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I have no idea how to go about this, except that we are supposed to use proof by contradiction to show that the square root of 3 is an irrational number. Any help or tips is appreciated. Thanks.
Assume 4 is an irrational. Then by definition, it cannot be written as a fraction. That is:Hehe! I just wanted to open almost a same kind of topic.
I don't really get proving by contradiction. What if I say, prove that square root of 4 is an irrational number. If my answer comes out as true, it is a rational number(contradiction assumption) then it is not an irrational.
But by using statements such as they are relative primes I can get every square number to be irrational. Or because they are perfect squares we know they are rational?
If someone can tell/prove me by using the proof of contradiction that the square root of 4 is not an irrational I will make him/her a chocolate cake which I'll eat.
<3,
Icelove
Now I got the aaaaaah!(not yawning, or pain or others things) moment. I love you. :)Assume 4 is an irrational. Then by definition, it cannot be written as a fraction. That is:
[tex]\forall u,v\in Z, (u,v)=1, 4\not=\frac{u}{v}.[/tex]
However,
[tex]4=\frac{4}{1}[/tex] thus contradicting the assumption that 4 is an irrational.
2^2 = 4. So 2 is the square root of 4, and this is rational. However I feel that you main difficulty lies in understanding why the usual proof that sqrt(2) is irrational doesn't show that sqrt(4) is irrational, so I'll show where the proof falls apart.If someone can tell/prove me by using the proof of contradiction that the square root of 4 is not an irrational I will make him/her a chocolate cake which I'll eat.
You cannot show that. The usual proof for 2 tries to write sqrt(2)=p/q with p and q relatively prime. You then manipulate it to get 2q^2 = p^2. Now key thing to note here is that you get 2|p^2, and since 2 is square-free and gcd(p,q)=1 you get 2|p. However this kind of logic doesn't work with 4 because 4|p^2 does not imply 4|p. Given a larger non-square number such as [itex]12 = 2^2 3[/itex] we still can't show 12|p^2 imply 12|p, but the proof only needs one factor so we instead use 12|p^2 imply 3 | p^2 imply 3|p imply 9|p^2, so 24|p^2.But by using statements such as they are relative primes I can get every square number to be irrational. Or because they are perfect squares we know they are rational?