# Proof of square root 3 irrational using well ordering

1. Feb 6, 2013

### bonfire09

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.

2. Feb 6, 2013

### tiny-tim

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. Feb 7, 2013

### bonfire09

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 $3t-s=s√3-s ⇔ 3t-s=s√3-t√3⇔3t-s=√3(s-t)⇔s√3-s= √3(s-t)⇔s(√3-1)=√3(s-t)$ 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?

4. Feb 7, 2013

### tiny-tim

hi bonfire09!

(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" )

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!

5. Feb 7, 2013

### bonfire09

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

6. Feb 7, 2013

### tiny-tim

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

(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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