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

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Homework Help Overview

The discussion revolves around proving the irrationality of the sum of the cubic root of 2 and the square root of 2. Participants explore the nature of irrational numbers and their sums, particularly focusing on the characteristics of the individual components.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Some participants attempt to reason through the properties of irrational numbers and their sums, questioning whether the presence of infinitely many decimal places in both components guarantees the irrationality of their sum. Others challenge this reasoning by providing counterexamples of rational sums.

Discussion Status

The discussion includes various viewpoints on the nature of irrational numbers and their sums. While some participants express confusion about the validity of certain arguments, others propose a potential proof, indicating a mix of exploration and clarification without reaching a consensus.

Contextual Notes

Participants are navigating the complexities of irrational numbers and their sums, with some expressing uncertainty about the implications of decimal representations. The original poster's attempt to establish a proof suggests a deeper inquiry into the properties of these mathematical constructs.

ehj
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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 :/.
 
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ehj said:
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.
 
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.
 
asleight said:
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...).
 
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.
 
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