Appendix: Retained Examinable Topics

Problem. State, with a reason, whether each is a binary operation on the set shown. (a) a * b = a + b on \mathbb{N} (b) a * b = a - b on \mathbb{N} (c) a * b = ab on \mathbb{Z} (d) a * b = \dfrac{a}{b} on \mathbb{Z}

Solution. (a) Yes. The sum of two natural numbers is a natural number, so * is closed.

(b) No. Take a = 3, b = 5: 3 - 5 = -2, which is not in \mathbb{N}. A single counter-example is enough to destroy closure.

(c) Yes. The product of two integers is an integer.

(d) No. 1 \div 2 = 0.5 \notin \mathbb{Z} (and division by 0 is not defined either).

Problem. Show that Z_4 = \{0, 1, 2, 3\} is an abelian group under addition modulo 4, and state its order.

Solution. The Cayley table above is the working. Take the four axioms in turn.

Closure. Every entry of the table is 0, 1, 2 or 3, so Z_4 is closed under +_4.

Associativity. Addition of integers is associative, and reducing modulo 4 does not change this: (a +_4 b) +_4 c and a +_4 (b +_4 c) are both the remainder of a+b+c on division by 4.

Identity. The row headed 0 reads 0, 1, 2, 3 — the heading row — and the column headed 0 likewise. So e = 0.

Inverse. The entry 0 occurs once in each row: 0 +_4 0 = 0, 1 +_4 3 = 0, 2 +_4 2 = 0, 3 +_4 1 = 0. Hence 0^{-1} = 0, 1^{-1} = 3, 2^{-1} = 2, 3^{-1} = 1.

So (Z_4, +_4) is a group. The table is symmetric about the leading diagonal, so it is abelian, and it is finite of order 4.

Problem. Show that G = \{1, -1, i, -i\} is an abelian group under multiplication of complex numbers, and find the inverse of each element.

Solution. Read the table in the figure.

Closure. Every product is again one of 1, -1, i, -i; for instance i \times i = i^2 = -1 and i \times (-i) = -i^2 = 1.

Associativity. Multiplication of complex numbers is associative.

Identity. e = 1, since 1 \times z = z \times 1 = z for each element.

Inverses. The 1s in the table give 1^{-1} = 1, (-1)^{-1} = -1, i^{-1} = -i, (-i)^{-1} = i.

The table is symmetric about the leading diagonal (equivalently, multiplication of complex numbers is commutative), so G is an abelian group of order 4.

Problem. Show that (\mathbb{Z}, +) is an infinite abelian group but (\mathbb{N}, +) is not a group.

Solution. For (\mathbb{Z}, +):

• closure: the sum of two integers is an integer;
• associativity: (a+b)+c = a+(b+c) for integers;
• identity: e = 0, because a + 0 = 0 + a = a;
• inverse: the inverse of a is -a \in \mathbb{Z}, since a + (-a) = 0;
• commutativity: a + b = b + a.

\mathbb{Z} has infinitely many elements, so (\mathbb{Z}, +) is an infinite abelian group.

For (\mathbb{N}, +) the identity 0 is not in \mathbb{N} (and even taking \mathbb{N} to contain 0, the element 3 would need the inverse -3, which is not a natural number). Axiom 4 fails, so (\mathbb{N}, +) is not a group.

Problem. Show that G = \{1, 2, 3, 4\} is a group under multiplication modulo 5, and find the inverse of each element.

Solution. The composition table (computed entry by entry, e.g. 2 \times_5 3 = 6 \equiv 1 modulo 5) is

Closure. No entry is 0 and every entry lies in G. (This is why 5 must be prime: modulo 6, 2 \times_6 3 = 0 \notin G.)

Associativity. Inherited from multiplication of integers.

Identity. The row and column headed 1 repeat the headings, so e = 1.

Inverses. The identity 1 appears once in each row:

The table is symmetric, so G is an abelian group of order 4.

Problem. Prove that in any group (G, *) the identity element is unique and each element has exactly one inverse.

Solution. Identity. Suppose e and e' are both identities. Treating e as the identity, e * e' = e'. Treating e' as the identity, e * e' = e. Hence e = e'.

Inverse. Suppose b and c are both inverses of a. Then

using in turn the identity axiom, a*c = e, associativity, b*a = e, and the identity axiom again. So b = c.

Problem. Using the table above, show that S_3 is a non-abelian group of order 6, and find (r \circ a)^{-1}.

Solution. Every entry is one of the six symbols, so S_3 is closed; composition of functions is always associative; the row and column headed e repeat the headings, so e is the identity; and e appears once in every row and column, so every element has an inverse. Hence S_3 is a group of order 6.

From the table, r \circ a = c but a \circ r = b, and b \ne c, so

and the group is not abelian. Finally (r \circ a)^{-1} = c^{-1}, and the row for c contains e in the column headed c (that is, c \circ c = e), so (r \circ a)^{-1} = c.

Problem. Verify Rolle's theorem for f(x) = x^2 - 4x + 3 on [1, 3] and find the value of c.

Solution. Check the three conditions.

(i) f is a polynomial, so it is continuous on [1, 3]. (ii) f is a polynomial, so it is differentiable on (1, 3). (iii) f(1) = 1 - 4 + 3 = 0 and f(3) = 9 - 12 + 3 = 0, so f(1) = f(3).

All three hold, so Rolle's theorem guarantees a c \in (1, 3) with f'(c) = 0.

Find c. f'(x) = 2x - 4, so

Since 1 < 2 < 3, the value c = 2 lies in the open interval and the theorem is verified.

Problem. Verify Rolle's theorem for f(x) = x^3 - 4x on [-2, 2].

Solution. f is a polynomial, hence continuous on [-2, 2] and differentiable on (-2, 2).

so f(-2) = f(2) and Rolle's theorem applies. Now f'(x) = 3x^2 - 4, so

Both values lie in (-2, 2), so there are two such points. (The theorem promises at least one; it does not say only one.)

Problem. Explain why Rolle's theorem does not apply to (a) f(x) = |x| on [-1, 1], (b) f(x) = \dfrac{1}{x} on [-1, 1].

Solution. (a) f(-1) = f(1) = 1 and f is continuous, but f is not differentiable at x = 0 — the graph has a corner there. Condition (ii) fails, and indeed f'(x) = \pm 1 is never zero.

(b) f is not continuous on [-1, 1]: it is not even defined at x = 0. Condition (i) fails. (f(-1) = -1 and f(1) = 1 are not equal either.)

Problem. Verify the Mean Value Theorem for f(x) = x^2 on [1, 3].

Solution. f is a polynomial, so it is continuous on [1, 3] and differentiable on (1, 3); the MVT applies.

f'(x) = 2x, so f'(c) = 4 gives

The tangent at x = 2 has gradient 4, the same as the chord.

Problem. Verify the Mean Value Theorem for f(x) = \sqrt{x} on [1, 4].

Solution. f is continuous on [1, 4] and differentiable on (1, 4) (the only bad point, x = 0, is outside the interval).

f'(x) = \dfrac{1}{2\sqrt{x}}, so f'(c) = \dfrac13 gives

and 1 < 2.25 < 4, so the theorem is verified.

Problem. Verify the Mean Value Theorem for f(x) = \ln x on [1, e].

Solution. \ln x is continuous on [1, e] and differentiable on (1, e).

f'(x) = \dfrac1x, so

Since 1 < 1.718 < 2.718, the value c = e-1 lies in (1, e).

Problem. Evaluate \displaystyle\int_1^2 \frac{dx}{x} by the trapezoidal rule with n = 4, correct to six decimal places, and state a bound for the error. Compare with the exact value \ln 2 = 0.693147.

Solution. h = \dfrac{2-1}{4} = 0.25. Make the ordinate table.

Error bound. f(x) = x^{-1}, so f''(x) = 2x^{-3}, and on [1, 2] this is largest at x = 1: \max|f''| = 2. Hence

Comparison. The actual error is 0.697024 - 0.693147 = 0.003877, comfortably inside the bound, and the estimate is too large — as expected, since 1/x is concave up and the chords lie above the curve.

Problem. Repeat with Simpson's rule, n = 4, and state the error bound.

Solution. The same five ordinates are used, now with the weights 1, 4, 2, 4, 1.

Error bound. f^{(4)}(x) = 24x^{-5}, largest at x = 1, so \max|f^{(4)}| = 24 and

The actual error is 0.693254 - 0.693147 = 0.000107 — about 36 times smaller than the trapezium error from exactly the same five function values.

Problem. Evaluate \displaystyle\int_0^6 \frac{dx}{1+x} by (a) the trapezoidal rule and (b) Simpson's rule, both with n = 6. The exact value is \ln 7 = 1.945910.

Solution. h = \dfrac{6-0}{6} = 1.

(a) Sum of the five interior ordinates = 0.500000 + 0.333333 + 0.250000 + 0.200000 + 0.166667 = 1.450000.

(b) Odd-numbered ordinates: 0.500000 + 0.250000 + 0.166667 = 0.916667. Even interior ordinates: 0.333333 + 0.200000 = 0.533333.

Errors: 2.021429 - 1.945910 = 0.075519 for the trapezium against 1.958730 - 1.945910 = 0.012820 for Simpson's rule. The strips are wide here (h = 1), so both errors are large, but Simpson's is nearly six times better.