Prove that the set of all rational numbers of the form 3 m 6 n , where m and n a

Question

Prove that the set of all rational numbers of the form 3m6n, where m and n are integers, is a group under multiplication.

Explanation / Answer

Assume that all rational numbers are in the form 3^m *6^n . Therefor 3^m*6^n = p/q where p and q are real numbers. Suppose that 3^m*6^n is a group under multiplication. Then this non-empty set must have the following 3 properties: must have an inverse, an identity and the assiociative property 3^m*6^n is equivalent to (3(sub1)+3(sub2)+3(sub3)+...+3(sub m))*(6(sub 1)+6(sub 2) +6(sub 3)+...+6(sub n)) On the side note, in an abstract group, 3^3=(3+3+3) and 5^3=(5+5+5). To show that 3^m*6^n has an inverse, I must show that 3^m*6^n =6^n*3^m. To be honest I not sure whether I have to show that 3^m*6^n=6^n *3^m or 3^m *6^n=3^-m*6^-n. Either way, either expression has to equal e, or einheit. I could supposed that b and c are inverses of the expression 3^m*6^n , where b*(3^m*6^n)=e and c*(3^m*6^n)=e => c*(3^m*6^n)= b*(3^m*6^n) => b=c To prove that 3^m*6^n has an identity property, I must show that (3^m*6^n)*e=e*(3^m*6^n)=(3^m*6^n) I then have to supposed that e and e' are identities of of the Group . Then I can conclude that (3^m*6^n)*e=(3^m*6^n) for all (3^m*6^n) forms in G and e'*(3^m*6^n)=(3^m*6^n) for all (3^m*6^n) forms in G.=> e'e=e' and ee'=e . Thus, e and e' are both equal to e'e and so are equal to each other.

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