Paint my testicles blue and call me a moron...

[Shroom Man 777]

Why the fuck is this! Number A is three times larger than number B. Okay? Now I add 3 to both A and B. Logically, since both A and B get the same amount added into them, A+3 and B+3 would have the same proportion as A and B. But they don't. Now B is more than half A's amount. I don't get it.

Say there's this dwarf. He's a third of your height. Both of you put on hats of equal size. Now he's more than half your height. I don't comprehend this. ARGH!!!

I am a moron.

[dworkin]

A=9, B=3.

The dwarf and you are wearing cartoon top hats.

[Quadlok]

1+3 and 3+3.

[Shroom Man 777]

I'm not asking what the numbers are, rather...I'm asking why. It makes no sense to me.

[Kuroneko]

It is trivial to verify that a/b is not the same thing as (a+n)/(b+n) except for very trivial cases; you own dwarf example illustrates this. To make an even more clear-cut example of why this should be so, if a building is 100 times taller than a six-foot person, then adding six feet to their respective heights obviously does not preserve the ratio, nor should it. Things will make much more sense once one realizes that what is important is not how large the hat is, but what fraction of the person's height it is. Thus, to preserve the hypothetical 3:1 ratio, the hats would have to follow the same ratio also. In the more apparent bulding example, if three feet are added to the person and three hundred feet to the bulding (note that the amount added is exactly half the respective heights), then the resulting ratio is still 100:1.

To make things slightly more general, the relation between a/b and (a+c)/(b+d) is actually very interesting. If the nonnegative rationals in lowest terms with numerators and denominators less than a given number N are arranged from lowest to highest, then any three consecutive entries will be a/b, (a+c)/(b+d), c/d. For example, if N = 4, then the "Farey sequence" is (0/1, 1/4, 1/3, 1/2, 2/3, 3/4, 1/1, 3/2, 2/1, 3/1, 4/1), and we between 1/4 and 1/2 we have (1+1)/(4+2) = 1/3, just as it should be, and everywhere else besides; if we make a special "4/0" symbol at the end, it even works for the last two: (3+1)/(1+0) = 4/1. This is actually a moderately useful property, since it can be exploited to allow computer arithmetic on fractions to be done on the hardware level using exactly the same instructions as those for ordinary integers without any special handling in regards to to such things as getting common denominators before adding fractions or getting the result in lowest terms when multiplying, etc. (Unfortunately, while making actual operations very easy, comparison of two rationals, which is necessary for conditionals, is made a bit more difficult, which can be out-of-range checking cumbersome.)

[Lone_Prodigy]

It's because you are adding an equal amount to both, instead of an amount proportioned to A and B's original values. In other words, 1+3 and 3+3 ruins the proportion, but 1+1 and 3+3 (tailored to be proportional to the variable's values) does not. Understand?

[SyntaxVorlon]

If A=3 and B=1
A + 3 = 6 and B + 3 = 4
(B+3)/(A+3) = 2/3 where B/A = 1/3

and 2/3 > 1/2

[Shroom Man 777]

Yes, I see now. Painfully obvious. Very much so.

So, who wants to smear my genitals with blue paint? Any takers?

[Boyish-Tigerlilly]

Don't feel bad. It's numbers, and numbers are a part of the international conspiracy to give people headaches.

[Shroom Man 777]

I know. Goddamit. It would be much simpler if those damned towelheads didn't bother inventing math.

[NecronLord]

I know. Goddamit. It would be much simpler if those damned towelheads didn't bother inventing math.

*Gives Shroom a straw hat and a pitchfork*

"Off to thy primative farming with ye!"

[Queeb Salaron]

I know. Goddamit. It would be much simpler if those damned towelheads didn't bother inventing math.

Damn, that's a whole new breed of inbred hillbilly racism!

"Damn those dune coons and their numbers!! We blue-blooded Americans don't need no stinkin' numbers! That's why we invented computers!!"

[Lancer]

I know. Goddamit. It would be much simpler if those damned towelheads didn't bother inventing math.

Damn, that's a whole new breed of inbred hillbilly racism!

"Damn those dune coons and their numbers!! We blue-blooded Americans don't need no stinkin' numbers! That's why we invented computers!!"

I find that terribly amusing, especially since my ENEE class seems to confirm that normal math (algebra as most know it, trig, & calculus) isn't very useful hardware side.

[Darth Wong]

Don't feel bad. It's numbers, and numbers are a part of the international conspiracy to give people headaches.

Incorrect. Numbers are part of the international conspiracy to give stupid people headaches.

[Shroom Man 777]

Damn, that's a whole new breed of inbred hillbilly racism!

"Damn those dune coons and their numbers!! We blue-blooded Americans don't need no stinkin' numbers! That's why we invented computers!!"

One, I'm a Chinese guy living in the Philippines.

Two, I'm not a racist, though what I did say might've come out as such. It was meant to be humorous or something. I know, I should've put a disclaimer or something, but meh, it was late.

[brianeyci]

It is trivial to verify that a/b is not the same thing as (a+n)/(b+n) except for very trivial cases; you own dwarf example illustrates this. To make an even more clear-cut example of why this should be so, if a building is 100 times taller than a six-foot person, then adding six feet to their respective heights obviously does not preserve the ratio, nor should it. Things will make much more sense once one realizes that what is important is not how large the hat is, but what fraction of the person's height it is. Thus, to preserve the hypothetical 3:1 ratio, the hats would have to follow the same ratio also. In the more apparent bulding example, if three feet are added to the person and three hundred feet to the bulding (note that the amount added is exactly half the respective heights), then the resulting ratio is still 100:1.

Yes. Some people think algebraically better including me. To me it seems that the best way to "see" ratios is to use multiplication. Take two quantities A and B.

A/B = (A(1/3))/(B(1/3))

The equation is worded "A divided by B is equal to one third of each of A and B" therefore making it obvious that if you want to add you have to add the same proportion.

To make things slightly more general, the relation between a/b and (a+c)/(b+d) is actually very interesting. If the nonnegative rationals in lowest terms with numerators and denominators less than a given number N are arranged from lowest to highest, then any three consecutive entries will be a/b, (a+c)/(b+d), c/d.

Strictly speaking, no. What if a = 1, b = 1, c = 2 and d = 2. Then (a/b) = 1, (a + c)/(b + d) = 1, (c/d) = 1. Then there is ambiguity and no lowest or highest number (I know I'm an asshole lol).

(Unfortunately, while making actual operations very easy, comparison of two rationals, which is necessary for conditionals, is made a bit more difficult, which can be out-of-range checking cumbersome.)

Ah, out of range checking. Array index out of bounds. This was the fundamental problem in first year computer science, at least I remember the checks and tests for special cases like 1/0 and upper and lower limits requiring the most code and thought (they used an autochecking program to test everything so you needed to be prepared for that). I actually do not every remember writing a program requiring checking of rationals, which is funny since comparing two rationals seems to be a trivial problem that would be covered in first year. How would you actually do it, a linked list? Glad I'm not into that now, pure math and English majors for me. Thanks for the refuge from cardinality of sets, back to that now.

Brian

[brianeyci]

Correction, "A divided by B is equal to one third of each of A and B" should be "A divided by B is equal to one third of each of A and B divided by each other".

Anyway the point is made that numbers are not your enemies... they're your friends. Imagine if Math was all words and you had to write long sentences like the above rather than using symbols. ShroomMan don't worry. When I first learned factoring, I kept constantly asking "how can they see it" when seeing others do factoring in a sinch. Now I laugh at myself. You're not a moron, just someone who didn't "see" it properly. Don't be obsessed with the "seeing" as you progress through your mathematical life, examples usually clear up the "seeing" and usually it is something small like not thinking of proportions for your ratios (be glad it is only this, when you do harder math "seeing" will be a big deal and may take you days to think of how to do a problem )

Brian

[sketerpot]

I actually do not every remember writing a program requiring checking of rationals, which is funny since comparing two rationals seems to be a trivial problem that would be covered in first year. How would you actually do it, a linked list?

If I wanted to compare rationals (and I wasn't using a language or library with built-in support for them) I would probably get them over a common denominator and compare numerators.

Where would a linked list come into play? Am I misunderstanding the question?

[Kuroneko]

Strictly speaking, no. What if a = 1, b = 1, c = 2 and d = 2. Then (a/b) = 1, (a + c)/(b + d) = 1, (c/d) = 1. Then there is ambiguity and no lowest or highest number (I know I'm an asshole lol).

Your situation can only occur when the numbers involved are not coprime, which was explictly excluded above, since that is what 'in lowest terms' means when referring to fractions. My statement was correct as far as I can see, the only minor ambiguity being that (a+c)/(b+d) is not generally in lowest terms and needs reducing; I clarified that in the subsequent example, but I probably should have stated it explicitly.

I actually do not every remember writing a program requiring checking of rationals, which is funny since comparing two rationals seems to be a trivial problem that would be covered in first year. How would you actually do it, a linked list? Glad I'm not into that now, pure math and English majors for me.

Comparing nonnegative fractions a/b and c/d is equivalent to comparing ad and bc in the same order; e.g., a/b<c/d iff ad<bc. This part is rather trivial, and it is not necessary to compute common denominators. The Farey-fraction scheme uses a mapping from the order-N fractions to a certain subset of the group Z_p, where p is a prime greater than 2N² that is morphic will respect to both addition and multiplication, but not order. For comparisons, the fractions would have to be 'unpacked' from their integer representation first. This is not fatal in itself--it would still be useful for applications that are computation-itensive but make very few comparisons and for which the checking of a result is relativily easy. This is, however, still a fairly serious limitation.

The scheme is as follows. Pick a prime p>2N², and map a fraction a/b of order N (i.e., 0≤|a|,b≤N, b nonzero and a,b comprime) into the integer ab' mod p, where ' denotes multiplicative inverse modulo p; i.e., b' is the integer for which b*b' = 1 mod p. This number can be computed straightforwardly from the extended Euclidean algorithm for gcd(b,m); the result is, of course, unity, but the algorithm also computes the numbers X,Y such that Xb+Ym=gcd(b,m)=1, in which case X = b'. The normal arithemtic operations of addition, subtraction, multiplication of fractions correspond precisely to the same integer operations modulo p, as long as the result is a fraction of order N, and division by c/d is only slightly less straightforward, since it is the multiplication by d/c, and dc' = (cd')', so it only involves another computation of inverse of modulo p.

In my previous post, I used N = 4, but let's use the slighly more interesting N = 6, for which p = 73 is the lowest choice of prime modulus. For example, 1/3 - 1/4 + 2/3 = 3/4, and it is trivial to compute 3' = 49 and 4' = 55 (check: 3*49 = 2*73+1, thus 49 is the correct inverse since 3*49 = 1 mod 73). Thus, the computation is 1*49 - 1*55 + 2*49 = 19 mod p, which is the correct representation of 3/4 (3*55 = 19 mod 73). In general, recovering the original fraction a/b is not overly difficult, but cumbersome enough to make comparisons fairly costly. The real bad news is that there is very little to indicate when we have gone out of range, i.e., when the result is not an order-N fraction. The good news is that incorrect intermediate results do not matter; for example, 1/3 - 1/4 = 1/12, which is not an order-6 fraction--indeed, 1*49 - 1*55 = 67 mod 73. This is the representation of -6, which is outright wrong, but its usage is completely unproblematic as long as the final result is an order-6 (or, in general, N) fraction, as in this case 3/4. This is still living dangerously, languages like C do not have overflow checking whatsoever in the first place that is visible to the programmer, only now it is even more diffucult to check, but can nevertheless be of use for computationally-intensive applications in which the result can be checked fairly easily compared to the computation itself. Besides which, the effective elimination of division means that the computation is free from roundoff error, and the order-N fraction representation for the initial data is superior besides, since for any given number x, there is a fraction a/b of order N that is within a distance 1/[b(N+1)] of x. This is frequently by far better than floating-point, and is actually on the same order as if the fractions were uniformly distributed.

It should be noted that 2^32-1 is a prime, and thus order-32767 fractions can be represented naturally on the typical home computer. So is 2^61-1, for which order-(2^30-1) fractions will do; computation modulo 2^61-1 is extremely easy on most computers, since most C compilers support 64-bit integers even on those machines that do not have it at the hardware level, so simply zeroing the top three bits after a 64-bit computation would suffice. In general, however, the extended Euclidean algorithm for very large primes becomes computationally intensive (if we were interested in the common divisor only, then there are superior algorithms for computer implementation, but alas, they do not compute what we really want), in which case it is convenient do so something very cute to improve computational efficiency. Instead of one prime modulus, we can apply the Chinese remainder theorem and do the computations modulo several distinct, smaller primes, which enables comparatively easy computation with fractions of very large orders. But let's not get into that now.

Thanks for the refuge from cardinality of sets, back to that now.

Quick, if A is the cardinality of the integers (aleph-null), rank the following in the proper order: A^2, 2^A, A, A^A, (2^A)^A, A+A, 3^A, A*A, (2^A)^2. Some of them are equal.

[Boyish-Tigerlilly]

Math seems cool and all, but I could never do it. I mess up adding with calculators. I would rather deal with literature and stuff. I don't know how anyone can understand numbers.

[brianeyci]

Quick, if A is the cardinality of the integers (aleph-null), rank the following in the proper order: A^2, 2^A, A, A^A, (2^A)^A, A+A, 3^A, A*A, (2^A)^2. Some of them are equal.

Good thing you asked that because I am having trouble understanding cardinality operations (moronic, but since you're here I might as well ask). In my notes operations involving cardinality of sets is defined like this. Let c1 = |S1| and c2 = |S2|.

1. c1 + c2 = |S1 U S2|
2. c1 x c2 = |S1 x S2|
3. c1^c2 = |{f : c2 -> c1}|

So,

A^2 = |{f : 2 -> A}| = ?
2^A = |{f : A -> 2}| = |{f : A -> {0,1}}| = c ?
A = |N|
(2^A)^A = c^A = |{f : A -> c}| = c ?
A + A = |N U N| = |N|
3^A = |{f : A -> 3}| = |{f : A -> {0,1}}| = c?
A ^ A = |{f : A -> A}| = |{f : {0,1} -> {0,1}| = c?
A * A = |N x N| = |N ^ 2| = A
(2^A)^2 = c ^ 2 = |{f : 2 -> c}| = ?

See the problem, I do not know what it means to map an infinite set onto another infinite set. I also do not know what it means to map a single number into an infinite set. I see my teacher uses the characteristic function. But why is 2 = |{0,1}|? Is not the |{0,1}| = |P(N)| = |R| = c? Isn't cardinality supposed to be a numerical value, and therefore how can you map a numerical value onto another numerical value in operation 3? I think I am understanding it all wrong.

Brian

[Kuroneko]

Your definitions are correct. The trouble seems to be is that you confuse the sets with their cardinal numbers. This is somewhat understandable, as cardinal numbers are typically defined in terms of sets, and it is this definitions that you are operating on (an alternative to this is, for example, to assume the axiom of choice, in which case the cardinal numbers are ordinal numbers for which there is no bijection between the ordinal and any lesser ordinal). The crucial part of the definitions of the arithmetic operations is that operations on cardinals are really operations on their representative sets. We say that 2 = |{0,1}| because the set {0,1} has two elements, and the whole point of cardinal numbers is to represent the sizes of sets. {0,1} is the typical representative set for these operatiorns, but any set of two elements will do just as well, be it {44,55} or even {Michael Wong, Robert Scott Anderson}. It does not map a numerical value into a set, but rather any set with cardinality that of the given numerical value into some other set.

As to why 2^x = |P(X)|, for any cardinal x for which |X| = x, that is actually fairly easy. Pick any set of size two, say {a,b}. By definition, 2^x = |{f:X→{a,b}}|, the cardinality of the set of functions between X and {a,b}. Now, P(X) is the set of all subsets of X. For any particular subset Y of X, any element of X can be either in the set or not in it: exactly two options. There is therefore a bijection between subsets and functions of the type f:X→{a,b}, since each f identifies a unique subset (say, by convection that if f(z) = a, then element z is in the subset, and otherwise it is is not), and vice versa. It is most common to use, instead of {a,b}, the set {0,1}, but this is not at all relevant. Perhaps its only utility is to make apparent why |{f:X→{a,b}}| should be 2^x; to see this, think of this as the number of binary strings of size |X|.

Now, try again. I'm curious, though, does your class assume the continuum hypothesis (or even its generalized version) or the axiom of choice? It seems you at least assume the continuum hypothesis, since you treat 2^a as equal to c, which is usually reserved for the cardinality of the continuum of real numbers.

[brianeyci]

Your definitions are correct. The trouble seems to be is that you confuse the sets with their cardinal numbers. This is somewhat understandable, as cardinal numbers are typically defined in terms of sets, and it is this definitions that you are operating on (an alternative to this is, for example, to assume the axiom of choice, in which case the cardinal numbers are ordinal numbers for which there is no bijection between the ordinal and any lesser ordinal). The crucial part of the definitions of the arithmetic operations is that operations on cardinals are really operations on their representative sets. We say that 2 = |{0,1}| because the set {0,1} has two elements, and the whole point of cardinal numbers is to represent the sizes of sets. {0,1} is the typical representative set for these operatiorns, but any set of two elements will do just as well, be it {44,55} or even {Michael Wong, Robert Scott Anderson}. It does not map a numerical value into a set, but rather any set with cardinality that of the given numerical value into some other set.

Very good Kuroneko. Thank you. It was moronic to think of sets as their cardinal numbers and think that it was mapping a cardinal number to a set rather than mapping a set of size 2 to another set lol.

As to why 2^x = |P(X)|, for any cardinal x for which |X| = x, that is actually fairly easy. Pick any set of size two, say {a,b}. By definition, 2^x = |{f:X→{a,b}}|,

So far so good. So if there's for example 3^x, by definition it would be equal to |{f:X -> {a,b,c}| or for n^x = |{f:X->{x1, x2, x3, ... , xn}| for any cardinal for which |X| = x?

the cardinality of the set of functions between X and {a,b}.

I don't understand what this means. How can you have a set of functions between one set and another set?

Now, P(X) is the set of all subsets of X. For any particular subset Y of X, any element of X can be either in the set or not in it: exactly two options. There is therefore a bijection between subsets and functions of the type f:X→{a,b}, since each f identifies a unique subset (say, by convection that if f(z) = a, then element z is in the subset, and otherwise it is is not), and vice versa.

This is too condensed for me to understand. For a bijection, you need to define a one-to-one and onto function that maps A -> B. How are you achieving this by noting simply that any element of X is in {0,1} or it is not?

It is most common to use, instead of {a,b}, the set {0,1}, but this is not at all relevant. Perhaps its only utility is to make apparent why |{f:X→{a,b}}| should be 2^x; to see this, think of this as the number of binary strings of size |X|.

Ok.

Now, try again. I'm curious, though, does your class assume the continuum hypothesis (or even its generalized version) or the axiom of choice? It seems you at least assume the continuum hypothesis, since you treat 2^a as equal to c, which is usually reserved for the cardinality of the continuum of real numbers.

This is what my notes says about the continuum hypothesis.

Continuum Hypothesis : There is no S satisfying (aleph-not) < |S| < c.
Godel Theorem (1940s) - The continuum hypothesis is consistent with the rest of (Zernado-Frankel) set theory.
Cohen (1965ish) The assertion that there exists S such that (aleph-not) < |S| < c is also consistent with (Zernado-Frankel) set theory.
Godel and Cohen => The continuum hypothesis is independent of set theory.

In other words, we consider it an axiom it seems (at least there is no proof except the above appeal to authority, and I don't think we proved it in class).

There is something in the supplemental textbook using Schroder-Bernstein Theorem to prove |P(N)| = |R| = c, I'm not going to type it and I don't completely understand it.

Anyway, here is my second attempt.

A ^ 2 = |{f : 2 -> A}| = |{f : {0,1} -> A}| = |P(N)| = c
2 ^ A = |{f : A -> 2}| = |{f : A -> {0,1}}| = |P(N)| = c
A = |N| = A
A ^ A = |N|^|N| = |{f : |N| -> |N|| = |N| = A
(2^A)^A = |{f : A -> 2^A}| = |{f : A -> P(N)| = |{f : A -> R}| = c
A + A = |N U N| = |N| = A
3 ^ A = |{f : A -> 3}| = |{f : A -> {0,1,2}}| = |{f : A -> P(N)}| = |{f : A -> R}| = c
A * A = |N * N| = |N^2| = |N| = A
(2 ^ A) ^ 2 = |{f : 2 -> 2^2}| = |{f : {0,1} -> R}| = c

I understand enough to use the definition to complete the first step. However, how do you find a bijective function that maps say {0,1} to R, or {0,1} to N, or N to {0,1}? Right now the steps like |{f : A -> {0,1}}| = |P(N)| don't make sense to me (if it's even correct).

Brian

[Bob the Gunslinger]

I know. Goddamit. It would be much simpler if those damned towelheads didn't bother inventing math.

They didn't invent math. They preserved it and added to it but they didn't invent it. I think you're going to have to blame a caveman for this invention.

You're always welcome to join that one south american tribe that doesn't have any concept of numbers.

[Darth Wong]

You're always welcome to join that one south american tribe that doesn't have any concept of numbers.

I didn't know that George W. Bush came from South America.