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Proof of square root 3 irrational using well ordering 
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#1
Feb613, 01:45 PM

P: 232

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√3s=s√3t√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.



#2
Feb613, 05:10 PM

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hi bonfire09!
√3 = s/t = 3t/s 3t  s = √3(s  t) but 3t s < s (because it's (√3  1)s, or about 0.7s) 


#3
Feb713, 12:02 AM

P: 232

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] 3ts=s√3s ⇔ 3ts=s√3t√3⇔3ts=√3(st)⇔s√3s= √3(st)⇔s(√31)=√3(st) [/itex] But this is a contradiction since√31<√3 and st< s so √3 is irrational. Would this be a better way of restating this part of the proof? 


#4
Feb713, 02:09 AM

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Proof of square root 3 irrational using well ordering
hi bonfire09!
(just got up ) what you're actually doing is "subtract one equation from the other" ) your equations should stop here … you have now proved that if the pair (s,t) is in S, then so is (3ts,st), because 3ts and st are obviously in Z also, it would be better if you used the word "ordering" somewhere! 


#5
Feb713, 10:31 AM

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Thanks the part where you said to subtract both equations instead of s was what I was confused about.



#6
Feb713, 10:39 AM

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ah, what i meant was:
you have two equations s = t√3if you subtract them you immediately get (3t  s) = (t s)√3and both brackets are clearly in Z (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 ) 


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