# Prove that it is irrational?

1. Jun 15, 2009

### 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?

2. Jun 15, 2009

### sutupidmath

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

3. Jun 15, 2009

### 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.

4. Jun 15, 2009

### 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.

5. Jun 15, 2009

### 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.

6. Jun 15, 2009

### Dragonfall

Which method? Why can't you use that one-line proof?

7. Jun 15, 2009

### sutupidmath

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

8. Jun 15, 2009

### sutupidmath

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

9. Jun 15, 2009

### Dragonfall

Yes but you can say this immediately. There's a lot of unnecessary stuff in your proof.

10. Jun 15, 2009

### JasonRox

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

11. Jun 16, 2009

### sutupidmath

yeah i got it! thnx for the input!