Date | May 2013 | Marks available | 7 | Reference code | 13M.2.hl.TZ0.5 |
Level | HL only | Paper | 2 | Time zone | TZ0 |
Command term | Show that | Question number | 5 | Adapted from | N/A |
Question
The set \(S\) consists of real numbers r of the form \(r = a + b\sqrt 2 \) , where \(a,b \in \mathbb{Z}\) .
The relation \(R\) is defined on \(S\) by \({r_1}R{r_2}\) if and only if \({a_1} \equiv {a_2}\) (mod2) and \({b_1} \equiv {b_2}\) (mod3), where \({r_1} = {a_1} + {b_1}\sqrt 2 \) and \({r_2} = {a_2} + {b_2}\sqrt 2 \) .
Show that \(R\) is an equivalence relation.
Show, by giving a counter-example, that the statement \({r_1}R{r_2} \Rightarrow r_1^2Rr_2^2\) is false.
Determine
(i) the equivalence class \(E\) containing \(1 + \sqrt 2 \) ;
(ii) the equivalence class \(F\) containing \(1 - \sqrt 2 \) .
Show that
(i) \({(1 + \sqrt 2 )^3} \in F\) ;
(ii) \({(1 + \sqrt 2 )^6} \in E\) .
Determine whether the set \(E\) forms a group under
(i) the operation of addition;
(ii) the operation of multiplication.
Markscheme
reflexive: if \({r_{}} = {a_{}} + {b_{}}\sqrt 2 \in S\) then \(a \equiv a(\bmod 2)\) and \(b \equiv b(\bmod 3)\)
\(( \Rightarrow rRr)\) A1
symmetric: if \({r_1}R{r_2}\) then \({a_1} \equiv {a_2}(\bmod 2)\) and \({b_1} \equiv {b_2}(\bmod 3)\) , and M1
\({a_2} \equiv {a_1}(\bmod 2)\) and \({b_2} \equiv {b_1}(\bmod 3)\) , (so that \({r_2}R{r_1}\) ) A1
transitive: if \({r_1}R{r_2}\) and \({r_2}R{r_3}\) then
\(2|{a_1} - {a_2}\) and \(2|{a_2} - {a_3}\) M1
\( \Rightarrow 2|{a_1} - {a_2} + {a_2} - {a_3} \Rightarrow 2|{a_1} - {a_3}\) M1A1
\(3|{b_1} - {b_2}\) and \(3|{b_2} - {b_3}\)
\( \Rightarrow 3|{b_1} - {b_2} + {b_2} - {b_3} \Rightarrow 3|{b_1} - {b_3}( \Rightarrow {r_1}R{r_3})\) A1AG
[7 marks]
consider, for example, \({r_1} = 1 + \sqrt 2 \) , \({r_2} = 3 + \sqrt 2 \) \(({r_1}R{r_2})\) M1
Note: Only award M1 if the two numbers are related and neither \(a\) nor \(b = 0\) .
\(r_1^2 = 3 + 2\sqrt 2 \) , \(r_2^2 = 11 + 6\sqrt 2 \) A1
the squares are not equivalent because \(2 \ne 6(\bmod 3)\) A1
[3 marks]
(i) \(E = \left\{ {2k + 1 + (3m + 1)\sqrt 2 } :k,m \in \mathbb{Z} \right\}\) A1A1
(ii) \(F = \left\{ {2k + 1 + (3m - 1)\sqrt 2 } :k,m \in \mathbb{Z} \right\}\) A1
[3 marks]
(i) \({(1 + \sqrt 2 )^3} = 7 + 5\sqrt 2 \) A1
\( = 2 \times 3 + 1 + (3 \times 2 - 1)\sqrt 2 \in F\) R1AG
(ii) \({(1 + \sqrt 2 )^6} = 99 + 70\sqrt 2 \) A1
\( = 2 \times 49 + 1 + (3 \times 23 + 1)\sqrt 2 \in E\) R1AG
[4 marks]
(i) \(E\) is not a group under addition A1
any valid reason eg \(0 \notin E\) R1
(ii) \(E\) is not a group under multiplication A1
any valid reason eg \(1 \notin E\) R1
[4 marks]
Examiners report
The majority of candidates earned significant marks on this question. However, many lost marks in part (a) by assuming that equivalence modulo \(2\) and \(3\) is transitive. This is a non-trivial true result but requires proof.
The majority of candidates earned significant marks on this question.
The majority of candidates earned significant marks on this question.
The majority of candidates earned significant marks on this question.
The majority of candidates earned significant marks on this question.