1. You want to integrate x^2 over  all points between 2 and 6.
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2. You want to integrate x^2 + y^2 over all points between y=0, x in [2,6].
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3. You want to integrate  x^2 + y^2 over all points on the straight line between (2,2) and (6,6)  (not intervals now--- vectors).
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4. You want to integrate x^2 + y^2 over all points on the curve  y = x^3 between (2,8) and 3,27).  This is the real stuff.
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5.  You want to integrate  x^2 + y^2 over all points in the area bounded by  (0,0) and (0,2) and (2,0) and (2,2).
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6. You want to integrate  x^2 + y^2 over all points in the area bounded by  y = x^3 and y  = log x (or some two curves that cross).
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==Numbers==
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*[https://en.wikipedia.org/wiki/Cistercian_numerals#:~:text=The%20medieval%20Cistercian%20numerals,%20or,were%20introduced%20to%20northwestern%20Europe Cistercian Numbers, 13th Century]:"The medieval Cistercian numerals, or "ciphers" in nineteenth-century parlance, were developed by the Cistercian monastic order in the early thirteenth century at about the time that Arabic numerals were introduced to northwestern Europe. They are more compact than Arabic or Roman numerals, with a single glyph able to indicate any integer from 1 to 9,999."
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==Proofs==
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*See [https://twitter.com/taz_chu/status/1763892059146948823 "A One-Sentence Proof . . .] (2024).
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==Slide Rules==
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{{Quotation|December 6, 2021: In late 2021, Russian video appeared that showed a heavy bomber in flight with one of the crew using s slide rule, apparently to calculate course and/or fuel consumption rate. Most pilot training, especially for the crews of long-range aircraft, includes instruction on how to use special slide rules for such calculations in the event of problems with the electronic navigation and flight management instruments that do this automatically. For generations, ever since long-range flight became possible, manual tools were used for these calculations, along with a special bubble sextant to obtain the location of an aircraft. On the surface the original sextant is used for this but in the air, there is no fixed horizon to base these calculations. The bubble-sextant creates an artificial horizon that enables aerial navigation that shows position within ten kilometers or less. Surface sextant navigation is even more accurate and electronic navigation, especially using GPS, is accurate enough for landing aircraft.
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Media tends to refer to EMP (Electromagnetic Pulse) created by a distant nuclear weapon detonation as the reason for these manual backups. }}
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Latest revision as of 12:19, 2 March 2024

Combinatorics

The way to think of it is this. There are 4*3*2*1 permutations of ABCD because there are 4 ways to do the first item, then for each of those 4 ways there are 3 remaining ways to do the second item, and for each of those there are 2 remaining ways to do the third item, which leaves only 1 way to do the last item.

If you are only picking a permutation of 3 from the 8 letters ABCDEFGH, then there are 8 ways to pick the first letter, and for each of those there are 7 ways to pick the second, and 6 to pick the third. So the number is 8*7*6. This is easier than to think of 8!/5!.

If you are picking a combination of 4 letters, there is only 1, despite the 24 permutations. But if you are picking a combination of 3 from the 8 letters, you start by realizing there are 8*7*6 permutations, but the number of combinations will be smaller. Each combination consists of 3 letters, and those 3 letters have 3*2*1 permutations. Thus, we must divide the number of permutations by 3*2*1 to get the number of combinations. So the number of combinations is 8*7*6/(3*2*1).


Equality

Miscellaneous


Erdos Numbers

  • My Erdos Number is 5, from
  • Rasmusen-Connell-Farb-Lubotzky-Alon-Erdos,
  • Rasmusen-Janssen-Sierksma-Doignon-Fishburn-Erdos
  • Rasmusen-Ayres-Rowat-Beardon-Lehner-Erdos

Geometry

Joyce has an amazingly good website on Euclid's Elements.


Math Education


Irrational Numbers

  • I read that Kronecker opposed the use of irrational numbers because they can't be constructed from integers by a finite number of steps. But you can construct pi as the ratio not only as an infinite series, but as the ratio of circumference to diameter. What's wrong with that?
  • If you don't use infinity, you can't derive the area of a circle, can you? Or the length of a curve?
  • Not using proof by contradiction is separate from not using infinity in proofs, right?
  • Theorem: The square root of 2 is irrational.
Proof: Suppose not. If the square root of 2 is rational, Sqrt(2) = a/b for some integers a and b. First, note that this implies that Sqrt(2) = x/y for some integers x and y that are not both even. That's because if Sqrt(2) = a/b and both a and b are even, then it is also true that a and b are divisible by 2, so Sqrt(2) = .5a/.5b with both .5a and .5b being integers. If .5a and .5b are both even, we can repeat the division by 2, and we can keep doing this till either the numerator or denominator is odd (or both).
So suppose Sqrt(2) = x/y, where x and y are not both even. We will show that this implies that both x and y are even, a contradiction.
First, square both sides so 2 = x^2/y^2. Then 2y^2 = x^2. This implies that x^2 is even. But then x must be even too, since if x is odd, x(x-1) is even because it multiplies an odd number (x) times an even one (x-1), and so x(x-1)+x is odd because it adds an even number (x(x-1)) to an odd number (x).
If x is even, we can write it as x = 2m for some number m. But then 2y^2 = (2m)^2 = 4m^2, so y^2 = 2m^2. But then y^2 is even, so by our earlier argument, y is even too. But that means both x and y are even, which contradicts our starting assumption. So it must be that Sqrt(2) != a/b for any integers a and b. QED.
  • The Real Numbers are Uncountable. Cantor proved this. His Diagonalization Argument is one way. The Wikipedia article on this is good. The basic idea is that you can convert all irrational number into a list of decimals if they're countable, and then contradict this by constructing a new decimal that is different from the first one on your list in the first decimal place, the second one in the second decimal place, the third one in the third decimal place, etc.



Line Integrals

I should set up a latex file in a rasmapedia directory to explain line integrals.

They are really "curve integrals".

1. You want to integrate x^2 over all points between 2 and 6.

2. You want to integrate x^2 + y^2 over all points between y=0, x in [2,6].

3. You want to integrate x^2 + y^2 over all points on the straight line between (2,2) and (6,6) (not intervals now--- vectors).

4. You want to integrate x^2 + y^2 over all points on the curve y = x^3 between (2,8) and 3,27). This is the real stuff.

5. You want to integrate x^2 + y^2 over all points in the area bounded by (0,0) and (0,2) and (2,0) and (2,2).

6. You want to integrate x^2 + y^2 over all points in the area bounded by y = x^3 and y = log x (or some two curves that cross).


Numbers

  • Cistercian Numbers, 13th Century:"The medieval Cistercian numerals, or "ciphers" in nineteenth-century parlance, were developed by the Cistercian monastic order in the early thirteenth century at about the time that Arabic numerals were introduced to northwestern Europe. They are more compact than Arabic or Roman numerals, with a single glyph able to indicate any integer from 1 to 9,999."

Proofs


Slide Rules

December 6, 2021: In late 2021, Russian video appeared that showed a heavy bomber in flight with one of the crew using s slide rule, apparently to calculate course and/or fuel consumption rate. Most pilot training, especially for the crews of long-range aircraft, includes instruction on how to use special slide rules for such calculations in the event of problems with the electronic navigation and flight management instruments that do this automatically. For generations, ever since long-range flight became possible, manual tools were used for these calculations, along with a special bubble sextant to obtain the location of an aircraft. On the surface the original sextant is used for this but in the air, there is no fixed horizon to base these calculations. The bubble-sextant creates an artificial horizon that enables aerial navigation that shows position within ten kilometers or less. Surface sextant navigation is even more accurate and electronic navigation, especially using GPS, is accurate enough for landing aircraft.

Media tends to refer to EMP (Electromagnetic Pulse) created by a distant nuclear weapon detonation as the reason for these manual backups.