Date | May 2007 | Marks available | 6 | Reference code | 07M.1.hl.TZ0.3 |
Level | HL only | Paper | 1 | Time zone | TZ0 |
Command term | Show that | Question number | 3 | Adapted from | N/A |
Question
Show that the set \(S\) of numbers of the form \({2^m} \times {3^n}\) , where \(m,n \in \mathbb{Z}\) , forms a group \(\left\{ {S, \times } \right\}\) under multiplication.
Show that \(\left\{ {S, \times } \right\}\) is isomorphic to the group of complex numbers \(m + n{\rm{i}}\) under addition, where \(m\), \(n \in \mathbb{Z}\) .
Markscheme
Closure: Consider the numbers \({2^{{m_1}}} \times {3^{{m_1}}}\) and \({2^{{m_2}}} \times {3^{{n_2}}}\) where M1
\({m_1},{m_2},{n_1},{n_2}, \in \mathbb{Z}\) . Then,
Product \( = {2^{{m_1} + {m_2}}} \times {3^{{n_1} + {n_2}}}\) which \( \in S\) A1
Identity: \({2^0} \times {3^0} = 1 \in S\) A1
Since \(({2^m} \times {3^n}) \times ({2^{ - m}} \times {3^{ - n}}) = 1\) and \({2^{ - m}} \times {3^{ - n}} \in S\) R1
then \({2^{ - m}} \times {3^{ - n}}\) is the inverse. A1
Associativity: This follows from the associativity of multiplication. R1
[6 marks]
Consider the bijection
\(f({2^m} \times {3^n}) = m + n{\rm{i}}\) (M1)
Then
\(f({2^{{m_1}}} \times {3^{{n_1}}}) \times ({2^{{m_2}}} \times {3^{{n_2}}}) = f({2^{{m_1} + {m_2}}} \times {3^{{n_1} + {n_2}}})\) M1A1
\( = {m_1} + {m_2} + ({n_1} + {n_2}){\rm{i}}\) A1
\( = ({m_1} + {n_1}{\rm{i}}) + ({m_2} + {n_2}{\rm{i}})\) (A1)
\( = f({2^{{m_1}}} \times {3^{{n_1}}}) + f({2^{{m_2}}} \times {3^{{n_2}}})\) A1
[6 marks]