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.