Proof about irrational numbers.

[cragar]

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.

 

[micromass]

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.

 

[cragar]

ok thanks for your response. on the second one I can't think of a counterexample off hand. maybe i should think about it more.

 

[micromass]

ok thanks for your response. on the second one I can't think of a counterexample off hand. maybe i should think about it more.

\sqrt[3]{4}*\sqrt[3]{2}

or

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

or

\sqrt{18}*\sqrt{2}

 

[rollcast]

ok thanks for your response. on the second one I can't 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.

 

[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.

 

[micromass]

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

Damn you Micro you're too quick.

Ya snooze, you 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??

 

[cragar]

Okay tell me if this works. let's assume we have a multiplication of primes to the nth root.
and let's 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.