IB DP Maths Topic 8.6 Proof that left-cancellation and right-cancellation by an element a hold, provided that a has an inverse. HL Paper 3

 

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Question

Given that p , q and r are elements of a group, prove the left-cancellation rule, i.e. \(pq = pr \Rightarrow q = r\) .

Your solution should indicate which group axiom is used at each stage of the proof.

[4]
a.

Consider the group G , of order 4, which has distinct elements a , b and c and the identity element e .

(i)     Giving a reason in each case, explain why ab cannot equal a or b .

(ii)     Given that c is self inverse, determine the two possible Cayley tables for G .

(iii)     Determine which one of the groups defined by your two Cayley tables is isomorphic to the group defined by the set {1, −1, i, −i} under multiplication of complex numbers. Your solution should include a correspondence between a, b, c, e and 1, −1, i, −i .

[10]
b.
Answer/Explanation

Markscheme

\(pq = pr\)

\({p^{ – 1}}(pq) = {p^{ – 1}}(pr)\) , every element has an inverse     A1

\(({p^{ – 1}}p)q = ({p^{ – 1}}p)r\) , Associativity     A1

Note: Brackets in lines 2 and 3 must be seen.

 

\(eq = er\), \({p^{ – 1}}p = e\), the identity     A1

\(q = r\), \(ea = a\) for all elements a of the group     A1

[4 marks]

a.

(i)     let ab = a so b = e be which is a contradiction     R1

let ab = b so a = e which is a contradiction     R1

therefore ab cannot equal either a or b     AG

(ii)     the two possible Cayley tables are

table 1

     A2

table 2

     A2

(iii)     the group defined by table 1 is isomorphic to the given group     R1

because

EITHER

both contain one self-inverse element (other than the identity)     R1

OR

both contain an inverse pair     R1

OR

both are cyclic     R1

THEN

the correspondence is \(e \to 1\), \(c \to – 1\), \(a \to i\), \(b \to – i\)

(or vice versa for the last two)     A2

Note: Award the final A2 only if the correct group table has been identified.

 

[10 marks]

b.

Examiners report

Solutions to (a) were often poor with inadequate explanations often seen. It was not uncommon to see \(pq = pr\)

\({p^{ – 1}}pq = {p^{ – 1}}pr\)

\(q = r\)

without any mention of associativity. Many candidates understood what was required in (b)(i), but solutions to (b)(ii) were often poor with the tables containing elements such as ab and bc without simplification. In (b)(iii), candidates were expected to determine the isomorphism by noting that the group defined by {1, –1, i, –i} under multiplication is cyclic or that –1 is the only self-inverse element apart from the identity, without necessarily writing down the Cayley table in full which many candidates did. Many candidates just stated that there was a bijection between the two groups without giving any justification for this.

a.

Solutions to (a) were often poor with inadequate explanations often seen. It was not uncommon to see \(pq = pr\)

\({p^{ – 1}}pq = {p^{ – 1}}pr\)

\(q = r\)

without any mention of associativity. Many candidates understood what was required in (b)(i), but solutions to (b)(ii) were often poor with the tables containing elements such as ab and bc without simplification. In (b)(iii), candidates were expected to determine the isomorphism by noting that the group defined by {1, –1, i, –i} under multiplication is cyclic or that –1 is the only self-inverse element apart from the identity, without necessarily writing down the Cayley table in full which many candidates did. Many candidates just stated that there was a bijection between the two groups without giving any justification for this.

b.

Question

The set \(S\) is defined as the set of real numbers greater than 1.

The binary operation \( * \) is defined on \(S\) by \(x * y = (x – 1)(y – 1) + 1\) for all \(x,{\text{ }}y \in S\).

Let \(a \in S\).

Show that \(x * y \in S\) for all \(x,{\text{ }}y \in S\).

[2]
a.

Show that the operation \( * \) on the set \(S\) is commutative.

[2]
b.i.

Show that the operation \( * \) on the set \(S\) is associative.

[5]
b.ii.

Show that 2 is the identity element.

[2]
c.

Show that each element \(a \in S\) has an inverse.

[3]
d.
Answer/Explanation

Markscheme

\(x,{\text{ }}y > 1 \Rightarrow (x – 1)(y – 1) > 0\)     M1

\((x – 1)(y – 1) + 1 > 1\)     A1

so \(x * y \in S\) for all \(x,{\text{ }}y \in S\)     AG

[2 marks]

a.

\(x * y = (x – 1)(y – 1) + 1 = (y – 1)(x – 1) + 1 = y * x\)     M1A1

so \( * \) is commutative     AG

[2 marks]

b.i.

\(x * (y * z) = x * \left( {(y – 1)(z – 1) + 1} \right)\)     M1

\( = (x – 1)\left( {(y – 1)(z – 1) + 1 – 1} \right) + 1\)     (A1)

\( = (x – 1)(y – 1)(z – 1) + 1\)     A1

\((x * y) * z = \left( {(x – 1)(y – 1) + 1} \right) * z\)     M1

\( = \left( {(x – 1)(y – 1) + 1 – 1} \right)(z – 1) + 1\)

\( = (x – 1)(y – 1)(z – 1) + 1\)     A1

so \( * \) is associative     AG

[5 marks]

b.ii.

\(2 * x = (2 – 1)(x – 1) + 1 = x,{\text{ }}x * 2 = (x – 1)(2 – 1) + 1 = x\)     M1

\(2 * x = x * 2 = 2{\text{ }}(2 \in S)\)     R1

Note:     Accept reference to commutativity instead of explicit expressions.

so 2 is the identity element     AG

[2 marks]

c.

\(a * {a^{ – 1}} = 2 \Rightarrow (a – 1)({a^{ – 1}} – 1) + 1 = 2\)     M1

so \({a^{ – 1}} = 1 + \frac{1}{{a – 1}}\)     A1

since \(a – 1 > 0 \Rightarrow {a^{ – 1}} > 1{\text{ }}({a^{ – 1}} * a = a * {a^{ – 1}})\)     R1

Note:     R1 dependent on M1.

so each element, \(a \in S\), has an inverse     AG

[3 marks]

d.

Examiners report

[N/A]

a.

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

c.

[N/A]

d.

Question

The function \(f\,{\text{: }}\mathbb{Z} \to \mathbb{Z}\) is defined by \(f\left( n \right) = n + {\left( { – 1} \right)^n}\).

Prove that \(f \circ f\) is the identity function.

[6]
a.

Show that \(f\) is injective.

[2]
b.i.

Show that \(f\) is surjective.

[1]
b.ii.
Answer/Explanation

Markscheme

METHOD 1

\(\left( {f \circ f} \right)\left( n \right) = n + {\left( { – 1} \right)^n} + {\left( { – 1} \right)^{n + {{\left( { – 1} \right)}^n}}}\)     M1A1

\( = n + {\left( { – 1} \right)^n} + {\left( { – 1} \right)^n} \times {\left( { – 1} \right)^{{{\left( { – 1} \right)}^n}}}\)     (A1)

considering \({\left( { – 1} \right)^n}\) for even and odd \(n\)     M1

if \(n\) is odd, \({\left( { – 1} \right)^n} = – 1\) and if \(n\) is even, \({\left( { – 1} \right)^n} = 1\) and so \({\left( { – 1} \right)^{ \pm 1}} =  – 1\)      A1

\( = n + {\left( { – 1} \right)^n} – {\left( { – 1} \right)^n}\)    A1

= \(n\) and so \(f \circ f\) is the identity function     AG

METHOD 2

\(\left( {f \circ f} \right)\left( n \right) = n + {\left( { – 1} \right)^n} + {\left( { – 1} \right)^{n + {{\left( { – 1} \right)}^n}}}\)      M1A1

\( = n + {\left( { – 1} \right)^n} + {\left( { – 1} \right)^n} \times {\left( { – 1} \right)^{{{\left( { – 1} \right)}^n}}}\)     (A1)

\( = n + {\left( { – 1} \right)^n} \times \left( {1 + {{\left( { – 1} \right)}^{{{\left( { – 1} \right)}^n}}}} \right)\)     M1

\({\left( { – 1} \right)^{ \pm 1}} =  – 1\)     R1

\(1 + {\left( { – 1} \right)^{{{\left( { – 1} \right)}^n}}} = 0\)     A1

\(\left( {f \circ f} \right)\left( n \right) = n\) and so \(f \circ f\) is the identity function     AG

METHOD 3

\(\left( {f \circ f} \right)\left( n \right) = f\left( {n + {{\left( { – 1} \right)}^n}} \right)\)      M1

considering even and odd \(n\)       M1

if \(n\) is even, \(f\left( n \right) = n + 1\) which is odd    A1

so \(\left( {f \circ f} \right)\left( n \right) = f\left( {n + 1} \right) = \left( {n + 1} \right) – 1 = n\)    A1

if \(n\) is odd, \(f\left( n \right) = n – 1\) which is even    A1

so \(\left( {f \circ f} \right)\left( n \right) = f\left( {n – 1} \right) = \left( {n – 1} \right) + 1 = n\)     A1

\(\left( {f \circ f} \right)\left( n \right) = n\) in both cases

hence \(f \circ f\) is the identity function     AG

[6 marks]

a.

suppose \(f\left( n \right) = f\left( m \right)\)     M1

applying \(f\) to both sides \( \Rightarrow n = m\)     R1

hence \(f\) is injective     AG

[2 marks]

b.i.

\(m = f\left( n \right)\) has solution \(n = f\left( m \right)\)       R1

hence surjective       AG

[1 mark]

b.ii.

Examiners report

[N/A]

a.

[N/A]

b.i.

[N/A]

b.ii.

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