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Square root 2 is irrational

  1. Aug 24, 2010 #1
    sorry, im using phone, owho i hope you can get it.

    suppose its rational. in the form a/b gcd(a,b)=1
    and so and so and so and then they concluded that a^2=2 is a contradiction.

    but i cannot see what it conradict the assumption. help
     
  2. jcsd
  3. Aug 24, 2010 #2

    Dick

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    a^2=2 isn't the contradiction. I'm guessing you didn't read the whole proof. Try that and then check back if you still don't get it.
     
  4. Aug 24, 2010 #3
    i'll denote a^2 is a2

    then a2=2b2
    hence b l a2
    if b>1 <skipped> we have contradiction. so b=1. hence a2=2, a contradiction. so squareroot 2 is not rational.

    we can't conclude, a2=2, a contradiction right? i cant concentrate on other lecture because this is bothering me. and of course i know how to prove it in other way
     
  5. Aug 24, 2010 #4

    Dick

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    Ok, so b | a^2. Now why is b>1 supposed to be a contradiction? Why did you write <skipped>?
     
  6. Aug 24, 2010 #5
    sorry i use phone, now i'm on computer,

    so if b>1, then there exist a prime number p such that p l b, so p l a^2, then p l a, hence gcd(a,b)=p>1, a contradiction, so b must be 1. so that prove is inconclusive right?

    btw, if you have time, can you response in this https://www.physicsforums.com/showthread.php?t=424018 [Broken]

    thanks ^^
     
    Last edited by a moderator: May 4, 2017
  7. Aug 24, 2010 #6

    Dick

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    That looks ok to me. Except you should write gcd(a,b)>=p if all you know is that p | a and p | b.
     
    Last edited by a moderator: May 4, 2017
  8. Aug 24, 2010 #7
    so do you mean a^2=2 is a contradiction?
     
  9. Aug 24, 2010 #8

    Dick

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    Well, yes. Contrary to what I said in post 2. I wasn't following your proof. a^2=2 is a contradiction because a is an integer and there is no integer whose square is 2. You are basically proving that the only integers who have rational square roots are the perfect squares.
     
  10. Aug 24, 2010 #9
    thankssss
     
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