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

## 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 $$x,{\text{ }}y \in S$$.


a.

Show that the operation $$*$$ on the set $$S$$ is commutative.


b.i.

Show that the operation $$*$$ on the set $$S$$ is associative.


b.ii.

Show that 2 is the identity element.


c.

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


d.

## 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.

[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.


a.

Show that $$f$$ is injective.


b.i.

Show that $$f$$ is surjective.


b.ii.

## 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.

[N/A]

a.

[N/A]

b.i.

[N/A]

b.ii.
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