## Homework Statement

Prove that $\sqrt{6}$ is irrational.

## 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.

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.

Sounds good.

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.

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

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

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

$\sqrt{4}*\sqrt{2}$

or

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

or

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

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

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

Damn you Micro you're too quick.

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.

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

Damn you Micro you're too quick.

Ya snooze, ya lose!! 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.

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

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.

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