Some Amazing Mathematical Thought

[Unknown Soldier]

Mathematicians can be amazing thinkers. For example, I recently was working on proving that log₂(3) is an irrational number. I tried the following proof by contradiction:

If log₂(3) is a rational number, then there are integers a > 0 and b > 0 where log₂(3) = a/b. And by the definition of a logarithm, 2^a/b= 3. And since 2^a/b = 3, then ln(2^a/b) = ln(3) so (a/b)ln(2) = ln(3) and a/b = ln(3)/ln(2). But this result is a contradiction because ln(3)/ln(2) is irrational. Therefore, since there is no rational number a/b where 2^a/b = 3, then log₂(3) is irrational.

Although I think I proved the proposition that log₂(3) is an irrational number, what bothered me about this proof is the statement I formatted in bold: "ln(3)/ln(2) is irrational." I don't know for sure if it has ever been proved true. So I checked YouTube and found this proof:

If log₂(3) is a rational number, then there are integers a > 0 and b > 0 where log₂(3) = a/b. And by the definition of a logarithm, 2^a/b= 3. Taking both sides to the power of b, we have (2^a/b)^b= 3^bor 2^a= 3^b. But since a and b are integers, then 2^a is always even while 3^b is always odd. Because assuming that log₂(3) is a rational number results in this contradiction, then log₂(3)) is irrational.

So a very simple and clear change in this proof from my proof to the alternate proof I found on YouTube solved the difficulty of my not knowing for sure that ln(3)/ln(2) is irrational. It just goes to show that genius involves simplicity.

 

[Swammerdami]

Theorem:

2^a = 3^b​

has no solution in the positive rational integers.​

Proof:

This fact follows immediately from the  Fundamental theorem of arithmetic.​

Conjecture Theorem:

p^a = q^b ± 1​

has only the obvious trivial solutions (in the positive rational integers) up to 3² = 2³ + 1.​

Proof:
This almost-ancient  Catalan's conjecture was finally proved in the 21st century.

 

[Unknown Soldier]

Theorem:

2^a = 3^b​

has no solution in the positive rational integers.​

But 2^a is an even integer if a is an integer, and 3^b is an odd integer if b is an integer. So 2^a = 3^b is a contradiction and so can never be true. So you don't need to cite the Fundamental Theorem of Arithmetic to know that 2^a = 3^b has no solution when a and and b are integers. The point I was masking in the OP is that it's best to take the simplest route when proving a theorem.

 

[steve_bank]

Don't know why you post in philosophy instead of math.

Here is one for you Soldier.

Math Problems

A ways back tere was a lebgthy thread with problems. Back when there were more people. An exponential probability density function is y(t) = k*e^(-t/u) where k is a constant. u represents the mean. The standard deviation of the distribution equals u. y is defined 0 to +infinity. For t = T what...

iidb.org

 

[AdamWho]

I was explaining to my stats students today that they all have more than the average number of legs......

 

[bilby]

I was explaining to my stats students today that they all have more than the average number of legs......

That's nothing. I have a friend who has an artificial leg, and a real, live, foot.

 

[Jimmy Higgins]

I didn't know you had a pet snail.

 

[steve_bank]

As this is a math thread and drawing on my deep and profound knowledge of mathematics I'd say

In the limit as population goes to infinity the average number of kegs goes to two.

Thank you, hank you no applause please.

 

[bilby]

In the limit as population goes to infinity the average number of kegs goes to two.

My beer fridge contains four kegs, so that's twice the expected number.

 

[steve_bank]

Seriously. What is more amazing then 1 + 1 = 2?

 

[bilby]

Seriously. What is more amazing then 1 + 1 = 2?

2+2=4

You're welcome.

 

[Gospel]

cat + box = alive cat + dead cat

Ok maybe that belongs in the all things cats thread.

 

[Alcoholic Actuary]

Ingenious, but simple:

You are in line to board an airplane. The seats are numbered 1-100 and passengers line up with ticket numbers 1-100 such that person 1 takes seat 1, and person 2 takes seat 2 etc... and unfortunately you are passenger 100. The following situation ensues - passenger 1 takes a random seat (like Southwest Airlines) instead of seat 1. The rest of the passengers boarding follow the same algorithm: if their seat number is available they take it and if someone is sitting in their seat, they take a different remaining seat at random. What is the probability that you (at seat 100) gets your assigned seat?

(for the purpose of this problem assume that taking a seat 'at random' is uniformly distributed among the remaining seats - no preference for window, aisle, front, exit row, or back of plane are considered).

Ans:
0.5 - either someone is in your seat or they aren't. Why?

aa

 

[steve_bank]

oops

 

[steve_bank]

Re framing the problem.

You have a box with 100 balls marked 1-100, 100 people line up with numbers 1-100 on their chest in sequence.

#1 picks a ball randomly and puts the ball on a table.

If any of the balls on the table is 2 then #2 randomly picks a ball and puts it on the table. If 2 is not on the table #2 takes 2 out of the box and puts it on the table.

If any of the balls on the table is 3 then #3 randomly picks a ball and puts it on the table. If 3 is not on the table #3 takes 3 out of the box and puts it on the table.

And so on.

It is similar to random sampling from a population without replacement which would be
(1/n) * (1/n-1) * (1/n-2)....

The probability for the first pick by #1 is 1/100 for any number. So the probability for 100 remaining for #2 to pick is 1 - 1/100, but the pick by #2 is conditional depending on whether o not #1 picks ball 2.

I don't know the notation to express the conditional probability, but I could code a simulation.

It would be something like
(1-1/n)*(1-1/n-1).... where the inclusion of each term depends on the previous sequence of picks.

That's my cut at it.

 

[Swammerdami]

Ingenious, but simple:

You are in line to board an airplane. The seats are numbered 1-100 and passengers line up with ticket numbers 1-100 such that person 1 takes seat 1, and person 2 takes seat 2 etc... and unfortunately you are passenger 100. The following situation ensues - passenger 1 takes a random seat (like Southwest Airlines) instead of seat 1. The rest of the passengers boarding follow the same algorithm: if their seat number is available they take it and if someone is sitting in their seat, they take a different remaining seat at random. What is the probability that you (at seat 100) gets your assigned seat?

(for the purpose of this problem assume that taking a seat 'at random' is uniformly distributed among the remaining seats - no preference for window, aisle, front, exit row, or back of plane are considered).

Ans:
0.5 - either someone is in your seat or they aren't. Why?

aa

This is one of those probability problems which looks ridiculously complicated. But your only asked about passenger #100 (the last passenger) and that allows the complexity to be short-circuited. The puzzle has a simple answer that can be explained in a short paragraph. aa's spoiler is a good start, but ingenuity is still needed.

I've seen the puzzle before, so should not be allowed to compete.

 

[Swammerdami]

Ingenious, but simple:

You are in line to board an airplane. The seats are numbered 1-100 and passengers line up with ticket numbers 1-100 such that person 1 takes seat 1, and person 2 takes seat 2 etc... and unfortunately you are passenger 100. The following situation ensues - passenger 1 takes a random seat (like Southwest Airlines) instead of seat 1. The rest of the passengers boarding follow the same algorithm: if their seat number is available they take it and if someone is sitting in their seat, they take a different remaining seat at random. What is the probability that you (at seat 100) gets your assigned seat?

If the passengers line up in RANDOM order, and we replace "passenger #1" and "passenger #100" with "passenger first to board" and "passenger last to board", the puzzle has the exact same solution as the stated puzzle.

However the strict (1,2,3,4,...) ordering Alcoholic Actuary imposes makes the puzzle somewhat easier to visualize.

Which seat(s) are certainly already taken when passenger #3 boards? When passenger #4 boards? When passenger #100 boards?

 

[Alcoholic Actuary]

Ingenious, but simple:

You are in line to board an airplane. The seats are numbered 1-100 and passengers line up with ticket numbers 1-100 such that person 1 takes seat 1, and person 2 takes seat 2 etc... and unfortunately you are passenger 100. The following situation ensues - passenger 1 takes a random seat (like Southwest Airlines) instead of seat 1. The rest of the passengers boarding follow the same algorithm: if their seat number is available they take it and if someone is sitting in their seat, they take a different remaining seat at random. What is the probability that you (at seat 100) gets your assigned seat?

If the passengers line up in RANDOM order, and we replace "passenger #1" and "passenger #100" with "passenger first to board" and "passenger last to board", the puzzle has the exact same solution as the stated puzzle.

However the strict (1,2,3,4,...) ordering Alcoholic Actuary imposes makes the puzzle somewhat easier to visualize.

Which seat(s) are certainly already taken when passenger #3 boards? When passenger #4 boards? When passenger #100 boards?

Actually, a good way to solve the problem is to start with a 2 seat airplane (you are seat 2). Then work on a 3 seat plane and maybe 4.

Then try to extend the results to an 'N' numbered seat plane to prove your results (this will save you the effort of attempting to write code for a 100 seat plane).

Spoiler: Passenger Last to Board

passenger last to board has the same probability on a 100 person aircraft and a 1000 person aircraft - as Swami noted.

aa

 

[steve_bank]

Ingenious, but simple:

You are in line to board an airplane. The seats are numbered 1-100 and passengers line up with ticket numbers 1-100 such that person 1 takes seat 1, and person 2 takes seat 2 etc... and unfortunately you are passenger 100. The following situation ensues - passenger 1 takes a random seat (like Southwest Airlines) instead of seat 1. The rest of the passengers boarding follow the same algorithm: if their seat number is available they take it and if someone is sitting in their seat, they take a different remaining seat at random. What is the probability that you (at seat 100) gets your assigned seat?

If the passengers line up in RANDOM order, and we replace "passenger #1" and "passenger #100" with "passenger first to board" and "passenger last to board", the puzzle has the exact same solution as the stated puzzle.

However the strict (1,2,3,4,...) ordering Alcoholic Actuary imposes makes the puzzle somewhat easier to visualize.

Which seat(s) are certainly already taken when passenger #3 boards? When passenger #4 boards? When passenger #100 boards?

Actually, a good way to solve the problem is to start with a 2 seat airplane (you are seat 2). Then work on a 3 seat plane and maybe 4.

Then try to extend the results to an 'N' numbered seat plane to prove your results (this will save you the effort of attempting to write code for a 100 seat plane).

Spoiler: Passenger Last to Board

passenger last to board has the same probability on a 100 person aircraft and a 1000 person aircraft - as Swami noted.

aa

Can you write out the sequence of probabilities?

 

[Swammerdami]

Can you write out the sequence of probabilities?

Spoiler

Last to board (Psgr #100) has a 1/2 chance of getting his correct seat.
2nd-to-last to board (Psgr #99) has a 2/3 chance of getting his correct seat.
3rd-to-last to board (Psgr #98) has a 3/4 chance of getting his correct seat.
4th-to-last to board has a 4/5 chance.
5th-to-last to board has a 5/6 chance.
And so on.