• Nov 7th 2007, 04:15 PM
somestudent2
• Nov 7th 2007, 05:36 PM
topsquark
Quote:

Originally Posted by somestudent2

a) 0 cannot be part of S because if it were then all three of $\displaystyle r \in S$, $\displaystyle -r \in S$, and $\displaystyle r = 0$ are true. But only one of them is allowed to be true.

b) I will simply note that the positive integers are closed under addition and multiplication. You can supply the rest of the details.

c) I would like to say the same thing about c) as I did about b): the positive rationals are closed under addition and multiplication. However, this set excludes the negative rationals, and I'm not sure exactly how to go about doing that. The exclusion of 0 from the set is going to be important, I can tell you that much.

-Dan
• Nov 7th 2007, 11:58 PM
Opalg
Notice first that if a∈S then a+a=2a∈S, and by induction na∈S for every positive integer n.

If –1∈S then (–1)(–1)=1∈S, and this contradicts the defining property of S. So –1∉S and therefore 1∈S. By the previous paragraph, n∈S for every positive integer n. Therefore n∉S for every negative integer n.

If a/b is a negative rational number, where a is a negative integer and b is a positive integer then, by the first paragraph, b(a/b)∈S ... .
• Nov 8th 2007, 05:51 AM
somestudent2
Thank You very much guys, this really helps:)