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[Kirian]

Came across this after some related Quora discussion:

http://www.maa.org/external_archive/devlin/devlin_06_08.html

I have big issues with this rant.... The basic claim is that multiplication is not defined through addition.

Someone call up all the computer engineers out there and tell them they've been building circuit chips wrong all this time.

[Tables]

It shocks the hell out of me that you can use the prime numbers to calculate pi.  How the hell can those possibly be related!?!?

You can also use, say, the odd numbers (ok, also 4)

In the list of all numbers of the form 2n+1 and also 4, I would say that 4 already is an odd number.

[Cuzz]

It shocks the hell out of me that you can use the prime numbers to calculate pi.  How the hell can those possibly be related!?!?

You can also use, say, the odd numbers (ok, also 4)

In the list of all numbers of the form 2n+1 and also 4, I would say that 4 already is an odd number.

That is interesting, but then again, so are all natural numbers.

[Witherweaver]

It shocks the hell out of me that you can use the prime numbers to calculate pi.  How the hell can those possibly be related!?!?

You can also use, say, the odd numbers (ok, also 4)

Maybe we can use the odd numbers to calculate 4.

[Cuzz]

It shocks the hell out of me that you can use the prime numbers to calculate pi.  How the hell can those possibly be related!?!?

You can also use, say, the odd numbers (ok, also 4)

Maybe we can use the odd numbers to calculate 4.

What sorcery.....?

[Witherweaver]

It shocks the hell out of me that you can use the prime numbers to calculate pi.  How the hell can those possibly be related!?!?

You can also use, say, the odd numbers (ok, also 4)

Maybe we can use the odd numbers to calculate 4.

What sorcery.....?

[qmech]

Came across this after some related Quora discussion:

http://www.maa.org/external_archive/devlin/devlin_06_08.html

I have big issues with this rant.... The basic claim is that multiplication is not defined through addition.

Someone call up all the computer engineers out there and tell them they've been building circuit chips wrong all this time.

I think that's a slight mischaracterisation of Devlin's position.  He's saying that, however it's defined, repeated addition is not what multiplication is.  This is an incredibly common pattern in mathematics.  You can define the natural numbers as {}, {{}}, {{},{{}}}, ... and the ordered pair (a,b) is {{a},{a,b}} and an integer as an equivalence class of ordered pairs of naturals, and a rational as an equivalence class of ordered pairs of integers, and a real as an equivalence class of Cauchy sequences (whatever they are) of rationals ... but that's not what a real number is.

A function from X to Y "is" a set of ordered pairs such that every x in X appears in exactly one pair (x,y) with y in Y.  But that's not how you think about a function, or how you interpret an occurrence of f(x).  A graph "is" a pair (V,E) where V is a set of vertices and E a set of 2-subsets of V, but you're not going to think in those terms if I ask you to draw a K_4 in the plane.  Mathematical objects have their own existence, and when we think about them we think about them in terms of properties they have.  The concrete definitions can convince somebody that a structure exists, but when you get to work with it you can throw away the scaffolding.  For multiplication the rules are that 0.x = 0, 1.x = x, x.y = y.x and (x+y).z = x.y + x.z.  These contain the inductive definition of multiplication on the naturals, but once you've checked that everything makes sense those particular cases are no longer all that special.

[Witherweaver]

Came across this after some related Quora discussion:

http://www.maa.org/external_archive/devlin/devlin_06_08.html

I have big issues with this rant.... The basic claim is that multiplication is not defined through addition.

Someone call up all the computer engineers out there and tell them they've been building circuit chips wrong all this time.

I think that's a slight mischaracterisation of Devlin's position.  He's saying that, however it's defined, repeated addition is not what multiplication is.  This is an incredibly common pattern in mathematics.  You can define the natural numbers as {}, {{}}, {{},{{}}}, ... and the ordered pair (a,b) is {{a},{a,b}} and an integer as an equivalence class of ordered pairs of naturals, and a rational as an equivalence class of ordered pairs of integers, and a real as an equivalence class of Cauchy sequences (whatever they are) of rationals ... but that's not what a real number is.

A function from X to Y "is" a set of ordered pairs such that every x in X appears in exactly one pair (x,y) with y in Y.  But that's not how you think about a function, or how you interpret an occurrence of f(x).  A graph "is" a pair (V,E) where V is a set of vertices and E a set of 2-subsets of V, but you're not going to think in those terms if I ask you to draw a K_4 in the plane.  Mathematical objects have their own existence, and when we think about them we think about them in terms of properties they have.  The concrete definitions can convince somebody that a structure exists, but when you get to work with it you can throw away the scaffolding.  For multiplication the rules are that 0.x = 0, 1.x = x, x.y = y.x and (x+y).z = x.y + x.z.  These contain the inductive definition of multiplication on the naturals, but once you've checked that everything makes sense those particular cases are no longer all that special.

Okay, I can get behind this a little more.. but, still, "is" is a bit subjective.  Sure, the set of natural numbers is something, but to get any handle on it we need to define it.  It, in a sense, "is" as much {}, {{}}, {{}, {{}}},  as it "is" anything else. 

But, also, 'multiplication' is not really anything.  Only multiplication together with sets A, B, C is a thing.   

[Cuzz]

Came across this after some related Quora discussion:

http://www.maa.org/external_archive/devlin/devlin_06_08.html

I have big issues with this rant.... The basic claim is that multiplication is not defined through addition.

Someone call up all the computer engineers out there and tell them they've been building circuit chips wrong all this time.

I think that's a slight mischaracterisation of Devlin's position.  He's saying that, however it's defined, repeated addition is not what multiplication is.  This is an incredibly common pattern in mathematics.  You can define the natural numbers as {}, {{}}, {{},{{}}}, ... and the ordered pair (a,b) is {{a},{a,b}} and an integer as an equivalence class of ordered pairs of naturals, and a rational as an equivalence class of ordered pairs of integers, and a real as an equivalence class of Cauchy sequences (whatever they are) of rationals ... but that's not what a real number is.

A function from X to Y "is" a set of ordered pairs such that every x in X appears in exactly one pair (x,y) with y in Y.  But that's not how you think about a function, or how you interpret an occurrence of f(x).  A graph "is" a pair (V,E) where V is a set of vertices and E a set of 2-subsets of V, but you're not going to think in those terms if I ask you to draw a K_4 in the plane.  Mathematical objects have their own existence, and when we think about them we think about them in terms of properties they have.  The concrete definitions can convince somebody that a structure exists, but when you get to work with it you can throw away the scaffolding.  For multiplication the rules are that 0.x = 0, 1.x = x, x.y = y.x and (x+y).z = x.y + x.z.  These contain the inductive definition of multiplication on the naturals, but once you've checked that everything makes sense those particular cases are no longer all that special.

Okay, I can get behind this a little more.. but, still, "is" is a bit subjective.  Sure, the set of natural numbers is something, but to get any handle on it we need to define it.  It, in a sense, "is" as much {}, {{}}, {{}, {{}}},  as it "is" anything else. 

But, also, 'multiplication' is not really anything.  Only multiplication together with sets A, B, C is a thing.

This reminds me of a discussion I recently had with a colleague. I remarked that my students were having a hard time understanding that a function as a concept is distinct from its graph, and he reminded me that, being a special type of relation, that is in fact exactly what a function is.

[Kirian]

Okay, I can get behind this a little more.. but, still, "is" is a bit subjective.

I'm guessing you did not have sex with that woman?

[Witherweaver]

Okay, I can get behind this a little more.. but, still, "is" is a bit subjective.

I'm guessing you did not have sex with that woman?

"woman" is a bit a subjective.  Oh, wait, that's not right... hold on...

[Kirian]

I just realized that joke makes almost no sense to non-Americans or people born before 1990-ish.

[Watno]

I'm European and born after 1990, and I think I understand what it refers too, and I don't believe this to be very unusual

[Tables]

I'm European and born after 1990, and I think I understand what it refers too, and I don't believe this to be very unusual

This. I can't quite remember if it was Clinton or Nixon, but I know it was definitely something to do with receiving Heads of State.

[Tables]

Fun thing I worked out today: Take a times table grid, any size. Draw a rectangle on it*. The average of all of the numbers in the rectangle is equal to the average of the four corner squares of the rectangle.

Equally fun thing: Take a times table grid, any size. The sum of all the numbers in this times table grid is equal to the average of the four corner squares, multiplied by the bottom right square.

I leave the proof of these statements for your own pleasure.

Hint: Doing it purely algebraically is doing it in hard mode. Use properties of the times table grid to cut the work you need to do significantly.

*A rectangle orthogonal to the grid and covering whole squares, for the pedants among you**

**Yes this definition is still probably too vague. Shut up.

[Polk5440]

Fun thing I worked out today: Take a times table grid, any size. Draw a rectangle on it. The average of all of the numbers in the rectangle is equal to the average of the four corner squares of the rectangle.

Equally fun thing: Take a times table grid, any size. The sum of all the numbers in this times table grid is equal to the average of the four corner squares, multiplied by the bottom right square.

I leave the proof of these statements for your own pleasure.

These are fun exercises to show. The second one is easy to show algebraically:

Using the formula 1 + 2 + ... + n =n(n+1)/2 (by induction or the brilliancy of Gauss), the sum of the elements in a multiplication table of size n is n(n+1)/2 + 2*n(n+1)/2 + ... + n*n(n+1)/2 = n(n+1)/2*n(n+1)/2 = n^2*(1 + n + n + n^2)/4. Noting the corner elements of a multiplication table are 1, n, n, and n^2, we are done.

For the first one, this can be brute forced in a similar fashion.

Intuitively, think about one row of the rectangle at a time. Because each row of the rectangle is an arithmetic sequence, the middle element (or average of the middle two elements) will be the average. This is the same as the average of the two end points. To visualize this, think about Gauss's intuitive proof for 1 + 2 + ... + n = n(n+1)/2. You've collapsed the rectangle to one column which is also an arithmetic sequence. The middle element (or average of the middle two elements) will be the average. This is the same as the average of the two end points. Therefore, all you need to do is take the average of the four corners of the rectangle.

Corollary: If your rectangle has odd numbers of rows and columns, you can also just pick the middle element. That's your average.

[Tables]

Yeah, that's more or less how I worked them out. I actually did the second one first, just like you. It came about from working out how you can add up all the squares on a multiplication grid. I'm not sure how I came up with thinking about the corners though.

I wouldn't consider your explanation for the first one there to be brute forcing it. Brute forcing it would be just setting it up as an algebra problem and then checking two equations match up - in particular this:

(n - m + 1)(p - q + 1)( nq + mq + np + mp ) / 4 = ( nq(n+1)(q+1) - mq(m-1)(q+1) - np(n+1)(p-1) + mp(m-1)(p-1) ) / 4

Left hand side is the number of boxes in your grid, multiplied by the average of the four corners (for a grid from m to n by p to q). Right hand side is the total of all the boxes using arithmetic series. These equations do indeed balance, but yeah that's just ugly and inelegant.

[liopoil]

Proof without hint:

Note that in each column c, the average value is c * m, where m is the mean value over all row numbers, or equivalently the average of the first and last row numbers. The average value of c is similarly the average of the first and last columns. Thus the average value of the whole table is the product of these two averages, which can be seen to be the average of the products (simply by exansion), or the average of the corners.

The equally fun thing follows from the the above proof and the fact that the bottom right corner represents the number of cells in the whole table.

PPE: Two posts I haven't read

[Polk5440]

Yeah, that's more or less how I worked them out. I actually did the second one first, just like you. It came about from working out how you can add up all the squares on a multiplication grid. I'm not sure how I came up with thinking about the corners though.

I wouldn't consider your explanation for the first one there to be brute forcing it.

Right. I didn't type my brute forced algebra for 1. I tried to outline how I was thinking about why it works.

[ehunt]

Kirian is probably referring to the result from https://en.wikipedia.org/wiki/Euler_product#Notable_constants



Edit: Looks like an explanation that generalizes requires more math than I know. An easier justification is at http://www.cut-the-knot.org/proofs/AfterEulerProduct.shtml. The idea is that you start with the Leibniz formula, and note that every odd number equal to 1 mod 4 must be the product of an even number of (not necessarily distinct) primes that are 3 mod 4.

If you're already happy with the Leibniz formula, then it follows formally from the following two observations:

1. 1 - 1/3 + 1/5 - 1/7 + 1/9 - ... is the value at s = 1 of the function sum_{n=0}^infinity f(n) n^{-s}, where f(n) is the function that 0 if n is even, 1 if n is in the set {1, 5, 9, ...} (i.e. is 1 mod 4), and -1 if n is in the sequence {3, 7, 11, ...} (i.e. is 3 mod 4).

2. If f is a multiplicative function (meaning f(xy) = f(x)f(y), as is satisfied for our f above), then

sum f(n) n^(-s) is the product over all primes p of (1-f(p)p^(-s))^(-1)

To prove this, first apply the geometric series formula from high school to every single term on the right hand side, then use the fact that every natural number factors uniquely into a product of primes.

[Kirian]

I was going to make a joke about abstract algebra, but I couldn't come up with one that had the right ring to it.

[scott_pilgrim]

It's important to make sure you're audience is in the right field; when you factor in the group you're telling it to, you can generate an ideal response.  I don't think a normal group would get an abstract algebra joke, but if you distribute it to people in the domain of math, they will associate certain words with their identity in algebra (and by extension, they will find your joke clever).

[pacovf]

I saw more fees (m).

...am I doing this right?

[Limetime]

Dominion opening probability question:
What is the probabilities of hitting 5 first shuffle if you open silver/wedding and wedding/(pay off debt)

[sudgy]

So, I saw this equation as (I think) a joke somewhere online:

y'' = y'^(y'*sin(x*y)) + (e^x)*tan(1/x)

It's way too advanced for me, and I can tell the solution is probably really nasty.  Can any of you solve it?  Wolframalpha wasn't giving it.