Probability Theory Primer: part I

Illustrative Example 1:

What is the chance of throwing a number greater than 4 with a fair die whose faces are numbered from 1 to 6?

Solution 1:

There are 6 possible ways in which the die can fall, and of these two are the favourable events required.

Hence, required chance is \frac{2}{6}=\frac{1}{3}.

Illustrative example 2:

From a bag containing 4 white and 5 black balls a man draws 3 at random; what are the odds against these being all black?

Solution 2:

The total number of ways in which 3 balls can be drawn is 9 \choose 3, and the number of ways of drawing 3 black balls is 5 \choose 3; therefore, the chance of drawing 3 black balls is equal to

\frac{{5 \choose 3}}{{9 \choose 3}}=\frac{5.4.3}{9.3.7}=\frac{5}{43}.

Thus, the odds against the event are 37 to 5.

Illustrative example 3:

Find the chance of throwing at least one ace in a single throw with two dice.

Solution 3:

In die games, there is no ace. We can pick up any number as an ace in this question. Once it is chosen, it is fixed.

So, the possible number of cases is 6 \times 6=36.

An ace on one die may be associated with any of the six numbers on the other die, and the remaining five numbers on the first die may each be associated with the ace on the second die; thus, the number of favourable cases is 11.

Therefore, the required probability is \frac{11}{36}.

Illustrative example 4:

Find the chance of throwing more than 15 in one throw with 3 dice.

Solution 4:

A throw amounting to 18 must be made up of 6, 6, 6 and this can occur in one way only; 17 can be made up of 6. 6. 5 which can occur in 3 ways; 16 may be made up of 6, 6, 4 and 6, 5, 5 each of whichΒ arrangementsΒ can occur in 3 ways.

Therefore, the number of favourable cases is 1+3+3+3, that is, 10.

And, the total number of cases possible is 6^{3}, that is, 216.

Hence, the required probability is \frac{10}{216}=5/108.

Illustrative example 5:

A has 3 shares in a lottery, in which there are 3 prizes and 6 blanks; B has 1 share in a lottery in which there is 1 prize and 3 blanks; show that A’s chance of success is to B’s as 16:7.

Solution 5:

A may draw 3 prizes in one way; A may draw 2 prizes and 1 blank in \frac{3.3}{1.2} \times 6 ways; A may draw 1 prize and 2 blanks in 3 \times \frac{6.5}{1.2} ways; the sum of these numbers is 64, which is the number of ways in which A can win a prize. Also, he can draw 3 tickets in \frac{9.8.7}{1.2.3} ways, that is, 84 ways.

Hence, A’s chance of success is \frac{64}{84} = \frac{16}{21}.

B’s chance of success is clearly \frac{1}{3}.

Therefore, the required ratio is \frac{\frac{16}{21}}{\frac{1}{3}} = \frac{16}{7}.

Tutorial questions:

  1. In a single throw with two dice, find the chances of throwing (a) a five (b) a six.
  2. From a pack of 52 cards, two cards are drawn at random, find the chance that one is a knave and other is a queen.
  3. A bag contains 5 white, 7 black and 4 red balls. Find the chance that 3 balls all drawn at random are all white.
  4. If four coins are tossed, find the chance that there are two heads and two tails.
  5. One of two events must happen; given that the chance of one is two-thirds that of the other, find the odds in favour of the other.
  6. If from a pack four cards are drawn, find the chance that they will be the four honours of the same suit.
  7. Thirteen persons take their places at a round table, show that it is five to one against two particular persons sitting together.
  8. There are three events A, B, C, out of which one must, and only one can happen; the odds are 8 is to 3 against A; 5 to 2 against B; find the odds against C.
  9. Compare the chance of throwing 4 with one die, 8 with two dice and 12 with three dice.
  10. In shuffling a pack of cards, four are accidentally dropped; find the chance that the missing cards should be one from each suit.
  11. A has 3 shares in a lottery containing 3 prizes and 9 blanks; B has 2 shares in a lottery containing 2 prizes and 6 blanks; compare their chances of success.
  12. Show that the chances of throwing six with 4, 3 or 2 dice respectively are as 1:6:16
  13. There are three works, one consisting of three volumes, one of four volumens, and the other of one volume. They are placed on a shelf at random; prove that the chance that the volumes of the same works are all together is \frac{3}{140}.
  14. A and B throw with two dice; if A throws 9, find B’s chance of throwing a higher number.
  15. The letters forming the word Clifton are placed at random in a row; what is the chance that the two vowels come together?
  16. In a hand, what is the chance that the four kings are held by a specified player?

Regards,

Nalin Pithwa.

 

 

 

 

 

A fourth primer: plane geometry question set including core theorems, preRMO and RMO

Hard core definitions of various special quadrilaterals:

  1. A quadrilateral is a plane figure bounded by four straight lines.
  2. A parallelogram is a quadrilateral whose opposite sides are parallel.
  3. A rectangle is a parallelogram which has one of its angles a right angle.
  4. A square is a rectangle which has two adjacent sides equal.
  5. A rhombus is a quadrilateral which has all its sides equal, but its angles are not right angles.
  6. A trapezium if a quadrilateral which has one pair of parallel sides.

Problem 1:

The straight lines which join the extremities of two equal and parallel straight lines towards the same parts are themselves equal and parallel. Prove this.

Problem 2:

The opposite sides and angles of a parallelogram are equal to one another, and each diagonal bisects the parallelogram.

Problem 3:

Corollary 1 of problem 2 above: If one angle of a parallelogram is a right angle, all its angles are right angles.

Corollary 2 of problem 2 above: All the sides of a square are equal; and all its angles are right angles.

Corollary 3 of problem 3 above: The diagonals of a parallelogram bisect one another.

Problem 4:

If the opposite sides of a quadrilateral are equal, then the figure is a parallelogram.

Problem 5:

If the opposite angles of a quadrilateral are equal, then the figure is a parallelogram.

Problem 6:

If the diagonals of a quadrilateral bisect each other, then the figure is a parallelogram.

Problem 7:

The diagonals of a rhombus bisect each other at right angles.

Problem 8:

If the diagonals of a parallelogram are equal, all its angles are right angles.

Problem 9:

In a parallelogram which is not rectangular, the diagonals are not equal.

Problem 10:

Any straight line drawn through the middle point of a diagonal of a parallelogram and terminated by a pair of opposite sides is bisected at that point.

Problem 11:

In a parallelogram, the perpendiculars drawn from one pair of opposite angles to the diagonal which joins the other pair are equal. Prove this.

Problem 12:

If ABCD is a parallelogram, and X, Y respectively the middle points of the sides AD, BC, show that the figure AYCX is a parallelogram.

Problem 13:

ABC and DEF are two triangles such that AB, BC are respectively equal to and parallel to DE, EF; show that AC is equal and parallel to DF.

Problem 14:

ABCD is a quadrilateral in which AB is parallel to DC, and AD equal but not parallel to BC; show that (i) angle A + angle C = 180 degrees = angle B + angle D; (ii) diagonal AC = diagonal BD (iii) the quadrilateral is symmetrical about the straight line joining the middle points of AB and DC.

Problem 15:

AP, BQ are straight rods of equal length, turning at equal rates (both clockwise) about two fixed pivots A and B respectively. If the rods start parallel but pointing in opposite senses, prove that (i) they will always be parallel (ii) the line joining PQ will always pass through a certain fixed point.

Problem 16:

A and B are two fixed points, and two straight lines AP, BQ, unlimited towards P and Q, are pivoted at A and B. AP, starting from the direction AB, turns about A clockwise at the uniform rate of 7.5 degrees a second; and BQ, starting simultaneously from the direction BA, turns about B counter-clockwise at the rate of 3.75 degrees a second. (i) How many seconds will elapse before AP and BQ are parallel? (ii) Find graphically and by calculation the angle between AP and BQ twelve seconds from the start. (iii) At what rate does this angle decrease?

Problem 17 (Intercept theorem or Basic Proportionality Theorem):

If there are three or more parallel straight lines, and the intercepts made by them on any transversal are equal, then the corresponding intercepts on any other transversal are also equal.

Prove the corollary: In a triangle ABC, if a set of lines Pp, Qq, Rr, \ldots, drawn parallel to the base, divide one side AB into equal parts they also divide the other side AC into equal parts.

Definition:

If from the extremities of a straight line AB perpendiculars AX, BY are drawn parallel to a straight line PQ of indefinite length, then XY is said to be the orthogonal projection of AB on PQ.

Problem 18:

Prove: The straight line drawn through the middle point of a side of a triangle parallel to the base bisects the remaining side.

Problem 19:

The straight line which joins the middle points of two sides of a triangle is equal to half the third side.

More later,

Nalin Pithwa.

 

 

 

 

A third primer: plane geometry question set for preRMO and RMO

Problem 1:

Prove: Two right angled triangles which have their hypotenuses equal, and one side of one equal to one side of the other, are equal in all respects. (PS: Please do not use “short-cut or the magic formula — Pythagoras’s theorem; if you do, you are requested to prove Pythagoras’s theorem using plane geometry!) (PS:2: Remember Euclid’s geometry builds up from “scratch”…a small step at a time….axioms, definitions, lemmas, …; you can only use the theorems used and proved so far in “class”).

Problem 2:

If two triangles have two sides of the one equal to two sides of the other, each to each, but the angle included by the two sides of one greater than the angle included by the corresponding sides of the other; then the base of that which has the greater angle is greater than the base of the other.

Problem 3:

Prove converse of the above: If two triangles have two sides of the one equal to two sides of the other, each to each, but the base of one greater than the base of the other, then the angle contained by the sides of that which has the greater base is greater than the angle contained by the corresponding sides of the other.

Problem 4:

(i) The perpendicular is the shortest line that can be drawn to a given straight line from a given point. (ii) Obliques which make equal angles with the perpendicular are equal. (iii) Of two obliques the less is that which makes the smaller angle with the perpendicular. Prove all these.

Problem 5:

If two triangles have two sides of the one equal to two sides of the other, each to each, and have likewise the angles opposite to one pair of equal sides equal, then the angles opposite to the other pair of equal sides are either equal or supplementary, and in the former case, the triangles are equal in all respects.

More later,

Cheers,

Nalin Pithwa

A Second primer: plane geometry for preRMO and RMO: basic questions set

Problem 1:

If from any point in the bisector of an angle a straight line is drawn parallel to either arm of the angle, the triangle thus formed is isosceles.

Problem 2:

From X, a point in the base BC of an isosceles triangle ABC. a straight line is drawn at right angles to the base, cutting AB in Y, and CA produced in Z, show the triangle AYZ is isosceles.

Problem 3:

If the straight line which bisects an exterior angle of a triangle is parallel to the opposite side, show that the triangle is isosceles.

Problem 4:

The straight line drawn from any point in the bisector of an angle parallel to the arms of the angle, and terminated by them, are equal, and the resulting figure is a quadrilateral having all its sides equal.

Problem 5:

AB and CD are two straight lines intersecting at D, and the adjacent angles so formed are bisected; if through any point X in DC a straight line XYZ is drawn parallel to AB and meeting the bisectors Y and Z, show that XY is equal to XZ.

Problem 6:

Two straight rods PA, QB revolve about pivots at P and Q, PA making 12 complete revolutions a minute, and QB making 10. If they start parallel and pointing the same way, how long will it be before they are again parallel (i) pointing opposite ways (ii) pointing the same way?

Problem 7:

Prove that: if two straight lines are perpendicular to two other straight lines, each to each, the acute angle between the first pair is equal to the acute angle between the second pair.

Problem 8:

Show that the only regular figures which may be fitted together so as to form a plane surface are (i) equilateral triangles (ii) squares (iii) regular hexagons.

Problem 9:

If one side of a regular hexagon is produced, show that the exterior angle is equal to the interior angle of an equilateral triangle.

Problem 10:

If a straight line meets two parallel straight lines, and the two interior angles on the same side are bisected, show that the bisectors meet at right angles.

Problem 11:

If the base of any triangle is produced both ways, show that the sum of the two exterior angles minus the vertical angle is equal to two right angles.

Problem 12:

In the triangle ABC, the base angles at B and C are bisected by BO and CO respectively. Show that the angle BOC is 90 degrees plus A/2.

Problem 13:

In the triangle ABC, the sides AB, AC are produced, and the exterior angles are bisected by BO and CO. Show that the angle BOC is 90 degrees minus A/2.

Problem 14:

Prove: the angle contained by the bisectors of two adjacent angles of a quadrilateral is equal to half the sum of the remaining angles.

Problem 15:

A is the vertex of an isosceles triangle ABC, and BA is produced to D, so that AD is equal to BA; if DC is drawn, show that BCD is a right angle. Prove this.

Problem 16:

Prove: The straight line joining the middle point of the hypotenuse at a right angled triangle to the right angle is equal to half the hypotenuse.

Problem 17:

If two triangles have two angles of one equal to two angles of the other, each to each, and any one side of the first equal to the corresponding side of the other, the triangles are equal in all respects. (ASA congruency test):

Problem 18:

Show that the perpendiculars drawn from the extremities of the base of an isosceles triangle to the opposite sides are equal.

Problem 19:

Prove: Any point on the bisector of an angle is equidistant from the arms of the angle.

Problem 20:

Through O the middle point of a straight line AB, any straight line is drawn, and perpendiculars AX and BY are dropped upon it from A and B: show that AX is equal to BY.

Problem 21:

If the bisector of the vertical angle of a triangle is at right angles to the base, the triangle is isosceles.

Problem 22:

If in a triangle the perpendicular from the vertex on the base bisects the base, then the triangle is isosceles. Prove.

Problem 23:

If the bisector of the vertical angle of a triangle also bisects the base, the triangle is isosceles. Prove this.

Problem 24:

The middle point of any straight line which meets two parallel straight lines, and is terminated by them, is equidistant from the parallels. Prove this.

Problem 25:

A straight line drawn between two parallels and terminated by them is bisected; show that any other straight line passing through the middle point and terminated by the parallels is also bisected at that point.

Problem 26:

If through a point equidistant from two parallel straight lines, two straight lines are drawn cutting the parallels, the portions of the latter thus intercepted are equal. Prove this.

Problem 27:

In a quadrilateral ABCD, if AB=CD, and BC=DC: show that the diagonal AC bisects each of the angles which it joins, and that AC is perpendicular to BD.

Problem 28:

A surveyor wishes to ascertain the breadth of a river which he cannot cross. Standing at a point A, near the bank, he notes an object B immediately opposite on the other bank. He lays down a line AC of any length at right angles to AB, fixing a mark at O the middle point of AC. From C he walks along a line perpendicular to AC until he reaches a point D from which O and B are seen in the same direction. He now measures CD: prove that the result gives him the width of the river.

More later,

Cheers,

Nalin Pithwa.

PS: There is no royal road to geometry —- Plato.

Towards Baby Analysis: Part I: INMO, IMO and CMI Entrance

\bf{Reference: \hspace{0.1in}Introductory \hspace{0.1in} Real Analysis: \hspace{0.1in} Kolmogorov \hspace{0.1in} and \hspace{0.1in} Fomin; \hspace{0.1in}Dover \hspace{0.1in }Publications}

\bf{Equivalence \hspace{0.1in} of \hspace{0.1in} Sets \hspace{0.1in} The \hspace{0.1in}Power \hspace{0.1in }of \hspace{0.1in }a \hspace{0.1in}Set}

\bf{Section 1}:

\bf{Finite \hspace{0.1in} and \hspace{0.1in} infinite \hspace{0.1in} sets}

The set of all vertices of a given polyhedron, the set of all prime numbers less than a given number, and the set of all residents of NYC (at a given time) have a certain property in common, namely, each set has a definite number of elements which can be found in principle, if not in practice. Accordingly, these sets are all said to be \it{finite}.\it{Clearly \hspace{0.1in} we \hspace{0.1in}can \hspace{0.1in} be \hspace{0.1in} sure \hspace{0.1in} that \hspace{0.1in} a \hspace{0.1in} set \hspace{0.1in}is \hspace{0.1in}finite \hspace{0.1in} without \hspace{0.1in} knowing \hspace{0.1in} the \hspace{0.1in} number \hspace{0.1in} of elements \hspace{0.1in}in \hspace{0.1in}it.}

On the other hand, the set of all positive integers, the set of all points on the line, the set of all circles in the plane, and the set of all polynomials with rational coefficients have a different property in common, namely, \it{if \hspace{0.1in } we \hspace{0.1in}remove \hspace{0.1in} one \hspace{0.1in} element \hspace{0.1in}from \hspace{0.1in}each \hspace{0.1in}set, \hspace{0.1in}then \hspace{0.1in}remove \hspace{0.1in}two \hspace{0.1in}elements, \hspace{0.1in}three \hspace{0.1in}elements, \hspace{0.1in}and \hspace{0.1in}so \hspace{0.1in}on, \hspace{0.1in}there \hspace{0.1in}will \hspace{0.1in}still \hspace{0.1in}be \hspace{0.1in}elements \hspace{0.1in}left \hspace{0.1in}in \hspace{0.1in}the \hspace{0.1in}set \hspace{0.1in}in \hspace{0.1in}each \hspace{0.1in}stage}. Accordingly, sets of these kind are called \it{infinite} sets.

Given two finite sets, we can always decide whether or not they have the same number of elements, and if not, we can always determine which set has more elements than the other. It is natural to ask whether the same is true of infinite sets. In other words, does it make sense to ask, for example, whether there are more circles in the plane than rational points on the line, or more functions defined in the interval [0,1] than lines in space? As will soon be apparent, questions of this kind can indeed be answered.

To compare two finite sets A and B, we can count the number of elements in each set and then compare the two numbers, but alternatively, we can try to establish a \it{one-\hspace{0.1in}to-\hspace{0.1in}one \hspace{0.1in}correspondence} between the elements of set A and set B, that is, a correspondence such that each element in A corresponds to one and only element in B, and vice-versa. It is clear that a one-to-one correspondence between two finite sets can be set up if and only if the two sets have the same number of elements. For example, to ascertain if or not the number of students in an assembly is the same as the number of seats in the auditorium, there is no need to count the number of students and the number of seats. We need merely observe whether or not there are empty seats or students with no place to sit down. If the students can all be seated with no empty seats left, that is, if there is a one-to-one correspondence between the set of students and the set of seats, then these two sets obviously have the same number of elements. The important point here is that the first method(counting elements) works only for finite sets, while the second method(setting up a one-to-one correspondence) works for infinite sets as well as for finite sets.

\bf{Section 2}:

\bf{Countable \hspace{0.1in} Sets}.

The simplest infinite set is the set \mathscr{Z^{+}} of all positive integers. An infinite set is called \bf{countable} if its elements can be put into one-to-one correspondence with those of \mathscr{Z^{+}}. In other words, a countable set is a set whose elements can be numbered a_{1}, a_{2}, a_{3}, \ldots a_{n}, \ldots. By an \bf{uncountable} set we mean, of course, an infinite set which is not countable.

We now give some examples of countable sets:

\bf{Example 1}:

The set \mathscr{Z} of all integers, positive, negative, or zero is countable. In fact, we can set up the following one-to-one correspondence between \mathscr{Z} and \mathscr{Z^{+}} of all positive integers: (0,1), (-1,2), (1,3), (-2,4), (2,5), and so on. More explicitly, we associate the non-negative integer $n \geq 0$ with the odd number 2n+1, and the negative integer n<0 with the even number 2|n|, that is,

n \leftrightarrow (2n+1), if n \geq 0, and n \in \mathscr{Z}
n \leftrightarrow 2|n|, if n<0, and n \in \mathscr{Z}

\bf{Example 2}:

The set of all positive even numbers is countable, as shown by the obvious correspondence n \leftrightarrow 2n.

\bf{Example 3}:

The set 2,4,8,\ldots 2^{n} is countable as shown by the obvious correspondence n \leftrightarrow 2^{n}.

\bf{Example 4}:    The set latex \mathscr{Q}$ of rational numbers is countable. To see this, we first note that every rational number \alpha can be written as a fraction \frac{p}{q}, with q>0 with a positive denominator. (Of course, p and q are integers). Call the sum |p|+q as the “height” of the rational number \alpha. For example,

\frac{0}{1}=0 is the only rational number of height zero,

\frac{-1}{1}, \frac{1}{1} are the only rational numbers of height 2,

\frac{-2}{1}, \frac{-1}{2}, \frac{1}{2}, \frac{2}{1} are the only rational numbers of height 3, and so on.

We can now arrange all rational numbers in order of increasing “height” (with the numerators increasing in each set of rational numbers of the same height). In other words, we first count the rational numbers of height 1, then those of height 2 (suitably arranged), then those of height 3(suitably arranged), and so on. In this way, we assign every rational number a unique positive integer, that is, we set up a one-to-one correspondence between the set Q of all rational numbers and the set \mathscr{Z^{+}} of all positive integers.

\it{Next \hspace{0.1in}we \hspace{0.1in} prove \hspace{0.1in}some \hspace{0.1in}elementary \hspace{0.1in}theorems \hspace{0.1in}involving \hspace{0.1in}countable \hspace{0.1in}sets}

\bf{Theorem1}.

\bf{Every \hspace{0.1in} subset \hspace{0.1in}of \hspace{0.1in}a \hspace{0.1in}countable \hspace{0.1in}set \hspace{0.1in}is \hspace{0.1in}countable}.

\bf{Proof}

Let set A be countable, with elements a_{1}, a_{2}, a_{3}, \ldots, and let set B be a subset of A. Among the elements a_{1}, a_{2}, a_{3}, \ldots, let a_{n_{1}}, a_{n_{2}}, a_{n_{3}}, \ldots be those in the set B. If the set of numbers n_{1}, n_{2}, n_{3}, \ldots has a largest number, then B is finite. Otherwise, B is countable (consider the one-to-one correspondence i \leftrightarrow a_{n_{i}}). \bf{QED.}

\bf{Theorem2}

\bf{The \hspace{0.1in}union \hspace{0.1in}of \hspace{0.1in}a \hspace{0.1in}finite \hspace{0.1in}or \hspace{0.1in}countable \hspace{0.1in}number \hspace{0.1in}of \hspace{0.1in}countable \hspace{0.1in}sets \hspace{0.1in}A_{1}, A_{2}, A_{3}, \ldots \hspace{0.1in}is \hspace{0.1in}itself \hspace{0.1in}countable.}

\bf{Proof}

We can assume that no two of the sets A_{1}, A_{2}, A_{3}, \ldots have any elements in common, since otherwise we could consider the sets A_{1}, A_{2}-A_{1}, A_{3}-(A_{1}\bigcup A_{2}), \ldots, instead, which are countable by Theorem 1, and have the same union as the original sets. Suppose we write the elements of A_{1}, A_{2}, A_{3}, \ldots in the form of an infinite table

\begin{array}{ccccc} a_{11} & a_{12} & a_{13} & a_{14} &\ldots \\ a_{21} &a_{22} & a_{23} & a_{24} & \ldots \\ a_{31} & a_{32} & a_{33} & a_{34} & \ldots \\ a_{41} & a_{42} & a_{43} & a_{44} & \ldots \\ \ldots & \ldots & \ldots & \ldots & \ldots \end{array}

where the elements of the set A_{1} appear in the first row, the elements of the set A_{2} appear in the second row, and so on. We now count all the elements in the above array “diagonally”; that is, first we choose a_{11}, then a_{12}, then move downwards, diagonally to “left”, picking a_{21}, then move down vertically picking up a_{31}, then move across towards right picking up a_{22}, next pick up a_{13} and so on (a_{14}, a_{23}, a_{32}, a_{41})as per the pattern shown:

\begin{array}{cccccccc}   a_{11} & \rightarrow & a_{12} &\hspace{0.1in} & a_{13} & \rightarrow a_{14} & \ldots \\     \hspace{0.1in} & \swarrow & \hspace{0.1in} & \nearrow & \hspace{0.01in} & \swarrow & \hspace{0.1in} & \hspace{0.1in}\\     a_{21} & \hspace{0.1in} & a_{22} & \hspace{0.1in} & a_{23} \hspace{0.1in} & a_{24} & \ldots \\     \downarrow & \nearrow & \hspace{0.1in} & \swarrow & \hspace{0.1in} & \hspace{0.1in} & \hspace{0.1in} & \hspace{0.1in}\\     a_{31} & \hspace{0.1in} & a_{32} & \hspace{0.1in} & a_{33} & \hspace{0.1in} & a_{34} & \ldots \\     \hspace{0.1in} & \swarrow & \hspace{0.1in} & \hspace{0.1in} & \hspace{0.1in} & \hspace{0.1in} & \hspace{0.1in} & \hspace{0.1in}\\    a_{41} & \hspace{0.1in} & a_{42} &\hspace{0.1in} & a_{43} &\hspace{0.1in} &a_{44} &\ldots\\     \ldots & \hspace{0.1in} & \ldots & \hspace{0.1in} & \ldots & \hspace{0.1in} & \ldots & \hspace{0.1in} \end{array}

It is clear that this procedure associates a unique number to each element in each of the sets A_{1}, A_{2}, \ldots thereby establishing a one-to-one correspondence between the union of the sets A_{1}, A_{2}, \ldots and the set \mathscr{Z^{+}} of all positive integers. \bf{QED.}

\bf{Theorem3}

\bf{Every \hspace{0.1in}infinite \hspace{0.1in}subset \hspace{0.1in}has \hspace{0.1in}a \hspace{0.1in}countable \hspace{0.1in}subset.}

\bf{Proof}

Let M be an infinite set and a_{1} any element of M. Being infinite, M contains an element a_{2} distinct from a_{1}, an element a_{3} distinct from both a_{2} and a_{1}, and so on. Continuing this process, (which can never terminate due to “shortage” of elements, since M is infinite), we get a countable subset A= \{ a_{1}, a_{2}, a_{3}, \ldots, a_{n}, \ldots\} of the set M. \bf{QED.}

\bf{Remark}

Theorem 3 shows that countable sets are the “smallest” infinite sets. The question of whether there exist uncountable (infinite) sets will be considered below.

\bf{Section3}

\bf{Equivalence \hspace{0.1in} of \hspace{0.1in} sets}

We arrived at the notion of a countable set M by considering one-to-one correspondences between set M and the set \mathscr{Z^{+}} of all positive integers. More generally, we can consider one-to-one correspondences between any two sets M and N.

\bf{Definition}

Two sets M and N are said to be \bf{equivalent} (written M \sim N) if there is a one-to-one correspondence between the elements of M and the elements of N.

The concept of equivalence is applicable both to finite and infinite sets. Two finite sets are equivalent if and only if they have the same number of elements. We can now define a countable set as a set equivalent to the set \mathscr{Z^{+}} of all positive integers. It is clear that two sets are equivalent to a third set are equivalent to each other, and in particular that any two countable sets are equivalent.

\bf{Example1}

The sets of points in any two closed intervals $[a,b]$ and $[c,d]$ are equivalent; you can “see’ a one-to-one correspondence by drawing the following diagram: Step 1: draw cd as a base of a triangle. Let the third vertex of the triangle be O. Draw a line segment “ab” above the base of the triangle; where “a” lies on one side of the triangle and “b” lies on the third side of the third triangle. Note that two points p and q correspond to each other if and only if they lie on the same ray emanating from the point O in which the extensions of the line segments ac and bd intersect.

\bf{Example2}

The set of all points z in the complex plane is equivalent to the set of all points z on a sphere. In fact, a one-to-one correspondence z \leftrightarrow \alpha can be established by using stereographic projection. The origin is the North Pole of the sphere.

\bf{Example3}

The set of all points x in the open unit interval (0,1) is equivalent to the set of all points y on the whole real line. For example, the formula y=\frac{1}{\pi}\arctan{x}+\frac{1}{2} establishes a one-to-one correspondence between these two sets. \bf{QED}.

The last example and the examples in Section 2 show that an infinite set is sometimes equivalent to one of its proper subsets. For example, there are “as many” positive integers as integers of arbitrary sign, there are “as many” points in the interval (0,1) as on the whole real line, and so on. This fact is characteristic of all infinite sets (and can be used to define such sets) as shown by:

\bf{Theorem4}

\bf{Every \hspace{0.1in} infinite \hspace{0.1in} set \hspace{0.1in}is \hspace{0.1in} equivalent \hspace{0.1in} to \hspace{0.1in}one \hspace{0.1in}of \hspace{0.1in}its \hspace{0.1in}proper \hspace{0.1in}subsets.}

\bf{Proof}

According to Theorem 3, every infinite set M contains a countable subset. Let this subset be A=\{a_{1}, a_{2}, a_{3}, \ldots, a_{n}, \ldots \} and partition A into two countable subsets A_{1}=\{a_{1}, a_{3}, a_{5}, \ldots \} and A_{2}=\{a_{2}, a_{4}, a_{6}, \ldots \}.

Obviously, we can establish a one-to-one correspondence between the countable subsets A and A_{1} (merely let a_{n} \leftrightarrow a_{2n-1}). This correspondence can be extended to a one-to-one correspondence between the sets A \bigcup (M-A)=M and A_{1} \bigcup (M-A)=M-A_{2} by simply assigning x itself to each element x \in M-A. But M-A_{2} is a proper subset of M. \bf{QED}.

More later, to be continued,

Regards,
Nalin Pithwa

Set Theory, Relations, Functions: preliminaries: part 10: more tutorial problems for practice

Problem 1:

Prove that a function f is 1-1 iff f^{-1}(f(A))=A for all A \subset X. Given that f: X \longrightarrow Y.

Problem 2:

Prove that a function if is onto iff f(f^{-1}(C))=C for all C \subset Y. Given that f: X \longrightarrow Y.

Problem 3:

(a) How many functions are there from a non-empty set S into \phi\?

(b) How many functions are there from \phi into an arbitrary set S?

(c) Show that the notation \{ X_{i} \}_{i \in I} implicitly involves the notion of a function.

Problem 4:

Let f: X \longrightarrow Y be a function, let A \subset X, B \subset X, C \subset Y and D \subset Y. Prove that

i) f(A \bigcap B) \subset f(A) \bigcap f(B)

ii) f^{-1}(f(A)) \supset A

iii) f(f^{-1}(C)) \subset C

Problem 5:

Let I be a non-empty set and for each i \in I, let X_{i} be a set. Prove that

(a) for any set B, we have B \bigcap \bigcup_{i \in I}X_{i}=\bigcup_{i \in I}(B \bigcap X_{i})

(b) if each X_{i} is a subset of a given set S, then (\bigcup_{i \in I}X_{i})^{'}=\bigcap_{i \in I}(X_{i})^{'} where the prime indicates complement.

Problem 6:

Let A, B, C be subsets of a set S. Prove the following statements:

(i) A- (B-C)=(A-B)\bigcup(A \bigcap B \bigcap C)

(ii) (A-B) \times C=(A \times C)-(B \times C)

πŸ™‚ πŸ™‚ πŸ™‚

Nalin Pithwa

Set Theory, Relations, Functions: Preliminaries: Part IX: (tutorial problems)

Reference: Introductory Real Analysis, Kolmogorov and Fomin, Dover Publications.

Problem 1:

Prove that if A \bigcup B=A and A \bigcap B=A, then A=B.

Problem 2:

Show that in general (A-B)\bigcup B \neq A.

Problem 3:

Let A = \{ 2,4, \ldots, 2n, \ldots\} and B= \{ 3,6,\ldots, 3n, \ldots\}. Find A \bigcap B and A - B.

Problem 4:

Prove that (a) (A-B)\bigcap (C)=(A \bigcap C)-(B \bigcap C)

Prove that (b) A \Delta B = (A \bigcup B)-(A \bigcap B)

Problem 5:

Prove that \bigcup_{a}A_{\alpha}-\bigcup_{a}B_{\alpha}=\bigcup_{\alpha}(A_{\alpha}-B_{\alpha})

Problem 6:

Let A_{n} be the set of all positive integers divisible by n. Find the sets (i) \bigcup_{n=2}^{\infty}A_{n} (ii) \bigcap_{n=2}^{\infty}A_{n}.

Problem 7:

Find (i) \bigcup_{n=1}^{\infty}[n+\frac{1}{n}, n - \frac{1}{n}] (ii) \bigcap_{n=1}^{\infty}(a-\frac{1}{n},b+\frac{1}{n})

Problem 8:

Let A_{\alpha} be the set of points lying on the curve y=\frac{1}{x^{\alpha}} where (0<x<\infty). What is \bigcap_{\alpha \geq 1}A_{\alpha}?

Problem 9:

Let y=f(x) = <x> for all real x, where <x> is the fractional part of x. Prove that every closed interval of length 1 has the same image under f. What is the image? Is f one-to-one? What is the pre-image of the interval \frac{1}{4} \leq y \leq \frac{3}{4}? Partition the real line into classes of points with the same image.

Problem 10:

Given a set M, let \mathscr{R} be the set of all ordered pairs on the form (a,a) with a \in M, and let aRb if and only if (a,b) \in \mathscr{R}. Interpret the relation R.

Problem 11:

Give an example of a binary relation which is:

  • Reflexive and symmetric, but not transitive.
  • Reflexive, but neither symmetric nor transitive.
  • Symmetric, but neither reflexive nor transitive.
  • Transitive, but neither reflexive nor symmetric.

We will continue later, πŸ™‚ πŸ™‚ πŸ™‚

PS: The above problem set, in my opinion, will be very useful to candidates appearing for the Chennai Mathematical Institute Entrance Exam also.

Nalin Pithwa

 

 

Set Theory, Relations, Functions: Preliminaries: part VIIIA

(We continue from part VII of the same blog article series with same reference text).

Theorem 4:

A set M can be partitioned into classes by a relation R (acting as a criterion for assigning two elements to the same class) if and only R is an equivalence relation on M.

Proof of Theorem 4:

Every partition of M determines a binary relation on M, where aRb means that “a belongs to the same class as b.” It is then obvious that R must be reflexive, symmetric and transitive, that is, R is an equivalence relation on M.

Conversely, let R be an equivalence relation on M, and let K_{a} be the set of all elements x \in M such that xRa (clearly, a \in K_{a}, since R is reflexive). Then, two classes K_{a} and K_{b} are either identical or disjoint. In fact, suppose that an element c belongs to both K_{a} and K_{b}, so that cRa and cRb. But by symmetry of R, being an equivalence relation, we can infer that aRc also and, further by transitivity, we say that aRb. If now, x \in K_{a} then we have xRa and hence, xRb (since we already have aRb and using transitivity).

Similarly, we can prove that x \in K_{b} implies that x \in K_{a}.

Therefore, K_{a}=K_{b} if K_{a} and K_{b} have an element in common. Therefore, the distinct sets K_{a} form a partition of M into classes.

QED.

Remark:

Because of theorem 4, one often talks about the decomposition of a set M into equivalence classes.

There is an intimate connection between mappings and partitions into classes, as illustrated by the following examples:

Example 1:

Let f be a mapping of a set A into a set B and partition A into sets, each consisting of all elements with the same image b=f(a) \in B. This gives a partition of A into classes. For example, suppose f projects the xy-plane onto the x-axis by mapping the point (x,y) into the point (x,0). Then, the preimages of the points of the x-axis are vertical lines, and the representation of the plane as the union of these lines is the decomposition into classes corresponding to f.

Example 2:

Given any partition of a set A into classes, let B be the set of these classes and associate each element a \in A with the class (that is, element of B) to which it belongs. This gives a mapping of A into B. For example, suppose we partition three-dimensional space into classes by assigning to the same class all points which are equidistant from the origin of coordinates. Then, every class is a sphere of a certain radius. The set of all these classes can be identified with the set of points on the half-line [0, \infty) each point corresponding to a possible value of the radius. In this sense, the decomposition of 3-dimensional space into concentric spheres corresponds to the mapping of space into the half-line [0,\infty).

Example 3:

Suppose that we assign all real numbers with the same fractional part to the same class. Then, the mapping corresponding to this partition has the effect of “winding” the real line onto a circle of unit circumference. (Note: The largest integer \leq x is called the integral part of x, denoted by [x], and the quantity x -[x] is called the fractional part of x).

In the next blog article, let us consider a tutorial problem set based on last two blogs of this series.

πŸ™‚ πŸ™‚ πŸ™‚

Nalin Pithwa

 

 

 

 

 

Set Theory, Relations, Functions: Preliminaries: Part VIII

SETS and FUNCTIONS:

Reference: Introductory Real Analysis by A. N. Kolmogorov and S V Fomin, Dover books.

Operations on sets:

Let A and B be any two sets. Then, by the sum or union of A and B, denoted by A \bigcup B, is meant the set consisting of all elements which belong to at least one of the sets A and B. More generally, by the sum or union of an arbitrary number (finite or infinite) of sets A_{\alpha} (indexed by some parameter \alpha), we mean the set, denoted by \bigcup_{\alpha}A_{\alpha} of all elements belonging to at least one of the sets A_{\alpha}.

By the intersection A \bigcap B of two given sets A and B, we mean the set consisting of all elements which belong to both A and B. For example, the intersection of the set of all even numbers and the set of all integers divisible by 3 is the set of all integers divisible by 6. By the intersection of an arbitrary number (finite or infinite) of sets A_{\alpha}, we mean the set, denoted by \bigcap_{\alpha}A_{\alpha} of all elements belonging to every one of the sets A_{\alpha}. Two sets A and B are said to be disjoint if A \bigcap B=\phi, that is, if they have no elements in common. More generally, let \mathscr{F} be a family of sets such that A \bigcap B=\phi for every pair of sets A, B in \mathscr{F}. Then, the sets in \mathscr{F} are said to be pairwise disjoint.

It is an immediate consequence of the above definitions that the operations \bigcup and \bigcap are commutative and associative, that is,

A \bigcup B=B \bigcup A and (A \bigcup B) \bigcup C= A \bigcup (B \bigcup C); A \bigcap B=B \bigcap A; and, (A \bigcap B)\bigcap C = A \bigcap (B \bigcap C).

Moreover, the operations \bigcup and \bigcap obey the following distributive laws:

(A \bigcup B)\bigcap C = (A \bigcap C) \bigcup (B \bigcap C)….call this I

(A \bigcap B)\bigcup C = (A \bigcup C) \bigcap (B \bigcup C)….call this II.

For example, suppose that x \in (A \bigcup B)\bigcap C, so that x belongs to the left-hand side of I. Then, x belongs to both C and A \bigcup B, that is, x belongs to both C and at least one of the sets A and B. But then x belongs to at least one of the sets A \bigcap C and B \bigcap C, that is, x \in (A \bigcap C)\bigcup (B \bigcap C), so that x belongs to the right hand side of I. Conversely, suppose that x \in (A \bigcap C)\bigcup (B \bigcap C). Then, x belongs to at least one of the two sets A \bigcap C and B \bigcap C. It follows that x belongs to both C and at least one of the two sets A and B, that is, x \in C and x \in A \bigcup B, or equivalently x \in (A \bigcup B)\bigcap C. This proves I, and II is proved similarly.

By the difference of two sets, A - B between two sets A and B (in that order), we mean the set of all elements of A which do not belong to B. Note that it is not assumed that A \supset B. It is sometimes convenient (e.g., in measure theory) to consider the symmetric difference of two sets A and B, denoted by A \delta B and defined as the union of the two differences A-B and B-A: A \delta B=(A-B)\bigcup (B-A).

We will often be concerned later with various sets which are all subsets of some underlying basic set R, for example, various sets of points on the real line. In this case, given a set A, the difference R-A is called the complement of A, denoted by A^{'}.

An important role is played in set theory and its applications by the following duality principle:

R - \bigcup_{\alpha}A_{\alpha}=\bigcap_{\alpha}(R-A_{\alpha})…call this III

R - \bigcap_{\alpha}A_{\alpha}=\bigcup_{\alpha}(R-A_{\alpha})…call this IV.

In words, the complement of a union equal the intersection of the complements; and, the complement of an intersection equals the union of the complements. According to the duality principle, any theorem involving a family of subsets of a fixed set R can be converted automatically into another, “dual” theorem by replacing all subsets by their complements, all unions by intersections and all intersections by unions. To prove III, suppose that x \in R - \bigcup_{\alpha}A_{\alpha}….call this V.

Then, x does not belong to the union \bigcup_{\alpha}A_{\alpha}. …call this VI.

That is, x does not belong to any of the sets A_{\alpha}. It follows that x belongs to each of the complements R - \bigcup_{\alpha}A_{\alpha}, and hence, x \bigcap_{\alpha}(R-A_{\alpha})….call this VII.

Conversely, suppose that VII holds, so that x belongs to every set R - A_{\alpha}. Then, x does not belong to any of the sets A_{\alpha}, that is, x does not belong to the union VI, or equivalently V holds true. This proves 3.

Homework: Prove IV similarly.

Remark:

The designation “symmetric difference” for the set A \Delta B is not too apt, since A \Delta B has much in common with the sum A \bigcup B. In fact, in A \bigcup B the two statements “x belongs to A” and “x belongs to B” are joined by the conjunction “or” used in the “either …or …or both…” sense, while in A \Delta B the same two statements are joined by “or” used in the ordinary “either…or…” sense (as in “to be or not to be”). In other words, x belongs to A \bigcup B if and only if x belongs to either A or B or both, while x belongs to A \Delta B if and only if x belongs to either A or B but not both. The set A \Delta B can be regarded as a kind of “modulo-two sum” of the sets A and B, that is, a sum of the sets A and B in which elements are dropped if they are counted twice (once in A and once in B).

Functions and mappings. Images and preimages:

(Of course, we have dealt with this topic in quite detail so far in the earlier blog series; but as presented by Kolmgorov and Fomin here, the flavour is different; at any rate, it helps to revise. Revision refines the intellect.) πŸ™‚ πŸ™‚ πŸ™‚

A rule associating a unique real number y=f(x) with each element of a set real numbers X is said to define a (real) function f on X. The set X is called the domain (of definition) of f, and the set Y of all numbers f(x) such that x \in X is called the range of f.

More generally, let M and N be two arbitrary sets. Then a rule associating a unique element b=f(a) \in N with each element a \in M is again said to define a function f on M (or a function f with domain M). In this more general context, f is usually called a mapping of f M into N. By the same token, f is said to map M into N(and a into b).

If a is an element of M, the corresponding element b=f(a) is called the image of a (under the mapping f). Every element of M with a given element b \in N as its image is called a preimage of b. Note that in general b may have several pre-images. Moreover, N may contain elements with no pre-images at all. If b has a unique pre-image, we denote this pre-image by f^{-1}(b).

If A is a subset of M, the set of all elements f(a) \in N such that a \in A is called the image of A, denoted by f(A). The set of all elements of M whose images belong to a given set B \subset N is called the preimage of B, denoted by f^{-1}(B). If no element of B has a preimage, then f^{-1}(B)=\phi. A function f is said to map M into N if f(M)=N, as is always the case, and onto N if f(M)=N. Thus, every “onto” mapping is an “into” mapping, but converse is not true.

Suppose f maps M onto N. Then, f is said to be one-to-one if each element b \in N has a unique preimage f^{-1}(b). In this case, f is said to establish a one-to-one correspondence between M and N, and the mapping f^{-1} associating f^{-1}(b) is called the inverse of f.

Theorem I:

The preimage of the union of two sets is the union of the preimages of the sets f^{-1}(A \bigcup B)=f^{-1}(A) \bigcup f^{-1}(B).

Proof of Theorem I:

If x \in f^{-1}(A \bigcup B), then f(x) \in A \bigcup B so that f(x) belongs to at least one of the sets A and B. But, then x belongs to at least one of the sets f^{-1}(A) and f^{-1}(B), that is, x \in f^{-1}(A) \bigcup f^{-1}(B).

Conversely, if x \in f^{-1}(A) \bigcup f^{-1}(B), then x belongs to at least one of the sets f^{-1}(A) and f^{-1}(B). Therefore, f(x) belongs to at least one of the sets A and B, that is, f(x) \in A \bigcup B. But, then x \in f^{-1}(A \bigcup B).

QED.

Theorem 2:

The preimage of the intersection of two sets is the intersection of the preimages of the sets:

f^{-1}(A \bigcap B)=f^{-1}(A) \bigcap f^{-1}(B).

Proof of Theorem 2:

If x \in f^{-1}(A \bigcap B), then f(x) \in A \bigcap B, so that f(x) \in A and (meaning, simultaneously) f(x) \in B. But, then x \in f^{-1}(A) and x \in f^{-1}(B), that is, x \in f^{-1}(A) \bigcap f^{-1}(B).

Conversely, if x \in f^{-1}(A) \bigcap f^{-1}(B), then x \in f^{-1}(A) and x \in f^{-1}(B). Therefore, f(x) \in A and f(x) \in B, that is, f(x) \in A \bigcap B. But, then x \in f^{-1}(A \bigcap B).

QED.

Theorem 3:

The image of the union of two sets equals the union of the images of the sets f(A \bigcup B) = f(A) \bigcup f(B).

Proof of theorem 3:

If y \in f(A \bigcup B), then y=f(x) where x belongs to at least one of the sets A and B. Therefore, y=f(x) belongs to at least one of the sets f(A) and f(B). That is, x \in f(A) \bigcup f(B).

Conversely, if y \in f(A) \bigcup f(B), then y=f(x). where x belongs to at least one of the sets A and B, that is, x \in A \bigcup B and hence, y=f(x) \in f(A \bigcup B).

QED.

Remark I:

Surprisingly enough, the image of the intersection of two sets is not so “well-behaved”. The image of the intersection of two sets does not necessarily equal the intersection of the images of the sets. For example, suppose the mapping f projects the xy-plane onto the x-axis, carrying the point (x,y) into the (x,0). Then, the segments 0 \leq x \leq 1 with y=0, and 0 \leq x \leq 1 with y=1 do not intersect, although their images coincide.

Remark 2:

In the light of above remark: consider the following: If a function is not specified on elements, it is important in general to check that f isΒ well-defined.Β That is, unambiguously defined. For example, if the set A is the union of two subsets A_{1} and A_{2}, and we are considering a function f: A \longrightarrow B, then one can try to specify a function from set A to the set B \{ 0,1\} by declaring that f is to map everything in A_{1} to 0 and is to map everything in A_{2} to 1. This unambiguously defines f unless A_{1} and A_{2} have elements in common in which case it is not clear whether these elements should map to 0 or to 1. Checking that this f is well-defined therefore amounts to checking that A_{1} and A_{2} have no intersection.

Remark 3:

Theorems 1-3 continue to hold for unions and intersections of an arbitrary number (finite or infinite) of sets A_{\alpha}:

f^{-1}(\bigcup_{\alpha})A_{\alpha}=\bigcup_{\alpha}f^{-1}(A_{\alpha})

f^{-1}(\bigcap_{\alpha}A_{\alpha})=\bigcap_{\alpha}f^{-1}(A_{\alpha})

f(\bigcup_{\alpha})A_{\alpha}=\bigcup_{\alpha}f(A_{\alpha})

Decomposition of a set into classes. Equivalence relations.

Decompositions of a given set into pairwise disjoint subsets play an important role in a great variety of problems. For example, the plane (regarded as a point set) can be decomposed into lines parallel to the x-axis, three dimensional space can be decomposed into concentric spheres, the inhabitants of a given city can be decomposed into different age groups, and so on. Any such representation of a given set M as the union of a family of pairwise disjoint subsets of M is called a decomposition or partition of M into classes.

A decomposition is usually made on the basis of a certain criterion, allowing us to assign the elements of M to one class or another. For example, the set of all triangles in the plane can be decomposed into classes of congruent triangles, or classes of triangles of equal area, the set of all functions of x can be decomposed into classes of functions all taking the same value at a given point x, and so on. Despite the great variety of such criteria, they are not completely arbitrary. For example, it is obviously impossible to partition all real numbers into classes by assigning the number b to the same class as number a if and only if b>a. In fact, if b>a, b must be assigned to same class as a but then a cannot be assigned to same class as b, since a<b. Moreover, since a is not greater than itself, a cannot be assigned to the class containing itself!! As another example, it is impossible to partition the points of the plane into classes by assigning two points to the same class if and only if the distance between them is less than 1. In fact, if the distance between a and b is less than 1, and if the distance between b and c is less than 1, it does not follow that distance between a and c is less than 1. (Hint: Think of triangle inequality). Thus, by assigning a to the same class as b and b to the same class as c, we may well find that two points fall in the same class even though the distance between them is greater than 1!

These examples suggest conditions that which must be satisfied by any criterion it it is to be used as the basis for partitioning a given set into classes. Let M be a set, and let certain ordered pairs (a,b) of elements of M be called “labelled.” If (a,b) is a labelled pair, we say that a is related to b by the binary relation R and we write it symbolically as aRb. For example, if a and b are real numbers, aRb might mean a<b, while if a and b are triangles, aRb might mean that a and b have the same area. A relation between elements of M is called a relation on M, if there exists at least one labelled pair (a,b) for every a \in M. A relation R on M is called an equivalence relation (on M) if it satisfies the following three conditions:

  • Reflexivity: aRa for every a \in M.
  • Symmetry: If aRb, then bRa.
  • Transititivity: If aRb and bRc, then aRc.

πŸ™‚ πŸ™‚ πŸ™‚

Nalin Pithwa, more later