Register to reply

Proof of square root 3 irrational using well ordering

by bonfire09
Tags: irrational, ordering, proof, root, square
Share this thread:
bonfire09
#1
Feb6-13, 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√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.
Phys.Org News Partner Science news on Phys.org
'Office life' of bacteria may be their weak spot
Lunar explorers will walk at higher speeds than thought
Philips introduces BlueTouch, PulseRelief control for pain relief
tiny-tim
#2
Feb6-13, 05:10 PM
Sci Advisor
HW Helper
Thanks
tiny-tim's Avatar
P: 26,148
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)
bonfire09
#3
Feb7-13, 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] 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?

tiny-tim
#4
Feb7-13, 02:09 AM
Sci Advisor
HW Helper
Thanks
tiny-tim's Avatar
P: 26,148
Proof of square root 3 irrational using well ordering

hi bonfire09!

(just got up )
Quote Quote by bonfire09 View Post
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)
(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
But this is a contradiction since√3-1<√3 and s-t< s so √3 is irrational.
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!
bonfire09
#5
Feb7-13, 10:31 AM
P: 232
Thanks the part where you said to subtract both equations instead of s was what I was confused about.
tiny-tim
#6
Feb7-13, 10:39 AM
Sci Advisor
HW Helper
Thanks
tiny-tim's Avatar
P: 26,148
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 )


Register to reply

Related Discussions
Square Root of an Irrational Number is Irrational Calculus & Beyond Homework 26
Proving square root of 2 is irrational with well ordering principle? Calculus & Beyond Homework 3
Square root of 3 is irrational General Math 2
Square root 2 is irrational Calculus & Beyond Homework 8
Proove that the cubic root of 2 + the square root of 2 is irrational Calculus & Beyond Homework 4