Proof that 3rd root of 3 is irrational?

[Solo177]

My goal is to prove that the 3rd root of 3 is irrational. I use the same method as proving sqrt(2) is irrational, however, I ran into a question.

When proving sqrt(2), we use a statement that because a^2 = 2b^2, a^2 is even and therefore a must also be. This is intuitive, but for 3rd root of 3, you get:

a^3 = 3b^3. Is it safe to use the logic in the aqrt(2) proof and say that a^3 is divisible by 3, so a must also be divisble by 3? If so, could someone supply me with the method of proving that statement? Thanks!

 

[reitz]

Check here.

 

[Ladies Man]

i've got a headache now

 

[AlphaIVT]

e=mc^2, hope that helps

 

[snakesnfrogs]

LMAO @ reitz

 

[Stealth1024]

if its an odd numbered problem its probably in the back of your math book

 

[Jothaxe]

Is it safe to use the logic in the aqrt(2) proof and say that a^3 is divisible by 3, so a must also be divisble by 3? If so, could someone supply me with the method of proving that statement? Thanks!

Yes the logic is like this:

if we have integers (a,b) s.t. a^3 = 3*b^3 --> a^3 is divisible by 3.

if a^3 is divisible by 3 (and a is an integer) then a is divisible by 3 also, because 3 is a prime, and by definition we cant multiply any number of other integers together to get a prime.

Now that I look at it, this logic seems a little bit shaky. Take it for what its worth!

 

[nomahe]

How do you spell that?

 

[MajesticMoose]

well my calculator says that 3^(1/3) = 1.4422495703074083823216383107801. And since no rational person would ever care to know that many decimal places when solving someting, it can be called irrational.

i hope that helped

m00se