Not quite, because a and b are constants not variables, so saying a=0 is not permissible. If you consider what type of number 3a is going to be, and what type 2b is going to be what can you deduce?
If I were to say 2 is even, what would you say about 2a? And 3? And 3b?
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:
In essence this is a solution. ba was said to be in it's simplest form, which it isn't if both are a multiple of 5 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? :/
Basically it's the exact same method, only the complex numbers introduce a greater probability of an algebraic slip up
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.
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 log32 is irrational
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.
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? :/