IB DP Maths Topic 8.6 Proofs of the uniqueness of the identity and inverse elements. HL Paper 3

 

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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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