Proove that the cubic root of 2 + the square root of 2 is irrational

[ehj]

How do you show that the cubic root of two + the square root of two is irrational? I can easily show that each of these numbers is irrational, but not the sum :/.

 

[asleight]

How do you show that the cubic root of two + the square root of two is irrational? I can easily show that each of these numbers is irrational, but not the sum :/.

If both of the numbers have infitely many decimals, there can never be a terminating digit for their sum.

 

[ehj]

Thats not a proof? And besides, doesn't 1 - sqrt(2) and 1 + sqrt(2) both have infinately many decimals, nevertheless their sum is rational, 2.

 

[Mark44]

If both of the numbers have infitely many decimals, there can never be a terminating digit for their sum.

But 1/3 = .333333... has infinitely many decimal places and it is rational as are 1/2 = .50000... (can also be written as .49999999...) and 1/5 = .200000 (can also be written as .1999999...).

 

[ehj]

I figured it out. Posting solution in case sombody might run into the same problem in the future :P

I assume 2^(1/3) + 2^(1/2) = a , where a is rational

=> 2=(a-2^(1/2))^3 <=> 2 = (a^3 + 6a) + sqrt(2)(-3a^2 -2)

Which is a contradiction since sqrt(2)(-3a^2 -2) is an irrational multiplied by a non-zero rational, which can be proved to always be irrational, and the sum of a rational (a^3 + 6a) and an irrational can be proved to always be irrational, and above cannot equal 2 since 2 is rational.