Methods of Proof

With

Focus on Mathematical Olympiads.

By

Prof. V.K. GroverDepartment of Mathematics,

Panjab University,

Chandigarh.

We cannot just list all the methods of proof, as new methods are developed from time to

time and this process is never ending. However we list here a few of these methods which are

commonly used in problems related to mathematical Olympiads irrespective of the branch of

mathematics to which the problem is related to.

Argument by Contradiction.

Mathematical induction.

No square is negative or completing squares.

Telescopic sums and products.

Arranging numbers in order.

Parity considerations.

Looking at extreme situation or solution.

Doing a problem in two different ways.

Relating two different situations

Pigeon hole principle.

Symmetric considerations.

Argument by contradictionIn this method first the statement is assumed to be false and then a sequence of

logical deductions yield a conclusion that contradicts either the hypothesis or a fact which

is known to be true.

Examples1. Prove that there are infinitely many prime numbers

Solution:- Assume to the contrary that there are only finitely many prime

numbers. List them all as p1=2 , p2=3 , p3=5 , ………. , pn. Consider the number

N=p1 . p2 . p3 …… pn+1Clearly this number N is greater than all the prime

numbers listed above hence cannot be one of these, so it must be a composite

number and hence divisible by one of the prime numbers, say pi as

p1 . p2 . p3 …… pn is divisible by pi this implies that 1 is also divisible by pi , a

contradiction and hence the fact is proved.

2. Let F={E1 , E2 ,………, E s } be a family of sets with r elements each. Show that if

the intersection of anyr+1 ( not necessarily distinct) sets in F is non empty then

the intersection of all the sets in F is non empty.

Solution:- Again assume the contrary , namely that the intersection of all sets in F

is empty. Consider the set E1= {x1 , x2 ,…… xr }Because none of x i for i=1,2 , … ..r

lies in the intersection of all the Ejs ( the intersection of all sets in F being empty),

it follows that for each i there we can find some set E jisuch that x i E ji. Then we

have

E1 ∩ E j1 ∩……∩ E jr=¿

Since, at the same time the intersection is included in E1 and does not contain any

of the elements ofE1. This contradicts the hypothesis and hence the result follows.

Mathematical Induction

We have following variations of the induction principle:

Given P (n ) , a property depending on an integer n,

(i) If P (n0 ) is true for some integer n0 and

(ii) If for every k ≥ n0, P (k ) is true implies that P (k+1 ) is true

Then P (n ) is true for all n ≥ n0 .

Given P (n ) a property depending on an integer n,

(i) If P(n0) is true for some integer n0 and

(ii) If for n0 ≤ i ≤ k, P (i ) is true implies P (k+1 ) is true

Then P (n ) is true for all integers n ≥n0.

Given P (n ), a property depending on an integer n,

(i) If P(m),P(m+1),………,P(m+r) are true for some integer m and

(ii) If truth of P(k),P(k+1),………,P(k+r)for k ≥ m implies the truth of

P(k+r+1)

Then P(n), is true for all integers n ≥ m.

We can also state backwards induction as follows

Given P(n), a property depending on an integer n,

(i) If P(n0) is true for some integer n0 and

(ii) If for every m≤ k ≤ n0, P(k) is true implies that P(k-1) is true

Then P(n) is true for all n = m-1,m,……..,n0.

Examples1. Prove that for any n ≥ 1, a 2n×2n chess board with one 1×1 square removed can be tiled

by Triminos, the blocks consisting of three 1×1 squares put together to make L-shape, of

the type shown in fig. 1

Fig.1

Solution:-The result is obvious for n = 1 as after removing one square from the 2×2

square we are left with just one trinimo. Now suppose the result is true for n-1, i.e. for the

2n-1×2n-1 squares, consider the 2n×2n square with one square removed, the following

diagram shows such square with n = 4

Fig.2

Divide this square in 4 equal parts ( here the division is shown by red lines in the

fig.2).Clearly one of the four parts will have the removed square. Place one trinimo at the

centre in such a way that it cover one square each from the remaining three of the four

parts as shown in the fig.3

Fig.3

2. For any natural number N, prove that

√2√3√4 √………… ..√(N−1¿¿¿¿¿)√N ¿¿¿¿¿ ¿3(¿)

Solution:-Here we use backward induction we will prove the following inequality

√m√(m+1)√(m+2)√………… ..√(N−1¿¿¿¿¿)√N ¿¿¿¿¿ ¿m+1¿

For all m = 2,3,……N.

For K= N , (**) becomes √ N < N+1 , which is clearly true for N > 1. Suppose the result is true

for k = m , for k = m-1 we need to prove

√(m−1)√m√(m+1)√………… ..√(N−1¿¿¿¿¿)√N ¿¿¿¿¿ ¿m

Using (**) we get

L.H.S. < √ (m−1 ) (m+1 ) < m , hence the result follows.

No square is negativeHere we use the fact that square of a real number is non negative.

Examples

1. Let a be a real number, prove that 4a – a4 ≤ 3.

Solution: - The inequality is equivalent to

(a2 – 1)2 +2(a -1)2 0,

Which is clearly true for real values of a.

2. Determine whether there exists a one to one function f : R R with the property that for

all x ,f ( x2 )−( f (x ))2≥ 14 .

Solution: - We will show that such a function does not exist. The idea is simple. We look

for two numbers that are equal to their squares, namely 0 and 1.

For x = 0, 1 we have f ( x2 )=f (x ). We first put x=0 in the given equation and

obtain f (0 )−(f (0))2≥ 14 , this implies ( f (0 )−1

2)

2

≤ 0 so we conclude that

( f (0 )−12)

2

=0, i.e. f (0 )=12 similarly we conclude that f (1 )=1

2 = 12 so that f ( x ) is not one

to one.

Telescopic Sums and ProductsThere are sums which can be put in the form

∑i=0

n

(F (i+1)−F (i)¿)¿

And there are products which can be put in the form

∏k =1

n F (k+1)F(k )

Or

∏k =1

n F(k )F (k+1)

In both the cases the in between terms cancel and we compute the value of the

sum or the product whichever the case may be.

Examples1. Evaluate

∑0

n 1(k+1 )√k+ k √(k+1)

Solution:- the kth term in the sum is

1(k+1 ) √k+ k √ (k+1 )

=(k+1 ) √k−k √( k+1 )

k (k+1)=( 1

√k− 1

√k+1 )So that the required sum becomes

∑0

n

( 1√k

− 1√k+1 )=¿1− 1

√n+1¿

2. Prove that

∏n=2

∞

(1− 1n2 )=1

2

Solution:- for large values of N

∏n=2

N

(1− 1n2 )=∏

n=2

N

(1−1n )∏

n=2

N

(1+ 1n )=∏

n=2

N

( n−1n )∏

n=2

N

( n+1n )=( 1

N )( N+12 )

Which 12 as N ∞.

Arranging Numbers in Order

We illustrate this method by using following examples

Examples

1. Given 7 distinct positive integers that add upto100 Prove that some three of them add

upto at least 50.

Solution:- we assume that the seven numbers are a, b, c,…….., g with a < b < c

<…….< g. We will show that e + f+ g 50.

If e > 15 then we get e + f + g 16 + 17 + 18 = 51.

If e ≤ 15 , then we get a + b + c + d ≤ 14 + 13 + 12 + 11 = 50 and hence e + f +

g 100 – (a + b + c + d) 50.

2 Prove that among any 50 distinct positive integers strictly less than 100 there are two that

are coprime.

Solution:- Order the numbers as x1< x2<……….<x50. If in this sequence there are

two consecutive integers then they are coprime and we are done. Otherwise we have

x50≥ x1+2.49≥ 99. Sincex50<100, implies that x50=99 and x1=1 , x2=3 , x3=5 , ……so

on. Among these 3 and 5 are coprime.

Parity considerationsSome problems are solved by just looking at the parity( even and odd property) of

the numbers under consideration.

Examples1. There are 100 soldiers in a detachment, and every evening three of them are on

duty. Can it happen that after a certain period of time each soldier has shared duty

with every other soldier exactly once.

Solution: - Suppose that what ever required in the problem can happen. Fix one of

the soldier S1(say) and during that period of time S1 has been on duty m times and

as no person on duty with him is repeated the number of persons with whom he

has shared his duty is 2m which is even however he need to share duty with 99

other soldiers once and as 99 is odd, this is not possible.

2. Forty five Points are chosen along line AB, all lying outside the segment AB.

Prove that the sum of the distances from these points to the point A is not equal to

the sum of distances of these points to the point B.

Solution:- for any point X lying out side segment AB, the difference

AX−BX=± AB.Let the given point be X1 , X2 ,………. X 45, then we obtain

∑i=1

45

A X i−¿∑i=1

45

B X i=∑i=1

45

(A X i−B X i¿)=∑i=1

45

± AB0¿¿

(As for the last sum to be zero we must have as many AB with positive sign as are

with negative sign. But the total number of terms is 45 which is odd. )

3. The product of 22 integers is 1. Show that their sum cannot be zero.

Solution:- Each of these integer must divide the product which is 1 so that the

values taken by these integers are ±1. To obtain the sum as zero we must have

equal number of +1’s as are -1’s. But then there are 11 -1’s, which is odd and the

product of these numbers must then be -1, a contradiction and hence the proof.

4. The numbers from 1 to 10 are written along a row . Can the signs + and – be

placed between them so that the value of the resulting expression is 0.

Solution:- Note that the sum of first 10 numbers is 45, which is an odd number.

Now denote by ‘a’ the sum of sum of numbers with positive sign and by ‘b’ the

sum of numbers with negative sign. Clearly the value of the expression is a - b If

this value happens to be zero this will imply that a = b, and hence

45 = a + b = 2a, an even number, a contradiction. This proves the result.

5. Consider the chess board with two corner squares along one of the diagonals

removed as shown in the Fig.4 Can this be covered by 2×1 rectangles as shown

in Fig.5? If possible to cover 31 such rectangles are required.

Fig.4

Fig.5

Solution:- Answer is No. We justify our answer by the following argument.

As shown in the Diagram, there are 32 black squares and 30 white

squares. If we pair each black square with a white square in whichever way we

get 30 pairs and two black squares are left unpaired. Also when we place any

block of size 2×1 horizontally or vertically on the board it covers equal areas in

black as well as in white. So complete covering is not possible

Looking at Invariants

In some of the problem we allow certain things or numbers to change, i.e. we replace a

set of numbers by another set of numbers etc. In order to solve such a problem we look out for

something which does not undergo any change during the whole process. This helps us in

solving the problem. The something which remains same during the process is called invariant.

We list below some examples.

Examples

1. Suppose for a positive odd integer n, the numbers 1,2,3,………….2n are written on a

black board.. We pick any two numbers a and b erase them and write instead ,|a−b|.

Prove that an odd number will remain at the end.

Solution:- Suppose S is the sum of all numbers on the board at any stage. ( Here S is a

variable quantity). Initially S=1+2+3+………….+2 n=n (2n+1 ), which is an odd

integer. Each step reduces S by 2min (a ,b ) which is an even number so that the parity of S

will remain odd. Hence S will also be odd at the end. But at the end we are left with only

one number, hence that number must be odd.

2. The numbers 1, 2, 3,…….., 20 are written on a blackboard. It is allowed to erase any two

numbers a and b and write the new numbera+b−1. What number will be on the

blackboard after 19 such operations?

Solution: - For any collection n of numbers on the black board, let

X=(∑ of thenumbers) – n. Suppose that we have transformed the collection of n

numbers to a collection of n−1 numbers as described. How would the value of X

change? If we have picked the numbers a and b and replaced these by a+b−1 and if we

denote the new value of X by Y then

Y=(∑ of the numbers ) – a−b+(a+b−1 )−(n−1 )=(∑ of the numbers ) – n=X . X is an

invariant. Initially the value of X=(1+2+… ..20 )−20=190. Therefore after 19 operations

only one number will be left and the value of X will again be 190. the number = X+1

=191.

Doing a problem in two different ways

In order to solve a problem, we use the method of approaching some other problem related to it in two different ways and the results lead to the solution of the given problem.

Examples

1. Prove that

2n=∑k=0

n

(nk) (1)

Solution:- here we give a proof without using binomial theorem.Let us count the number of subsets of the set S= {1, 2, 3, ……,n} consisting of n elements. Clearly the number of subsets is same as number of ways of constructing a subset as different ways will give rise to distinct subsets. Consider any element say ‘k’ there are two ways of considering this element while forming a subset either to include this or to exclude this. Same is true for each of the n elements. So by multiplicative rule of permutations and combinations total number of ways of forming subsets and hence the total number of subsets of S is 2n.

Now we count the number of subsets having k elements for 0 ≤ k ≤ n, which is

clearly (nk), now by additive property the total number of subsets of S is ∑k=0

n

(nk ). Hence

(1) is established.2. Prove that there does not exist an equilateral triangle in the xy-cartisian plane whose

vertices have integral co-ordinates.Solution:- If there is such a triangle, with vertices ( x1 , y1 ) , ( x2 , y2 ) , ( x3 , y3 ) with x i and y i integers. Now we compute the area of this triangle in two ways,Using determinants

Area of the ∆=12|x1 y1 1

x2 y2 1x3 y3 1|,

this is a rational number. On the other hand

Area of the ∆=14

side2

√3=(a rational number)√3,

Which √3 is a rational number, a contradiction, which proves the result.

Relating two different situations

Here firstly we count the number of binary strings of length n with r zeros and n−rones. Which is equal to the number of permutations of n things where r are of one

kind and n – r of second kind. This number is(nr ).Now let us count the number of paths from A to C where along the following grid horizontal left to right and vertical upwards motions are allowed.

D

A Fig.7 B

C

To reach from A to C one need to make 6 vertical steps and 9 horizontal steps. (Total 15 steps) While constructing a path we write 1 if horizontal step is taken and we write 0 if vertical step is taken, this way we obtain a binary sequence of length 15 with 6 zeros there is unique such sequence associated to a given path. Also from a binary sequence of length 15 with 6 zeros we can construct an unique path. So that

Number of such paths = Number of binary sequences of length 15 with 6 zeros =(156 ).

Pigeon Hole Principle

This Principle states that if we have more than n+1 objects and n boxes and if we distribute the objects into the boxes randomly or applying some rule, the conclusion is that there must be some box containing at least two objects

Or

More generally if we have more than nk+1 objects and n boxes and if we distribute the objects into the boxes randomly or applying some rule, the conclusion is that there must be some box containing at leastk+1 objects.

Examples

1. In any group of n persons there are two who have the same number of acquaintances.Solution:- Each of the persons may have 0,1,2,3,……., n - 1 acquaintances. If any one of these has 0 acquaintances then none will have n−1 acquaintances and if any one of these has n - 1 acquaintances then none of them will have 0 acquaintance. So that the number of acquaintances for all n persons are coming from the set {0,1,2,3,……., n – 2} or {1,2,3,……., n – 1} We consider boxes numbered 0 to n -2 or 1 to n – 1,as the case may be. Now we allot box numbered i to a person having i acquaintances. As the number of persons is n and number of boxes is n−1, there must be two associated to the same box so the result follows.

2. Let A be a subset consisting of 101 numbers chosen from. {1,2,3,……., 200} Show that there exists a , b A such that a|b.Solution:- We make 100 boxes indexed by first 100 odd integers i.e. 1,3,5,…….,199. Now we pick any element say c of A and write it as c=2k c1, where c1is odd. We put this element in the box indexed c1, this is possible as c1 is odd and less than 200. Now there are 101 elements and 100 boxes one of the boxes must contain at least 2 elements. Let a=2k cand b=2l cbe two such elements in the box indexed c with k<l. Then clearly a|b

3. Prove that one of the numbers 1,11,111,1111,………. Is divisible by 2009.

Solution:- Let a i=111… .1⏟i׿¿

, The given sequence becomes a1, a2 , a3 ,……… .., let ri be

the remainder obtained on dividing ai by 2009. Now as the sequence is infinite and the set of remainders consists of 2009 numbers ( 0 to 2008) there must be two numbers in the sequence a j∧ak , with j<k such that their remainders are same but then ak−a j=11111 ……...0000⏟

withk− j 1' s∧ j 0 ' s, is divisible by 2009 and hence a j−k is divisible by 2009.

Symmetric considerationsWe illustrate this method by the following examples

Examples

1. There are two piles of balls as shown in the fig.8. At each turn, a player may take away as many balls as he chooses, but only from one of the piles. The winner is the player who takes away last ball. Who can win the game?. Explain the strategy.

Fig.8

Solution:- There are two cases(i) Heaps are of equal size.

(ii) Heaps are of unequal size.

In case (i) the second player can win and in the case (ii) first player can win.

The strategy is that the wining player maintains the symmetry and the next

player is forced to disturb the symmetry. In the first case when heaps are of

equal sizes at the beginning the symmetry is already there so first player will

be disturbing the symmetry and the second player will be maintaining it so

second player is the winner. In the second case the first player will make

heaps equal by removing balls from the heap containing more balls so he is

maintaining the symmetry while the second player will be disturbing it. In the

figure second heap contains one ball less than the first heap in this case first

player can win if he adopts this strategy.

2. Two players take turns putting one rupee coins of equal size on a round table,

without piling one coin on top of other and also without overlapping. The player

who cannot place a coin is the looser (A player can place exactly one coin on his

turn). Who can win and with what strategy?

Solution:- In this game first player can win no matter how big the table may be.

To do so, he must place the first coin so that its centre coincides with the center of

the table. After this he replies to each move of the second player by placing a coin

in a position symmetric to the coin placed by the second player, with respect to

the centre of the table. Notice that in such a strategy the positions of the two

players are symmetric after each move of the first player. It follows that if there is

a possible turn for the second player, and then there is a possible response for the

first player, who will therefore win.

q

p

C f b a Fig.6

.