how to prove that log_2(3) is irrational by using contradiction

Показать описание

Here we will show that the log base 2 of 3 is irrational by using contradiction.

Use "WELCOME10" for 10% off

-----------------------------
If you find my channel helpful and would like to support it 💪, then you can
-----------------------------
"Just Algebra" (by blackpenredpen) is dedicated to helping middle school, high school, and community college students who need to learn algebra. Topics include how to solve various equations (linear equations, quadratic equations, square root equations, rational equations, exponential equations, logarithmic equations, and more), factoring techniques, word problems, functions, graphs, Pythagorean Theorem, and more. We will also cover standardized test problems such as the SAT. Feel free to leave your questions in the comment!
-----------------------------
#justalgebra

bprp math basics

Рекомендации по теме

Комментарии

That's a good one. Simple and nicely illustrates the point!

michaeledwardharris

Also, notice the prime factorizations of 3^n and 2^m are different. Taking them equal contradicts the fundamental theorem of arithmetic which requires factorization to be unique.

williamperez-hernandez

Thank you so much, i had an argument with my friend and sent this, i instantly won!

eutopian

“The input is bigger than the base“ is not the reasom for the number to be positive, it is “the input is greater than 1“. “The input is bigger than the base“ is the reason for m>n.

WolfgangKais

so 0/0 solves it, if we don't take into account it would destroy the universe as we know it

ZReChannel

I went about it slightly differently, but same result nonetheless.
Used COB to get Log 3 / Log 2 = m/n
cross multiply to get n log 3 = m log 2
moved the variables to the exponent to get log (3^n) = log (2^m)
logs are equal means inputs are equal (or just raise both sides to 10^LHS = 10^RHS to get rid of logs, 3^n = 2^m, same contradiction

JeffCox