Assignment-2
CHAPTER 2: Proof and Strategies solution
Direct Proof :
1. Use a direct proof to show that sum of two odd integers is even .
Solution : Let s = 2m+1 and t=2n+1 are two odd integers .
Then ,
s+t = 2m+2n+2
s+t = 2(m+n+1)
Here s+t (sum of two odd integers ) is multiple of 2 i,e 2(m+n+1) is even .
Hence we proved .
2. Proof that sum of two even integers is even .
Solution: Let s= 2m and s=2t are two even integers.
Then,
S+t = 2m+2t
=2(m+t)
Here s+t (sum of two even integers ) is multiple of 2 again i,e 2(m+t) is even .
Hence we proved .
3. Use a direct proof to show that every odd integers is the difference of two squares .
Solution:
Let n is odd integer ,then we can write
n= 2s+1 for some integer s ,
difference of two squares is ,
(s+1)2- s2 = s2+1+2s-s2 = 2s+1 =n .
Hence we proved
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5) Use a proof by contradiction to prove that the sum of an irrational number and a rational number
is irrational .
Solution : suppose r is rational and i is irrational .
And s= i + r , is rational .
S+(-r) = i is rational which is a contradiction .
6) Prove that if x is irrational than 1/x is irrational .
Solution
proof by contraposition : If 1/x is rational , then by definition we have
1/x= p/q for some integers p and q where q =\ 0.
Now x= 1/(1/x) = 1/p/q = q/p . Hence x can be written as quotient of two
integers with the non zero denominator .Thus x is rational .
Since we proved the contra positive to be true ,then the original statement
must be true.
Exhaustive proof and Existence Proof:
1) Prove that the only consecutive positive integers not exceeding 100 that are perfect powers
are 8 and 9 .
Solution : The squares of positive integers not exceeding 100 are: 1, 4, 9, 16, 25,
36, 49, 64, 81, 100.
2) The Cubes of positive integers not exceeding 100 are: 1, 8, 27, 64
3) The 4th powers of positive integers not exceeding 100 are: 1, 16, 81
4) The 5th powers of positive integers not exceeding 100 are: 1, 32
5) The 6th powers of positive integers not exceeding 100 are: 1
23 = 8 and 32 = 9 , proof holds.
2) Prove that if x and y are real numbers then max(x,y)+min(x,y)=x+y
Solution :
We can proof by two cases ,
Case 1: when x<y
Max(x,y)+min (x,y)= y+x
= x+y
Case 2: when x>=y
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Max(x,y)+min(x,y)= x+y
There are only two cases ,Hence the equality always holds .
3) Prove that there is a positive integer that equals the sum of the positive integers not
exceeding it .Is your proof constructive or not constructive .
Solution :
If i=10 is a positive integer , then we can write i as a sum of positive integers not
greater than 10 .
i= 6+4, i=9+1 ,
Another example .if k=100 is positive integer , then we can write k as a sum of
positive integers not greater than 100 .
K=40+50 , k=10+50+40 .
Hence constructive
4) Prove that there exists a pair of consecutive integers such that one of this integers is perfect
square and one is a perfect cube .
Solution : 8 = 23, and 9 = 32
Uniqueness proof:
1) Suppose if a and b are odd integers with a =/ b . show that there is a unique integer c such
that
|a-c|=|b-c|.
Solution: equation |a-c| = |b-c| is equivalent to disjunction of two equations
a-c = b-c
which is , a=b which contradicts the assumption made in the given problem
so the equation is equivalent to a-c = -b+c
by solving , a-c+b+c=-b+c+b+c
a+b=2c
c=a+b/2 which is unique solution for c .
further more this c is an integer because sum of odd integers a and b is even
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2) Show that if n is odd integer .Then there is unique integer k such that n is the sum of k-2 and
k+3.
(hint :solve the equation k-2+k+3 for value of k).
Solution: n = k-2+k+3
Solve the above for k
n= 2k+1
k=(n-1)/2
This is the one and only one value of k .Hence n is odd ,then n-1 is even .so k
is an integer
Tilings :
1) Can we tile a board obtained by removing one of the four corner squares of standard checker
board .
Solution: Standard checker board = 8 x 8 = 64. If we remove one corner square
then 64 -1 = 63. As we know that one domino has two tile and we can not
overlap while tiling. Therefore we can not tile a board having odd number
squares i.e 63.
2) Prove or disprove that you can use dominoes to tile the standard checkboard with two adjacent
corners removed .
Solution: Suppose that the upper left and upper right corners of the board are
removed. Place 3 dominoes horizontally to fill the remaining portion of the
first row, and fill each of the other seven rows with 4 horizontal dominoes.
3)Prove or disprove that you can use dominoes to tile the standard checkboard with all four
corners removed .
Solution :
.yes, we can tile… the standard checkboard with all four corners removed .
64-4=60. Therefore 30 horizontally and 30 vertically
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3) Prove or disprove that you can use dominoes to tile a 4*4 checkboard with three corners
removed .
Solution: A domino will cover two adjacent squares, one white and one black.
So, in order to tile the board, there must be the same number of black
squares as there are white squares.
But if you remove three (black) corner squares, there will be 5 black squares
and 8 white squares left. Hence, no matter how hard you try, there will
always be at least two white squares left over that cannot be covered with
one domino.
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