Go to the thread and click "report" on the post you want to correct and type in a request to the mods to re-open the thread for you to do so.rickhev said:I wanted to correct the answer to a thread, but it has been closed. What should I do? (The thread is https://www.physicsforums.com/threa...-2-the-square-root-of-2-is-irrational.270791/.)
Thanks, I followed your suggestion.phinds said:Go to the thread and click "report" on the post you want to correct and type in a request to the mods to re-open the thread for you to do so.
Why not?BobG said:You want to correct an error that was made seven years ago?
What do you mean? I've already found it.mfb said:Try to find the error.
;)
Thanks, it's mostly fixed. The phrase 'irrational multiplied by a rational' should also be changed to 'irrational multiplied by a non-zero rational'. Would you please make that change too?mfb said:Try to find it in the old thread. I found one and fixed it - was just a typo I guess.
rickhev said:Thanks, it's mostly fixed. The phrase 'irrational multiplied by a rational' should also be changed to 'irrational multiplied by a non-zero rational'. Would you please make that change too?
ehj said: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.
Looks good to me. Thanks very much.berkeman said:I made the change, and added some whitespace to make it more readable. Does this look right now?
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