1. Jan 21, 2012

### cragar

1. The problem statement, all variables and given/known data
Prove that $\sqrt{6}$ is irrational.
3. The attempt at a solution

Would I just do a proof by contradiction and assume that $\sqrt{6}$ is rational and then get that $6q^2=p^2$ which would imply that p is even so I put in p=2r
and then multiply it out. then this would imply that q is also even and this is a contradiction because they would have factors in common. I know I skipped some of the steps. Could I also make an argument that $\sqrt{6}$ is $\sqrt{3}\sqrt{2}$ and then say that an irrational times an irrational is an irrational as long as its not the same irrational.

2. Jan 21, 2012

### micromass

Staff Emeritus
Sounds good.

Uuh, that isn't true. There are many counterexamples.

3. Jan 21, 2012

### cragar

ok thanks for your response. on the second one I cant think of a counterexample off hand. maybe i should think about it more.

4. Jan 21, 2012

### micromass

Staff Emeritus
$\sqrt[3]{4}*\sqrt[3]{2}$

or

$\sqrt{2}*\frac{1}{\sqrt{2}}$

or

$\sqrt{18}*\sqrt{2}$

5. Jan 21, 2012

### rollcast

$\sqrt{81} = \sqrt[]{3}\sqrt[]{27}$

Damn you Micro you're too quick.

6. Jan 21, 2012

### cragar

ya but all the counterexample have common factors under the radical. I was saying that there, well I was thinking that there were no common factors under the radical.
What if i said a prime number that is square rooted times a different prime that is square rooted will be irrational.

7. Jan 21, 2012

### micromass

Staff Emeritus
Ya snooze, ya lose!!

Then the statement is probably true. But did you prove the statement??

8. Jan 21, 2012

### cragar

Okay tell me if this works. lets assume we have a multiplication of primes to the nth root.
and lets assume that it is rational and that they have no common factors .
$(P_1P_2P_3.......P_r)^{\frac{1}{n}}=\frac{x}{y}$
then we take both sides to then power of n and then multiply the y^n over
and we get $y^n(P_1P_2P_3.......P_r)=x^n$
therefor this implies that x^n is divisible by a prime. so we will now write x=aP
and we get that $y^n(P_1P_2P_3.......P_r)=(aP)^n$
and this would imply that y is divisible by some prime in our list. and this would imply that x and y share a common factor which is a contradiction.
not sure if my last line of reasoning is valid.

Last edited: Jan 21, 2012