SOLUTION: Using proof by contradiction, show that (3+√2)/3 is irrational.

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Question 1130147: Using proof by contradiction, show that (3+√2)/3 is irrational.
Answer by math_helper(2461) About Me  (Show Source):
You can put this solution on YOUR website!
I will assume that we accept +sqrt%282%29+ as irrational (proof provided at the end, for completeness):
Assume +%283+%2B+sqrt%282%29%29+%2F+3+=+p%2Fq+ where p,q are relatively prime integers, both greater than zero

++3p%2Fq+=+3%2Bsqrt%282%29+
Write 3p as P, where P is obviously also an integer:
++P%2Fq+=+3%2Bsqrt%282%29+

Re-arrange:
++q%2Asqrt%282%29+=+P-3q+

The RHS is integer minus integer, which must produce another integer (integers are closed under subtraction). Call it R:
++q%2Asqrt%282%29+=+R+

+++sqrt%282%29+=+R%2Fq+ where R and q are integers, a contradiction.


Therefore, +%283%2Bsqrt%282%29%29%2F3+ is irrational.
——

Proof that +sqrt%282%29+ is irrational:
+sqrt%282%29+=+p%2Fq+ p,q relatively prime integers

++2+=+p%5E2%2Fq%5E2+

+2q%5E2+=+p%5E2+

LHS is even —> RHS must also be even, but this contradicts p,q being relatively prime. Thus sqrt%282%29 is irrational.