Solved Math Assignments and Essays - Desklib

The article contains solved assignments for MATH 1061/7861. It covers topics like unique prime factorization, Euclidean algorithm, irrational numbers, perfect squares, and mathematical induction. The article also provides counterexamples and proofs for various mathematical statements.

MATH 1061/7861
Assignment 2
Student Name
[Pick the date]

Question 1
(a) Method: Unique prime factorisation
 gcd ( 270 , 924 )∧lcm ( 270 , 924 )
2 270
5 135
3 27
3 9
3
2 924
2 462
11 231
7 21
3
270=2∗5∗33
924=22∗11∗7∗3
gcd ( 270 , 924 )=21∗50∗31∗110∗70=2∗3=6
lcm ( 270 , 924 ) =22∗5∗11∗7∗33=41580
(b) Method: Euclidean algorithm
gcd ( 132153 , 4263 )
4263 132153 31
132153
0
132153=4263∗31+ 0
gcd ( 132153 , 4263 )=4263
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Question 2
(a)The given statement is false as can be established from the following counterexample.
Let x = 2.5, y =-2.5
L.H.S. = I2.5-2.5I = 0
R.H.S. = 2.5 + 2.5 = 5
Clearly, the two are not equal and hence the given statement is false.
(b)The given statement is false as can be established from the following counterexample.
Let x = 2.5, y =-2
L.H.S. = I2.5-2I = 0.5
R.H.S. = 2.5 + 2 =4.5
Clearly, the two are not equal and hence the given statement is false.
(c) “For all irrational numbers x, the number 10x+3 is also termed as irrational.”
Let assume 10x+3 is rational and hence,
10 x+ 3= p
q
Where,
p , q ∈ z' ∧q ≠ 0
10 x=( p
q )−3
10 qx= p−3 q
x= 1
10
( p−3 q )
q
2

Here, x comes out to be rational for the contraction in the assumption that 10+x is rational and
hence, it is proved that 10+x would be irrational for being x as irrational.
(d) For all a , b , d ∈ z' if d |ab , thend|a∨d∨b
Let’s assume
d∨a
Hence, the greatest common divisor of a and d would be equal, which mean d' ( ≥ 1 )
gcd ( a , d )=d' d' <a
a x0 +d y0=d' d' < d
It is because the gcd is always represented as a linear combination. It can also be understood with
the help of the example shown below.
G |4∗3∨6|12
But
6∨4∧6∨3
Also, one needs to consider d to be the prime hence,
d' =1
gcd ( a , d )=1
a x0 +d y0=d'
a x0 +d y0=1
b (a x0+ d y0 )=b
ab x0 +db y0 =b
Thus,
d |ab∧d |db y0
So,
d∨ab x0+ db y0
3

Hence,
d∨b
(e) For all a , b , d ∈ z' if d |a+b ,∧d|a , thend ∨b
Based on the above conditions the following equations can be derived.
( a+b ) =d k 1
a=d k2
Thus,
( a+b )−a=d k 1−d k2
a+ b−a=d k 1−d k2
b=d k1−d k 2
b=d ( k ¿¿ 1−k2 )¿
d
b
Proved
Question 3
(a) The infinite perfect squares are shown below.
1 , 4 , 9 , 16 , 25 ,36 , 49 … … … … … …. .
It is apparent that there are only four possible remainders which the perfect square is divided by
7.
Hence,
When 1 is divided by 7 then the remainder would be 1.
When 4 is divided by 7 then the remainder would be 4.
When 9 is divided by 7 then the remainder would be 2.
4

When 25 is divided by 7 then the remainder would be 2.
When 36 is divided by 7 then the remainder would be 1.
When 49 is divided by 7 then the remainder would be 0.
Therefore, the remainder (r) would be 0, 1, 2, 4.
(b) The case when 2kis divided by 7 then,
20=1 , r=1
21=2 , r=2
22=4 , r=4
23=8 ,r =1
24 =16 , r=2
25=32 , r=4
26=64 , r=1
Therefore, the remainders (r) would be 1, 2, 4.
(c) The following conclusion can be drawn based on the above two parts.
n2 results remainder 0 ,1 , 2, 4
2m results remainder 1 , 2 , 4
Hence, if one will add the above two 2m +n2 then any of the two remainders, then there would be
no zero as shown below.
0+1 , 0+2 , 0+4
1+1 ,1+2 , 1+4
2+1 , 2+ 2, 2+4
4 +1 , 4+ 2, 4 + 4
Thus, it can be seen from the above that we will never get a zero or multiple of 7 and thereby, 7
does not divide 2m +n2i.e.
5

2m +n2 ≠ 0 ( Mod 7 )
Thus, no integer solution is exists for this.
Question 4
(a) The general term of the sequence is given by (-2)k where k≥0
Hence, first term = (-2)0 = 1
Second term = (-2)1 = -2
Third term = (-2)2 = 4
Fourth term = (-2)3 = -8
Hence, the first four terms of the given sequence are 1,-2,4 and -8.
(b) Mathematical expression needs to be proved through induction method is shown below.
1−3 ∑
k=0
n
(−2)k=(−2)n+1
Initial step: Let’s put n= 0 in the expression and prove that LHS and RHS are same.
Put n=0
1−3 ∑
k=0
0
(−2)k=(−2)0+1
1−3 (−2 )0 = (−2 )1
1−3 ( 1 ) =−2
−2=−2( Proved∧satisfied )
Inductive step: Assume there is n=m, which is a positive integer for which the expression is true
and hence, the second step in the induction in that it is also held to be true for n=m+1.
Put n= m (Positive integer). Let us assume that for n= m, the following equation is satisfied
1−3 ∑
k=0
m
(−2)k=(−2)m+1
6

Let’s put n= m+1 (Positive integer)
LHS=1−3∑
k=0
m+1
(−2)k
¿ 1−3 ∑
k=0
m
(−2 )k−3 (−2 )m +1
¿ (−2 )m +1−3 (−2 )m +1
¿−2∗( −2 ) m+1
RHS= ( −2 ) ( m+1 ) +1= ( −2 ) m +2=−2∗( −2 ) m +1
Hence, it is true as LHS= RHS.
7