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

[sutupidmath]

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?

 

[sutupidmath]

i don't know what's wrong with latex?

 

[MathematicalPhysicist]

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.

 

[Dragonfall]

Actually you can use the unique factorization thus:

$$a^2=30b^2$$

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

 

[sutupidmath]

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.

 

[Dragonfall]

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?

 

[sutupidmath]

Actually you can use the unique factorization thus:

$$a^2=30b^2$$

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.

 

[sutupidmath]

Which method?

Here i was referring to mathematical physicists's post. I know how to show that a nr is irrational using that methodology.

 

[Dragonfall]

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.

 

[JasonRox]

Just look at the prime factors on both sides... done.

 

[sutupidmath]

yeah i got it! thnx for the input!