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CM115A, CM115B Numbers and Functions: Assignment 4
Solutions to this assignment must be handed in at the end of the tutorial you attend
during week 5 of the term. Assignments handed in late will not normally be marked. It
is essential that you write your NAME, STUDENT NUMBER and GROUP NUMBER on
your work.
Some harder exercises are included in this sheet; they are marked with an asterisk *
and are not compulsory. Also included at the end are some additional exercises; these are
for practice only and not for submission to your tutor.
1. Solve the following inequalities for real π > 0:
(a)
(b)
1
10π3
3
β
π π
β€7
<5
(c) 2βπ β€ 20
(d) 2π β€ 20
(e) ( 12 )5π β₯ 1/8
(f) ( 12 )5π β€ 1/8
2. Solve the following equations and inequalities for π₯ β β:
(a) β£π₯ β 3β£ = 1
(b) β£π₯ + 1β£ β€ 3
(c) β£π₯ + 2β£ > 5
(d) β£π₯ + 10β£ < 1.
3. For each given set π, state whether it is bounded below and bounded above. If the
set π is bounded above, ο¬nd the set π+ of all upper bounds of π; if π is bounded
below, ο¬nd the set πβ of all lower bounds for π. Find max π and min π if they exist.
No proof is required.
(a) π = [β10, β1) βͺ (2, 5);
(b) π = {1} βͺ (2, 3);
(c) π = { π12 β£ π β β};
π
(d) π = { (β1)
β£ π β β};
π
(e) π = {(β1)π β
1
π
β£ π β β}.
4. Prove that:
(a) π = [0, 1] βͺ (3, β) is not bounded above;
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(b) The set β of rational numbers is not bounded below;
(c) π = {(β2)π β£ π β β} is not bounded above;
(d) π = βͺβ
π=1 [2π β 1, 2π] is not bounded above.
5. Prove that:
(a) min π does not exist, where π = (0, 1);
(b) min π does not exist, where π = {2π β£ π β β€};
(c) min π does not exist, where π = { log1 π β£ π β β, π β₯ 2}.
6. Prove that π is bounded if and only if βπ
β β such that βπ₯ β π one has β£π₯β£ β€ π
.
Note that you have two statements to prove: βifβ and βonly ifβ.
β
7. Prove that 5 is irrational.
Additional exercises:
8.* Two irrational numbers
β π₯ β= 0βand π¦ β= 0 are called rationally independent, if π₯/π¦ is
irrational. Prove that 2 and 5 are rationally independent.
9. Prove that:
(a) if π₯ and π¦ are rational numbers, then π₯ + π¦ and π₯π¦ are rational numbers;
(b) if π₯ is rational and π¦ is irrational, then π₯ + π¦ is irrational;
(c) if π₯ β= 0 is rational and π¦ is irrational, then π₯π¦ is irrational;
(d)* there exist irrational numbers π₯ and π¦ such that π₯ + π¦ is rational;
(e)* there exist irrational numbers π₯ β= 0 and π¦ β= 0 such that π₯π¦ is rational.
10. Prove that for any π > 0:
(a) there exists a rational number π₯ in the interval (0, π);
(b)* there exists an irrational number π¦ in the interval (0, π).
11.** Using the previous exercise, prove that for any two real numbers π and π such that
π < π:
(a) there exists a rational number π₯ in the interval (π, π);
(b) there exists an irrational number π¦ in the interval (π, π).
Conclude that between any two real numbers there are inο¬nitely many rational numbers and inο¬nitely many irrational numbers.
12.* Let π = {π₯ β β β£ π₯2 β€ 2}. Using the result of the previous exercise, prove that
max π does not exist.
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Hints to challenging exercises:
β β
8: Assume that 5/ 2 = π/π with integer π, π which have no common factors. Square this and following the method of the proof of
Theorem 3.1 from the
β lecture
β notes, obtain a contradiction.
9d: Consider 2 + (β 2).
β
β
9e: Consider 2 × (1/ 2).
β
10b: by the result of Exercise 10c, for any rational π β= 0, the number π 2 is irrational.
1 , 2 , . . . Try to make sure that one of these points is in the
11a: Choose π very large. Consider the grid of rational points . . . β1
, 0, π
π
π
interval (π, π).
β
β
β
11b: Choose π very large. Consider the grid of irrational points . . . βπ 2 , 0, π2 , 2 π 2 , . . . Try to make sure that one of these points is in
the interval (π, π).
12: Use proof by contradiction and the fact that between any two real numbers there is a rational number.
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