Is sqrt{30} Irrational? Proving its Irrationality through Unique Factorization

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Discussion Overview

The discussion revolves around the proof of the irrationality of sqrt{30} using unique factorization into primes. Participants explore various methods and reasoning related to the proof, including comparisons to the proof of the irrationality of sqrt{2}.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant presents a proof using unique factorization but expresses uncertainty about its correctness.
  • Another participant references the proof of sqrt{2}'s irrationality as a similar method that could be applied here.
  • Some participants suggest that the unique factorization approach leads to a contradiction due to differing counts of prime factors on each side of the equation.
  • There is a suggestion that the proof presented contains unnecessary steps and could be simplified.
  • Participants discuss the importance of using unique factorization specifically, which leads to confusion about the applicability of alternative methods.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correctness of the initial proof or the necessity of the unique factorization approach. Multiple competing views on the proof's structure and validity remain present.

Contextual Notes

Some participants express confusion regarding the application of unique factorization and its relation to other methods of proving irrationality. There are mentions of potential unnecessary complexity in the proof provided.

Who May Find This Useful

Readers interested in proofs of irrationality, unique factorization, and mathematical reasoning may find this discussion relevant.

sutupidmath
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Well, there is a problem, i have solved/proved it, but i am not sure whether it is correct.
THe problem is this:

Using unique factorization into primes prove that there are no integers a and b such that a^2=30b^2, and thus show that sqrt{30} is irrational.

Proof:using unique factorization of any integer greater than 1 or less than -1, we can factor any such integer into the product of powers of distinct primes, or simply into a product of primes.

a^2=30b^2=>b^2|a^2=>b|a=>\exists k,a=kb

Let:

a=p_1p_2...p_r; b=q_1q_2...q_s

a^2=30b^2=>30=\left(\frac{b}{a}\right)^2=k^2=>\sqrt{30}=k

Now from the unique factorization theorem again:\sqrt{30}=k=d_1d_2...d_n=>30=d_1^2d_2^2...d_n^2

=>

2*3*5=d_1^2d_2^2...d_n^2=>2|d_1^2d_2^2...d_n^2=>2|d_i^2=>2=d_i

but this would contradict the unique factorization theorem, and thus this contradiction shows that such a, and b do not exist.

Is this about correct, or there is another way around it?
 
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Recall how do you prove sqrt(2) is irrational: a^2=2b^2 (where gcd(a,b)=1) now 2 divides a^2 and thus 2 divides a and thus 4 divides a^2, so a=2k, 4k^2=2b^2 => b^2=2k^2 so also 2 divides b^2 and thus divides b, which means that gcd(a,b)>1 which is a contradiction, the same method is used here as well.
 
Actually you can use the unique factorization thus:

<br /> a^2=30b^2<br />

There are an even number of 2's on the left, but an odd number on the right. Contradiction.
 
well, yah i thought about this one, but since they asked to use the unique factorization of a number into primes, that part through me off, and i didn't know whether the same method is applied here.
 
sutupidmath said:
well, yah i thought about this one, but since they asked to use the unique factorization of a number into primes, that part through me off, and i didn't know whether the same method is applied here.

Which method? Why can't you use that one-line proof?
 
Dragonfall said:
Actually you can use the unique factorization thus:

<br /> a^2=30b^2<br />

There are an even number of 2's on the left, but an odd number on the right. Contradiction.

THis is pretty much what my proof eventually shows, that we will have more 2's in one side than on the other.
 
Dragonfall said:
Which method?

Here i was referring to mathematical physicists's post. I know how to show that a nr is irrational using that methodology.
 
sutupidmath said:
THis is pretty much what my proof eventually shows, that we will have more 2's in one side than on the other.

Yes but you can say this immediately. There's a lot of unnecessary stuff in your proof.
 
  • #10
Just look at the prime factors on both sides... done.
 
  • #11
yeah i got it! thnx for the input!
 

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