GRE test 8767 #66

Forum for the GRE subject test in mathematics.
Krutika
Posts: 5
Joined: Thu Oct 16, 2014 5:08 am

GRE test 8767 #66

Postby Krutika » Thu Oct 16, 2014 5:17 am

http://www.math.ucla.edu/~cmarshak/GRE3.pdf

Hi!

I was attempting this paper and the answer #66 suggests that option (III) is incorrect. Can anyone please explain why it shouldnt be true?

Thanks

Austin
Posts: 26
Joined: Fri Jun 01, 2012 9:32 am

Re: GRE test 8767 #66

Postby Austin » Thu Oct 16, 2014 11:44 am

Imagine that the set was defined with \sqrt{-1} in place of \sqrt{5} (pun most definitely intended). Then this set would be the closed unit disc in the complex plane. It should then be pretty clear that we won't have closure under addition.

Alternatively, you can just cook up an easy counterexample, such as a=1,b=0 and a=0,b=1. Adding these together should put you outside of the set.

Krutika
Posts: 5
Joined: Thu Oct 16, 2014 5:08 am

Re: GRE test 8767 #66

Postby Krutika » Thu Oct 16, 2014 12:11 pm

So the same should also be applicable to a+ b (root 2), right? That way (i) wouldn't be a subring either.

Austin
Posts: 26
Joined: Fri Jun 01, 2012 9:32 am

Re: GRE test 8767 #66

Postby Austin » Thu Oct 16, 2014 12:56 pm

No, in (i), a and b aren't subject to the restriction a^2+b^2\leq 1; they just have to be rational. Since \mathbb{Q} is closed under addition, so is the set in (i). Of course, you still need to check that it's closed under multiplication, but this is easy.




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