#### Unknown Soldier

##### Senior Member

_{2}(3) is an irrational number. I tried the following proof by contradiction:

If log

_{2}(3) is a rational number, then there are integers

*a*> 0 and

*b*> 0 where log

_{2}(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 log

_{2}(3) is irrational.

Although I think I proved the proposition that log

_{2}(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 log

_{2}(3) is a rational number, then there are integers

*a*> 0 and

*b*> 0 where log

_{2}(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

^{b}or 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 log

_{2}(3) is a rational number results in this contradiction, then log

_{2}(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.