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Proof of square root 3 irrational using well ordering

  1. Feb 6, 2013 #1
    The part I dont understand is how they show there exists a smaller element. They assume s=t√3 is the smallest element of S={a=b√3: a,b€Z} . Then what they do is add s√3 to both sides and get s√3-s=s√3-t√3. I don't get how they thought of that or why it works.I know there exists an element smaller than S but the way they prove is confusing.
     
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  3. Feb 6, 2013 #2

    tiny-tim

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    hi bonfire09! :smile:

    √3 = s/t = 3t/s

    3t - s = √3(s - t)

    but 3t -s < s (because it's (√3 - 1)s, or about 0.7s) :wink:
     
  4. Feb 7, 2013 #3
    So do I have it right?

    Since t=s√3 we can rewrite as t√3=s ⇔ 3t=s√3. So we subtract s from both side and we get [itex] 3t-s=s√3-s ⇔ 3t-s=s√3-t√3⇔3t-s=√3(s-t)⇔s√3-s= √3(s-t)⇔s(√3-1)=√3(s-t) [/itex] But this is a contradiction since√3-1<√3 and s-t< s so √3 is irrational. Would this be a better way of restating this part of the proof?
     
  5. Feb 7, 2013 #4

    tiny-tim

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    hi bonfire09! :smile:

    (just got up :zzz:)
    (why do you say "subtract s from both sides"?

    what you're actually doing is "subtract one equation from the other" :wink:)

    your equations should stop here …

    you have now proved that if the pair (s,t) is in S, then so is (3t-s,s-t), because 3t-s and s-t are obviously in Z
    the first part is a little unclear … you haven't specifically said what √3 - 1 has to do with it!

    also, it would be better if you used the word "ordering" somewhere! :smile:
     
  6. Feb 7, 2013 #5
    Thanks the part where you said to subtract both equations instead of s was what I was confused about.
     
  7. Feb 7, 2013 #6

    tiny-tim

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    ah, what i meant was:

    you have two equations

    s = t√3

    3t = s√3​

    if you subtract them you immediately get

    (3t - s) = (t -s)√3​

    and both brackets are clearly in Z :smile:

    (your way, which is to subtract s from both sides of 3t = s√3, gives you (3t -s) = s(√3 - 1), which is correct, but the RHS isn't obviously an integer times √3, so you have to waste time proving that it is :wink:)
     
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