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Original post by joostan
Not quite, because a and b are constants not variables, so saying a=0 is not permissible. :redface:
If you consider what type of number 3a3^a is going to be, and what type 2b2^b is going to be what can you deduce? :smile:


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Original post by Robbie242

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If I were to say 2 is even, what would you say about 2a2^a?
And 3? And 3b3^b?
Original post by joostan
Dunno what he's after, but yeah, Felix says he's getting somewhere. :tongue:


Well, alternatively:

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Also:

Original post by joostan
If I were to say 2 is even, what would you say about 2a2^a?
And 3? And 3b3^b?


I believe that Robbie's idea of prime numbers is actually a better way of tackling the question, because it can be extended to cases where both numbers are odd or both even. for example:

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(edited 10 years ago)
Original post by joostan
If I were to say 2 is even, what would you say about 2a2^a?
And 3? And 3b3^b


2^a is also even
3^b is not even ?

I'm trying to raise 3 to powers and it keeps coming out odd, is this the contradiction?
(edited 10 years ago)
Original post by Robbie242
2^a is also even
3^b is not even ?

I'm trying to raise 3 to powers and it keeps coming out odd, is this the contradiction?


Precisely :smile:
Original post by joostan
Precisely :smile:


Got there in the end ahah! its good I'm getting into these problems though, the alevel ones often guide you a lot
Original post by DJMayes
Well, alternatively:

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Basically it's the exact same method, only the complex numbers introduce a greater probability of an algebraic slip up :s-smilie:
Original post by Robbie242
Got there in the end ahah! its good I'm getting into these problems though, the alevel ones often guide you a lot


There's a nice induction one with irrationals if you're interested :tongue:
Original post by joostan
There's a nice induction one with irrationals if you're interested :tongue:


Go on then :tongue:, I'll at least get past the basis case and the assumption step xd
Reply 3829
Original post by joostan
In essence this is a solution.
ab\dfrac{a}{b}
was said to be in it's simplest form, which it isn't if both are a multiple of 5 :smile:
This is a contradiction, so therefore the cubed root of 5 is irrational.


OOHH, so basically in there type of questions you prove that a/b isn't its simplest form, this is means its a contradiction and therefore is irrational? Ahh, so I kind of solved it? :/
Original post by joostan
Basically it's the exact same method, only the complex numbers introduce a greater probability of an algebraic slip up :s-smilie:


I'm not saying it's any better (Although it did actually feel marginally quicker, possibly because I knew what I was going for) but it is an alternative approach using complex numbers which might've been the one Mayman was thinking of.
Original post by Felix Felicis
Ahh, muy bien :wink:

3 hours later on that xy=z2+1xy = z^{2} + 1 problem, I think I'm onto something :s-smilie:


If it took you that much. That means I could never do it:tongue:
(edited 10 years ago)
Original post by Robbie242
Go on then :tongue:, I'll at least get past the basis case and the assumption step xd


You've already got your basis case.
Prove that:
Unparseable latex formula:

5^{\frac{1}{3^n}} \not\in \matbb{Q}\ \forall n \in \mathbb{N}

Original post by DJMayes
Well, alternatively:

Spoiler



Also:



I believe that Robbie's idea of prime numbers is actually a better way of tackling the question, because it can be extended to cases where both numbers are odd or both even. for example:

Spoiler



Thanks DJ! I will probably use this one as it seems a bit more universal,
i.e. I'd say for my one, no matter how many times you multiply 2 by itself, it will never gain a factor 3, therefore this is a contradiction and hence log32log_{3} 2 is irrational
(edited 10 years ago)
Original post by DJMayes
I'm not saying it's any better (Although it did actually feel marginally quicker, possibly because I knew what I was going for) but it is an alternative approach using complex numbers which might've been the one Mayman was thinking of.


Yeah, I guess :smile:
Original post by joostan
You've already got your basis case.
Prove that:
Unparseable latex formula:

5^{\frac{1}{3^n}} \not\in \matbb{Q}\ \forall n \in \mathbb{N}



Explain this notation Lol
Original post by tigerz
OOHH, so basically in there type of questions you prove that a/b isn't its simplest form, this is means its a contradiction and therefore is irrational? Ahh, so I kind of solved it? :/


Yep :biggrin:
Original post by Robbie242
Explain this notation Lol


N is the set of natural numbers.
Q is the set of rationals.
The \in means in :smile:

Apologies for misleading you with the primes, you're right it works better :redface:
Original post by Robbie242
Explain this notation Lol


Not entirely sure what all the inequalities are there for, but the final part is saying "is not rational for all natural numbers n".

Also, I would question how appropriate this is for this thread, given that the question you're being asked is in fact Q3 of STEP II, 2003...
Original post by joostan
N is the set of natural numbers.
Q is the set of rationals.
The \in means in :smile:

Apologies for misleading you with the primes, you're right it works better :redface:


What about the upsidedown A?

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