Proof about irrational numbers.

  • Thread starter cragar
  • Start date
  • #1
2,552
3

Homework Statement


Prove that [itex] \sqrt{6} [/itex] is irrational.

The Attempt at a Solution



Would I just do a proof by contradiction and assume that [itex] \sqrt{6} [/itex] is rational and then get that [itex] 6q^2=p^2 [/itex] 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 [itex] \sqrt{6} [/itex] is [itex] \sqrt{3}\sqrt{2} [/itex] and then say that an irrational times an irrational is an irrational as long as its not the same irrational.
 

Answers and Replies

  • #2
22,129
3,297
Would I just do a proof by contradiction and assume that [itex] \sqrt{6} [/itex] is rational and then get that [itex] 6q^2=p^2 [/itex] 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 [itex] \sqrt{6} [/itex] is [itex] \sqrt{3}\sqrt{2} [/itex] 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.
 
  • #3
2,552
3
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
22,129
3,297
ok thanks for your response. on the second one I cant think of a counterexample off hand. maybe i should think about it more.

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

or

[itex]\sqrt{2}*\frac{1}{\sqrt{2}}[/itex]

or

[itex]\sqrt{18}*\sqrt{2}[/itex]
 
  • #5
409
0
ok thanks for your response. on the second one I cant think of a counterexample off hand. maybe i should think about it more.

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

Damn you Micro you're too quick.
 
  • #6
2,552
3
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
22,129
3,297
[itex]\sqrt{81} = \sqrt[]{3}\sqrt[]{27}[/itex]

Damn you Micro you're too quick.

Ya snooze, ya lose!! :biggrin:

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??
 
  • #8
2,552
3
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 .
[itex] (P_1P_2P_3.......P_r)^{\frac{1}{n}}=\frac{x}{y}[/itex]
then we take both sides to then power of n and then multiply the y^n over
and we get [itex] y^n(P_1P_2P_3.......P_r)=x^n[/itex]
therefor this implies that x^n is divisible by a prime. so we will now write x=aP
and we get that [itex] y^n(P_1P_2P_3.......P_r)=(aP)^n[/itex]
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:

Related Threads on Proof about irrational numbers.

  • Last Post
Replies
5
Views
3K
  • Last Post
Replies
6
Views
1K
Replies
3
Views
6K
Replies
8
Views
4K
Replies
1
Views
13K
  • Last Post
Replies
4
Views
5K
Replies
14
Views
8K
Replies
3
Views
7K
Replies
2
Views
7K
Top