Proof of square root 3 irrational using well ordering

In summary, the part I don't 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.
  • #1
bonfire09
249
0
The part I don't 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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  • #2
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:
 
  • #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?
 
  • #4
hi bonfire09! :smile:

(just got up :zzz:)
bonfire09 said:
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" :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
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! :smile:
 
  • #5
Thanks the part where you said to subtract both equations instead of s was what I was confused about.
 
  • #6
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:)
 

1. What is the concept of "Proof of square root 3 irrational using well ordering"?

The proof of square root 3 irrational using well ordering is a mathematical proof that shows the irrationality of the square root of 3 by using the well-ordering principle. This principle states that every non-empty set of positive integers has a smallest element. By assuming that the square root of 3 is rational, this proof shows that there would be a contradiction to the well-ordering principle, thus proving that the square root of 3 is indeed irrational.

2. Why is it important to prove that the square root of 3 is irrational?

The proof of the irrationality of the square root of 3 is important because it is a fundamental concept in mathematics. It helps us understand the properties of irrational numbers and provides a deeper understanding of number theory. This proof also has applications in other areas of mathematics, such as in the proof of the irrationality of other square roots.

3. What is the well-ordering principle and how is it used in this proof?

The well-ordering principle states that every non-empty set of positive integers has a smallest element. In this proof, the well-ordering principle is used to show that if we assume the square root of 3 is rational, then we can find a contradiction to the principle. This contradiction proves that our assumption is false, and therefore the square root of 3 must be irrational.

4. How does this proof differ from other proofs of the irrationality of the square root of 3?

There are several different proofs of the irrationality of the square root of 3, but this proof using well ordering is unique in that it relies on the well-ordering principle. Other proofs may use methods such as contradiction, continued fractions, or the Euclidean algorithm. However, all of these proofs ultimately lead to the same conclusion – the square root of 3 is irrational.

5. Can this proof be applied to other irrational numbers?

Yes, the proof of the irrationality of the square root of 3 using well ordering can be applied to other square roots. It can also be modified to prove the irrationality of other numbers, such as pi or e. However, the specific details of the proof may vary depending on the irrational number in question.

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