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OT: Content Warning: Mathematics

pioneerlion83

pioneerlion83

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Woodpecker said:
I agree that it is not suspension-worthy but we must be wary of such a slippery slope. I mean, if we're not careful. @step.eng69 will launch into a lengthy analysis of relative tensile strengths of different concrete formulations.o_O
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...and which ones would be best for the Beaver Stadium reno!!!!
 
LionJim

LionJim

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LionJim said:
Today marks the 178th birthday of Georg Cantor. My topology professor at Tennessee, Bob Daverman, once told our class that when called upon to talk to laymen about mathematics, he brings up Cantorian Set Theory. “It’s easy to explain and it’s fascinating as hell.” Without further ado:

Consider the set A = {2, -7, 184, pi/2, 23/125, 0, -176, 2.8}. The cardinality of A is 8; we say this because when we point to each element of A in turn and say “one, two, three, four, five, six, seven, eight,” we are establishing a one-to-one correspondence between the elements of A and the elements of B= {1, 2, 3, 4, 5, 6, 7, 8}. Similarly, C={-8, 3, -34} has cardinality 3 because the elements of C can be put into a one-to-one correspondence with D={1, 2, 3}. Simple idea, not original with Cantor. It’s fairly painless to prove that, for example, the elements of B and D cannot be put into an one-to-one correspondence with each other. B and D are examples of finite sets.

Consider the set of N of natural numbers, N={1, 2, 3, 4, 5, ... }. It's easy to show that the elements of N cannot be put into an one-to-one correspondence with the elements of any finite set; N is infinite. The question is: what subsets F of the real numbers can have its elements put into an one-to-one correspondence with the elements of N? To do this, all you need is some algorithm which can match each element of F with a unique element of N, and vice versa. For example, {7, 8, 9, 10, ... }, {1/2, 1, 3/2, 2, 5/2, 3, ...} and Z={0, 1, -1, 2, -2, 3, -3, ...} can each be put into an one-to-one correspondence with N. (And, it follows, with each other.) Note that in each example you can figure exactly what the next element will be. Use Z to convince yourself that the order of the elements is irrelevant. (Z is standard notation: zahlen means "count" in German.)

Consider Q, the set of all rational numbers, positive and negative fractions, plus zero. (Q: "quotient".) Since, for example, 5/1 is a rational number, N is a proper subset of Q. Indeed, one can readily envision millions upon untold millions of elements of Q, rational numbers, which are themselves not elements of N. Example: -2/532826721 is in Q, but not in N. Here is Cantor's proof that the elements of Q can be put into a one-to-one correspondence with the elements of N; it's a sweet idea.

For each k>1, let G(k) be the set of all positive rational a/b, where a/b is reduced, that is, a and b have no common prime factors, such that a+b=k, along with -a/b. For example, G(2)={1/1, -1/1}, G(3)={2/1, -2/1, 1/2, -1/2}, G(4)={3/1, -3/1, 1/3, -1/3} (note that 2/2=1/1 is in G(2), not in G(4): 2/2 is not reduced), G(5)={4/1, -4/1, 3/2, -3/2, 2/3, -2/3, 1/4, -1/4}. It's easy to see that each plus or minus a/b is either in G(a+b) if a/b is reduced or in some G(k) where k<a+b if a/b is not reduced: -13/25 is an element of G(38), and 12/20=3/5 is an element of G(8). Each G(k) is finite and the G(k) are mutually disjoint: each a/b is in exactly ONE G(k). Here's how Cantor listed the elements of Q:

Q={0, G(1), G(2), G(3), ...} that is, he lists the elements of each G(k) in turn (which is possible because each G(k) is finite). In effect:
Q={0, 1/1, -1/1, 2/1, -2/1, 1/2, -1/2, 3/1, -3/1, 1/3, -1/3, 4/1, -4/1, 3/2, -3/2, 2/3, -2/3, 1/4, -1/4, ...}. Since each nonzero rational a/b is in exactly one G(k), we know that a/b is going to show up in this list, eventually, and exactly once. Done.

Of course, this wasn't all Cantor did; if this thread doesn't self-destruct I'll later explain his very simple proof that the real numbers CANNOT be put into an one-to-one correspondence with N.

I don't know how the layman will react to this but this was a very big deal back in Cantor's day, and he got a whole lot of grief over it; some very prominent mathematicians were very harsh about these ideas. To me it's one of the greatest things ever. Cormac McCarthy has a shrink ask a mathematician, "Is it worth it?" "Like nothing else on earth."
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Alrighty, there are two things I’d like to prove before getting into more Cantor. I’m mentioning these results later, and I don’t like waving my hands.

1. Consider x=.999999… (“point 9 hat” or “point 9 bar”). (The standard notation is x=.9 with a bar over the 9. You’ve seen it.) You know what it is, the decimal expansion of x is an endless run of 9s. I claim that x=1.

Proof: Think about it: if x<1, then there has to be some y strictly between x and 1: x<y<1. What could the decimal expansion of y possibly be? Since y is greater than x, y has to be greater than each of .9, .99, .999, .9999, and so on. Since y<1, this means that the only possible decimal expansion of y is y=.9999999…, that is, y=x. Since we were assuming that x<y, this is a contradiction. Thus our original assumption that x<1 is false and so its converse must be true: either x>1 or x=1. Since clearly we can’t have x larger than 1, we must conclude that x=1. (I like this proof.)

2. If p>1 is prime, the square root of p, sqrt(p), is not a rational number. (Meaning, you can’t find natural numbers a, b such that sqrt(p)=a/b.)

Proof: You all know the Fundamental Theorem of Arithmetic (FTA): every natural number x>1 is either prime or x factors uniquely into a product of primes. (By uniquely, I mean uniquely up to order, for example, 6=2*3=3*2.) We’ll assume this without proof; most math majors first see this proof in 400-level algebra.

Because of FTA, given any natural number x>1, we can define N(x) to be the number of primes needed to multiply into x. For example, N(17)=1 because 17 is prime, while N(60)=4 because 60=2*2*3*5; you need four primes to multiply into 60. Easy peasy. This N has the property that N(xy)=N(x)+N(Y): see: N(8)=3, since 8=2*2*2. Now, 480=8*60, so 480=(2*2*2)*(2*2*3*5), which gives N(480)=7=3+4=N(8)+N(60). This means that the N of a square is always even: N(x*x)=N(x)+N(x)=2*N(x). N(4)=2, N(9)=2, N(16)=4, and so on.

Suppose that sqrt(p)=a/b. By squaring we have p=(a^2)/(b^2). Clearing the fraction gives p*(b^2)=(a^2). Now, N(p*(b^2))=N(p)+N(b^2)=1+N(b^2), since p is prime. Since N(b^2) is even, N(p*(b^2))=1+N(b^2) is odd. But N(a^2) is even, not odd, and thus we cannot have p*(b^2)=(a^2). Thus we have no a,b with sqrt(p)=a/b. Done.
 
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Bkmtnittany1

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Ok
LionJim said:
Alrighty, there are two things I’d like to prove before getting into more Cantor. I’m mentioning these results later, and I don’t like waving my hands.

1. Consider x=.999999… (“9 hat” or “9 bar”). (The standard notation is x=.9 with a bar over the 9. You’ve seen it.) You know what it is, the decimal expansion of x is an endless run of 9s. I claim that x=1.

Proof: Think about it: if x<1, then there has to be some y strictly between x and 1: x<y<1. What could the decimal expansion of y possibly be? Since y is greater than x, y has to be greater than each of .9, .99, .999, .9999, and so on. Since y<1, this means that the only possible decimal expansion of y is y=.9999999…, that is, y=x. Since we were assuming that x<y, this is a contradiction. Thus our original assumption that x<1 is false and so its converse must be true: either x>1 or x=1. Since clearly we can’t have x larger than 1, we must conclude that x=1. (I like this proof.)

2. If p>1 is prime, the square root of p, sqrt(p), is not a rational number. (Meaning, you can’t find natural numbers a, b such that sqrt(p)=a/b.)

Proof: You all know the Fundamental Theorem of Arithmetic (FTA): every natural number x>1 is either prime or x factors uniquely into a product of primes. (By uniquely, I mean uniquely up to order, for example, 6=2*3=3*2.) We’ll assume this without proof; most math majors first see this proof in 400-level algebra.
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You need x=3 beers and y= 45 minutes on the weather thread!
 
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IrishHerb

IrishHerb

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LionJim said:
Alrighty, there are two things I’d like to prove before getting into more Cantor. I’m mentioning these results later, and I don’t like waving my hands.

1. Consider x=.999999… (“point 9 hat” or “point 9 bar”). (The standard notation is x=.9 with a bar over the 9. You’ve seen it.) You know what it is, the decimal expansion of x is an endless run of 9s. I claim that x=1.

Proof: Think about it: if x<1, then there has to be some y strictly between x and 1: x<y<1. What could the decimal expansion of y possibly be? Since y is greater than x, y has to be greater than each of .9, .99, .999, .9999, and so on. Since y<1, this means that the only possible decimal expansion of y is y=.9999999…, that is, y=x. Since we were assuming that x<y, this is a contradiction. Thus our original assumption that x<1 is false and so its converse must be true: either x>1 or x=1. Since clearly we can’t have x larger than 1, we must conclude that x=1. (I like this proof.)

2. If p>1 is prime, the square root of p, sqrt(p), is not a rational number. (Meaning, you can’t find natural numbers a, b such that sqrt(p)=a/b.)

Proof: You all know the Fundamental Theorem of Arithmetic (FTA): every natural number x>1 is either prime or x factors uniquely into a product of primes. (By uniquely, I mean uniquely up to order, for example, 6=2*3=3*2.) We’ll assume this without proof; most math majors first see this proof in 400-level algebra.

Because of FTA, given any natural number x>1, we can define N(x) to be the number of primes needed to multiply into x. For example, N(17)=1 because 17 is prime, while N(60)=4 because 60=2*2*3*5; you need four primes to multiply into 60. Easy peasy. This N has the property that N(xy)=N(x)+N(Y): see: N(8)=3, since 8=2*2*2. Now, 480=8*60, so 480=(2*2*2)*(2*2*3*5), which gives N(480)=7=3+4=N(8)+N(60). This means that the N of a square is always even: N(x*x)=N(x)+N(x)=2*N(x). N(4)=2, N(9)=2, N(16)=4, and so on.

Suppose that sqrt(p)=a/b. By squaring we have p=(a^2)/(b^2). Clearing the fraction gives p*(b^2)=(a^2). Now, N(p*(b^2))=N(p)+N(b^2)=1+N(b^2), since p is prime. Since N(b^2) is even, N(p*(b^2))=1+N(b^2) is odd. But N(a^2) is even, not odd, and thus we cannot have p*(b^2)=(a^2). Thus we have no a,b with sqrt(p)=a/b. Done.
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#1 I've seen somewhere before. Nice proof though.
#2 is neat!

Really enjoy seeing nice clean proofs like this.
 
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LionJim

LionJim

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IrishHerb said:
#1 I've seen somewhere before. Nice proof though.
#2 is neat!

Really enjoy seeing nice clean proofs like this.
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Yeah, #2 is so much cleaner than the “standard” proof. That’s what kept me alive at Maryland, I never had much mathematical imagination but I always could write a good proof.
 
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s1uggo72

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Woodpecker said:
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me a dumb *** frosh listening to 2 smart guys talking....
Gauss' law says the Flux across any closed bound surface is....
any?
any!
any?
any!!!
they move on, I left.
I still remember that conversation 47 yrs later, and God that guy was a DB!
 
LionJim

LionJim

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LionJim said:
Alrighty, there are two things I’d like to prove before getting into more Cantor. I’m mentioning these results later, and I don’t like waving my hands.

1. Consider x=.999999… (“point 9 hat” or “point 9 bar”). (The standard notation is x=.9 with a bar over the 9. You’ve seen it.) You know what it is, the decimal expansion of x is an endless run of 9s. I claim that x=1.

Proof: Think about it: if x<1, then there has to be some y strictly between x and 1: x<y<1. What could the decimal expansion of y possibly be? Since y is greater than x, y has to be greater than each of .9, .99, .999, .9999, and so on. Since y<1, this means that the only possible decimal expansion of y is y=.9999999…, that is, y=x. Since we were assuming that x<y, this is a contradiction. Thus our original assumption that x<1 is false and so its converse must be true: either x>1 or x=1. Since clearly we can’t have x larger than 1, we must conclude that x=1. (I like this proof.)

2. If p>1 is prime, the square root of p, sqrt(p), is not a rational number. (Meaning, you can’t find natural numbers a, b such that sqrt(p)=a/b.)

Proof: You all know the Fundamental Theorem of Arithmetic (FTA): every natural number x>1 is either prime or x factors uniquely into a product of primes. (By uniquely, I mean uniquely up to order, for example, 6=2*3=3*2.) We’ll assume this without proof; most math majors first see this proof in 400-level algebra.

Because of FTA, given any natural number x>1, we can define N(x) to be the number of primes needed to multiply into x. For example, N(17)=1 because 17 is prime, while N(60)=4 because 60=2*2*3*5; you need four primes to multiply into 60. Easy peasy. This N has the property that N(xy)=N(x)+N(Y): see: N(8)=3, since 8=2*2*2. Now, 480=8*60, so 480=(2*2*2)*(2*2*3*5), which gives N(480)=7=3+4=N(8)+N(60). This means that the N of a square is always even: N(x*x)=N(x)+N(x)=2*N(x). N(4)=2, N(9)=2, N(16)=4, and so on.

Suppose that sqrt(p)=a/b. By squaring we have p=(a^2)/(b^2). Clearing the fraction gives p*(b^2)=(a^2). Now, N(p*(b^2))=N(p)+N(b^2)=1+N(b^2), since p is prime. Since N(b^2) is even, N(p*(b^2))=1+N(b^2) is odd. But N(a^2) is even, not odd, and thus we cannot have p*(b^2)=(a^2). Thus we have no a,b with sqrt(p)=a/b. Done.
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I haven’t quite decided exactly how I plan to finish this, as I’m struggling to decide which approach is the easiest to understand, and to explain. So here’s something that I might or might not use. It’s not something that is strictly required for me to do, just something that might make it easier to explain what I want to explain.

I’m going to prove here that there are infinitely many prime numbers. We’ll do this by proving that if p is a prime number then there is a prime q larger than p.


This is something you already know: for x a natural number, 2x+1 is never divisible by 2 (the remainder will be 1). Similarly, 7x+1 is not divisible by 7, and 83x+1 is not divisible by 83, you get the idea. Going further, 2*3*7*13+1 is not divisible by 2, 3, 7, or 13.

You know what n! is, it’s the product of every whole number less than or equal to n. For example, 5!=5*4*3*2. Now, this means that t=5!+1 is not divisible by any of 5, 4, 3, 2, so if q>1 divides t then q must be larger than 5. Let p be some prime number and set t=p!+1. By the Fundamental Theorem of Arithmetic there is a prime number q>1 which divides t (it’s possible that t might be prime, so in this case q=t). By our discussion, we must have q>p. In other words, if p is a prime number then there is a prime q larger than p. So there is no largest prime number, which means that the set of primes must be infinite. Done.
 
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IrishHerb

IrishHerb

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LionJim said:
I haven’t quite decided exactly how I plan to finish this, as I’m struggling to decide which approach is the easiest to understand, and to explain. So here’s something that I might or might not use. It’s not something that is strictly required for me to do, just something that might make it easier to explain what I want to explain.

I’m going to prove here that there are infinitely many prime numbers. We’ll do this by proving that if p is a prime number then there is a prime q larger than p.


This is something you already know: for x a natural number, 2x+1 is never divisible by 2 (the remainder will be 1). Similarly, 7x+1 is not divisible by 7, and 83x+1 is not divisible by 83, you get the idea. Going further, 2*3*7*13+1 is not divisible by 2, 3, 7, or 13.

You know what n! is, it’s the product of every whole number less than or equal to n. For example, 5!=5*4*3*2. Now, this means that t=5!+1 is not divisible by any of 5, 4, 3, 2, so if q>1 divides t then q must be larger than 5. Let p be some prime number and set t=p!+1. By the Fundamental Theorem of Arithmetic there is a prime number q>1 which divides t (it’s possible that t might be prime, so in this case q=t). By our discussion, we must have q>p. In other words, if p is a prime number then there is a prime q larger than p. So there is no largest prime number, which means that the set of primes must be infinite. Done.
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Wow! You really have your thinkers on @LionJim .
 
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PSUAXE70

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LionJim said:
I haven’t quite decided exactly how I plan to finish this, as I’m struggling to decide which approach is the easiest to understand, and to explain. So here’s something that I might or might not use. It’s not something that is strictly required for me to do, just something that might make it easier to explain what I want to explain.

I’m going to prove here that there are infinitely many prime numbers. We’ll do this by proving that if p is a prime number then there is a prime q larger than p.


This is something you already know: for x a natural number, 2x+1 is never divisible by 2 (the remainder will be 1). Similarly, 7x+1 is not divisible by 7, and 83x+1 is not divisible by 83, you get the idea. Going further, 2*3*7*13+1 is not divisible by 2, 3, 7, or 13.

You know what n! is, it’s the product of every whole number less than or equal to n. For example, 5!=5*4*3*2. Now, this means that t=5!+1 is not divisible by any of 5, 4, 3, 2, so if q>1 divides t then q must be larger than 5. Let p be some prime number and set t=p!+1. By the Fundamental Theorem of Arithmetic there is a prime number q>1 which divides t (it’s possible that t might be prime, so in this case q=t). By our discussion, we must have q>p. In other words, if p is a prime number then there is a prime q larger than p. So there is no largest prime number, which means that the set of primes must be infinite. Done.
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Without citing the Fundamental Theorem of Arithmetic, consider all the prime numbers. If there is a largest prime number Pn, multiply it by all the other smaller primes. Add 1 to that product. That number should be a prime number since it will not be divisible by any known prime. If it is not a prime, it must be divisible by prime numbers not listed above. Either way there will be a larger prime than Pn. Therefore there is no limit to prime numbers.
 
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LionJim

LionJim

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PSUAXE70 said:
Without citing the Fundamental Theorem of Arithmetic, consider all the prime numbers. If there is a largest prime number Pn, multiply it by all the other smaller primes. Add 1 to that product. That number should be a prime number since it will not be divisible by any known prime. If it is not a prime, it must be divisible by prime numbers not listed above. Either way there will be a larger prime than Pn. Therefore there is no limit to prime numbers.
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This is the “standard” proof, the one I normally used until I Wiki’d. You actually do use the FTA: “If it is not a prime, it must be divisible by prime numbers not listed above,” this the existence part of the FTA.
 
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Woodpecker

Woodpecker

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I kinda feel bad for numbers that the MAA deems as only Choice or Select
 
IrishHerb

IrishHerb

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A bit off point, but similar to the false proof that 2 = 1.

Prove that 0/0 = 2

0/0 = (4 - 4)/(4 - 4) = (2^2 - 2^2)/(2^2 - 2^2) = [(2 - 2)(2 + 2)]/[2(2 - 2)] = (2 + 2)/2 = 4/2 = 2. there you go.
 
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s1uggo72

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IrishHerb said:
A bit off point, but similar to the false proof that 2 = 1.

Prove that 0/0 = 2

0/0 = (4 - 4)/(4 - 4) = (2^2 - 2^2)/(2^2 - 2^2) = [(2 - 2)(2 + 2)]/[2(2 - 2)] = (2 + 2)/2 = 4/2 = 2. there you go.
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So where’s the flaw?? Pretty early Saturday for me
 
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PrtLng Lion

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IrishHerb said:
A bit off point, but similar to the false proof that 2 = 1.

Prove that 0/0 = 2

0/0 = (4 - 4)/(4 - 4) = (2^2 - 2^2)/(2^2 - 2^2) = [(2 - 2)(2 + 2)]/[2(2 - 2)] = (2 + 2)/2 = 4/2 = 2. there you go.
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Pretty cool. I seem to remember reading somewhere that you can make all kinds of paradoxes with showing equivalence to some number over 0, since that's "undefined".
 
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PrtLng Lion

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You guys ever checkout Numberphile on YouTube? They've got some fascinating content with number theory, finding primes, proofs, etc. One of the absolute best channels.

Anyhow, they've had more than one episode on Cantor's work (or extensions thereof), explaining the Cardinals, Ordinals, comparing infinite sets, Hilbert's hotel, etc.

Can anyone how how that proof that 1+2+3+4+5... = -1/12 is flawed?
 
LionJim

LionJim

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PrtLng Lion said:
You guys ever checkout Numberphile on YouTube? They've got some fascinating content with number theory, finding primes, proofs, etc. One of the absolute best channels.

Anyhow, they've had more than one episode on Cantor's work (or extensions thereof), explaining the Cardinals, Ordinals, comparing infinite sets, Hilbert's hotel, etc.

Can anyone how how that proof that 1+2+3+4+5... = -1/12 is flawed?
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I freaking love explaining Hilbert’s Hotel. Coolest thing ever.
 
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LionJim

LionJim

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LionJim said:
I freaking love explaining Hilbert’s Hotel. Coolest thing ever.
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Hilbert’s Hotel has infinitely many rooms, each numbered by a whole number: Room 1, Room 2, Room 3, …, Room 45389974556, and so on.

Joe Blow shows up at the hotel and asks for a room. “Well, every room is occupied, at the moment we have no empty rooms.” Solution: have the person in Room 1 move to Room 2, the person in Room 2 move to Room 3, the person in Room N move to Room N+1. So now Room 1 is empty and Joe Blow has a room, and nobody already there is kicked out.

There are more complicated scenarios, as easily explained as this one.
 
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