Question

Prove that log_2(3) is irrational. @ganeshie8 @myininaya @UnkleRhaukus @RadEn @robtobey @chmvijay @Luigi0210 @eliassaab @genius12 @Vincent-Lyon.Fr @Preetha @Kainui @ehuman @wolfe8 @lucaz @INeedHelpPlease? @khadeeja @thadyoung @A_clan @tester97 @Andras @Confusionist @linh412986 @BlackLabel @leozap1 @JMark @Microrobot @zacharyf @Rubio101 @NaomiBell1997 @MeganEdward @theballer225 @link,zoro @Goldenmoon @CrayolaCrayon_

Answer 1

Assume log(base 2)3 is rational

This means that log(base 2)3 = a/b where a and b are integers

2^(a/b) = 3
2^a = 3^b

Note that 2^a MUST be even if 'a' is an integer
Note that 3^b MUST be odd if 'b' is an integer

Hence a and b cannot both be integers

Hence a/b is irrational

Hence, using proof by contradiction, log(base 2)3 is irrational

Answer 2

To add to @Yttrium 's proof, we have that equality occurs when a = b = 0. And so it is better if we had initially defined that log_2(3)=a/b, where a and b are coprime positive integers. Since 2^a = 3^b is never true for positive integers a and b as the two sides are always of opposite parity, we reach a contradiction. Hence log_2(3) is irrational. @FutureMathProfessor