Basic math solution

Basic Math
Solution
For third semester
It is prepared for only self study .
9/22/2074

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Chapter 1
Elementry group theory
Exercise 3.1
1.A binary operation * is defined on the set of integers by
a.
Soln:
(i) m = 3, n = 5
Given, m * n = m + n
So, 3 * 5 = 3 + 5 = 8
(ii) m = 2, n = - 5
So, 2 * (-5) = 2 + (-5) = 2 – 5 = - 3.
b.
(i) m = 3, n = 5
Given, m * n = m – n
So, 2 * (-5) = 2 + (-5) = 2 – 5 = - 3.
(ii) m = 2, n = - 5
Given, m + n = m – n.
So, 2 * (-5) = 2 – (-5) = 2 + 5 = 7.
c.
(i) m = 3, n = 5
Given, m * n = nm + m + n
So, 3 * 5 = 3 * 5 + 3 + 5 = 23.
(ii) m = 2, n = - 5
Given, m * n = mn + m + n
So, 2 * (-5) = 2 * (-5) + 2 + (-5) = - 10 + 2 – 5 = - 13.

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2.
Soln:
a.
S = {-1,0,1}
Consider 1,1 ԑ S.
Here, a * b = a + b
So, 1 * 1 = 1 + 1 = 2 ∉ S.
Thus, a * b = a + b Is not a binary operation on S = {-1,0,1}
b.
S = {1,2,4}
Consider 2,4 ԑ S.
Here, a * b = a.b
So, 2 * 4 = 2 * 4 = 8 ∉ S.
Thus, a * b = abIs not a binary operation on S = {1,2,4}
c.
S = {2,4,6,8,10,…}
Here, a * b = a + b
We know that the addition of two even numbers is always an even number which
belong to the set S.
So, a * b = a + b is a binary operation on the set S = {2,4,6,8,10,….}
d.
S = Set of integers.
Given, a * b = a – b
Since, the difference of two integers always yields an integer.
Thus, for all,a,b ԑ S, a * b = a – b ԑ S.
So, a * b = a – b is a binary operation on the set ‘S’ of integers.
3.
Soln:

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Cayley’s table.
X -1 1
-1 1 -1
1 -1 1
From the above table,
(-1) * (-1) = 1 ԑ S, (-1) * 1 = - 1 ԑ S.
1 * (-1) = - 1 ԑ S, 1 * 1 = 1 ԑ S.
So, multiplication is a binary operatio on set a S.
4.
Soln:
The set of positive integers is denoted by Z
a.
Any m, n ԑ Z+Z+à m + n ԑ Z+Z+.
So, Z+Z+ is closed under addition.
b.
Any m,n ԑ Z+Z+àmn = nm
So, Z+Z+ is commutative under multiplication.
c.
Consider, 2,3,4 ԑ Z+Z+.
2 – (3 – 4) = 2 – (-1) = 2 + 1 = 3.
And (2 – 3) – 4 = - 1 – 4 = - 5
Thus, 2 – (3 – 4) ≠ (2 – 3) – 4.
SO, Z+Z+ is not associative under subtraction.
5.
Soln:
a.
For closure,
Consider 1,2 ԑ Z.

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Then , m * n = 1212(m – n)
So, 1 * 2 = 1212(1 – 2) = −12−12∉ Z.
So, Z is not closed.
For Associative,
Consider 2,4,8 ԑ Z.
= 2 * (4 * 8) = 2 * {12(4−8)}{12(4−8)}
= 2 * (-2) = 1212 {2 – (-2)} = 2
And (2 * 4) * 8 = {12(2−4)}{12(2−4)} * 8
= (-1) * 8 = 1212 {(-1) – 8} = 1212 (-9) = −92−92∉ Z.
Hence, the operation * defined on Z is now associative,
For Commutative,
For, m , n ԑ Z, m * n = 1212(m – n)
= −12−12(n – m) = - (n * m).
So, the operation * is not commutative on Z.
b.
For closure,
For, m,n ԑ Z, m * n = n ԑ Z.
So, the operation ‘*’ is closed on Z.
For associative,
Since, for m,n and p ԑ Z.
(m * n) * p = n * p = p ԑ Z.
And m * (n * p) = m * p = p ԑ Z.
So, (m * n) * p = m * (n * p)
So, the operation ‘*’ is associative.
For commutative.
Since, for m,n ԑ Z, m * n = n ԑ Z.
And n * m = m ԑ Z.
But m ≠ n; so the operation * is not commutative on Z/
c.
For closure
Since, for m,n ԑ Z, m * n = m + n + 1 ԑ Z.
So, * is closed on Z,
For associative

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Since, for n,m,p ԑ Z.
(m * n) * p = (m + n + 1) * p
= (m + n + 1) + p + 1 = m + n + p + 2.
And m * (n * p) = m * (m + p + 1)
= m + n (n + p + 1) + 1 = m + n + p + 2.
So, (m * n) * p = m * (n * p), for all m,n,p ԑ Z.
So, * is associative on Z.
For commutative.
For m,n ԑ Z.
m * n = m + n + 1.
And n * m = n + m + 1 = m + n + 1
So, m * n = n * m, for all m,n ԑ Z.
So, * is commutative on Z.
Exercise 3.2
1.
Soln:
Since, 0 + 40 = 0
1 + 40 = 0 + 41 = 1
2 + 40 = 0 + 42 = 3
3 + 40 = 0 + 43 = 3.
So, 0 is the identity element.
Since, 2 + 42 = 0
So, the inverse element of 2 is 2.
And 3 + 41 + 1 + 43 = 0
So, the inverse element of 3 is 1.
2.
Soln:
The composition table for G = {0,1,2} under multiplication modulo 3 is
X3 0 1 2
0 0 0 0
1 0 1 2

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2 0 2 1
Since, 1 * 31 = 1
So, identity element of 1 is 1.
And, 2 * 31 = 2 = 1 * 32.
Now, for inverse elements,
1 * 31 = 1
So, inverse of 1 is 1.
2 * 32 = 1.
3.
Soln:
Since, 1 + 0 = 0 + 1 = 1.
And 2 + 0 = 0 + 2 = 2.
So, 0 is the identity element of both 1 and 2.
Since, 1 + (-1) = (-1) + 1 = 0
Since, - 1 is the inverse element of 1.
And, 2 + (-2) = (-2) + 2 = 0
So, - 2 is the inverse element of 2.
4.
Soln:
Since, (-1).1 = 1. (-1) = - 1
And 1.1 = 1
So, 1 is the identity element of both 1 and -1.
Now, (-1) * (-1) = 1
And 1 * 1 = 1
So, the inverse elements of – 1 and 1 are -1 and 1 respectively.
5.
a.
Soln:
Let e be the identity element of 3 under binary operation defined by m*n = m + 1 + 1.

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Then 3 * e = 3
Or, 3 + e + 3 [m * n = m + n + 1]
So, e = -1.
Again, let e’ be the identity element of – 2.
Then (-2) * e’ = - 2.
Or, (-2) + e’ + 1 = -2
Or, e’ + 1 = 0
So, e’ = - 1.
So, - 1is the required element of both 3 and – 2.
Again, le t ‘a’ be the inverse element of 3 under the given binary operation *
Then, 3 * a = e
Or, 3 + a + 1 = - 1 [m * n = m + n + 1, e = -1]
Or, a = - 1 – 4 = - 5.
So, - 5 is the inverse element of 3,
And let a’ be the inverse element of – 2 under ‘*’.
Then (-2) * a’ = e’
Or, (-2) + a’ + 1 = - 1.
Or, - 1 + a’ = - 1.
So, a ‘ = 0
So, 0 Is the inverse element o f – 2.

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Exercise 3.3
1.
a.
Soln:
It is a false statement because the order of group G = Number elements in G = 4,
which is finite.

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b.
It is a true statement because the group contains two elements, so the order of given
group is 2.
c.
Soln:
If is a false statement because it is not closed, for e.g.
(- 2).2 = - 4 ∉ {-2,-1,0,1,2}
d.
Soln:
It is false statement because it is not closed,
For e.g. 4 + 8 = 12 ∉ {2,4,6,8,10}
2.
Soln:
Here, 1 + 1 = 2 ∉ S where, 1 , 1 ԑ S.
So, S is not closed under addition.
Thus, S is not a group under addition.
3.
Soln:
Existence of Identity: Let a be any natural number and if it exists, let e be the identity
element of a, then a + e = a. à e = a – a = 0 which is not a natural number,
So, identity element under addition doesn’t exist in the set of natural numbers,
Hence, the set of natural numbers does not form a group under the addition
operation,
4.

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Soln:
Closure property: From the multiplication table, we see that T is closed under
multiplication, i.e.
Multiplication Table.
X -1 1
-1 1 -1
1 -1 1
(-1) * (-1) = 1, (-1) * 1 = -1
1 * (-1) = - 1, 1 * 1 = 1
So, a * b ԑ T for all a,b ԑ T.
Associative property: Here, the elements of T are – 1 and 1. We know that all
integers under multiplication obey associative law. So, the element of T being
integers satisfy associative law under multiplication.
i.e. a * b(b * c) = (a * b) * c for all a,b,c, ԑ T.
For e.g. (-1) * {1 * (-1)} = (-1) * (-1) = 1.
And {(-1) * 1} * (-1) = (-1) * (-1) = 1
So, (-1) * {1 * (-1} = {(-1) * 1} * (-1) and so on.
Existence of Identity element,
Since (-1) * 1 = (-1) = 1 * (-1)
And 1 * 1 = 1
So, 1 is the multiplicative identity in T.
Existence of inverse
Each element of T is an inverse of itself since,
1 * 1 = 1 and (-1) * (-1) = 1.
Hence, T = {-1,1} forms a group under multiplication.
5.
a.
Soln:
Multiplication Table.
X 1 -1 i -i
1 1 -1 i -i
-1 -1 1 -i i

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i I -i -1 1
-i -i I 1 -1
From above table, we see that for al a,b ԑ G, a * b ԑ G. So G is closed under
multiplication.
b.
Soln:
Since all the complex numbers satisfy associative law under multiplication. So all the
elements of G being complex numbers also satisfy associative law.
Ie. a * (b * c) = (a * b) * c, for all a,b,c ԑ G.
c.
Soln:
For all a ԑ G, a * 1 = 1 * a = a.
So, 1 is the multiplication identity in G.
Since, 1 * 1 = 1
(-1) * (-1) = 1
i * (-i) = -i2
= 1
and (-i) * I = -i2
= 1
So, the inverse of 1, -1, i and – I and 1, - 1, - i and i , respectively. So ,every element
of G possesses an inverse element in G. Hence, identity element and inverse exists.
d.
Soln:
Yes, G forms a group under multiplication as (G,x) is closed, associative, and the
identity and inverse exist in G.
6.
Soln:
Let a,b ԑ Z.
Since, a,b ԑ Z à a + b ԑ Z.
So, closure property is satisfied.
If a,b,c ԑ Z, then

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a + (b + c) = a + b + c ԑ Z.
Also, (a + b) + c = a + b + c ԑ Z.
So, a + (b + c) = (a + b) + c.
Hence, associative property is satisfied. 0 is an integer and
0 + a = a + 0 = a ԑ Z.
0 is an identity element.
Also, if a ԑ Z then – a ԑ Z.
Or, a + (- a) = (-a) + a = 0.
So, - a is the inverse element of a. Above relations are true for all elements of Z.
Hence, the set of integers Z forms a group under addition.
7.
Soln:
* A b c
a A b c
B B c a
C C a b
From the table, we see that the operation defined on any two elements of G gives
and element of G itself.
So, G is closed under the operation *.
a * (b * c) = a * a = a
(a * b) * c = b * c = a
So, a * (b * c) = (a *b) * c
So, * satisfies associative property.

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Since, a * a = a, a * b = b * a= b
And a * c = c * a = c, so a is an identity element.
a * a = a, so a is the inverse of a,b * c = c * b = a, so b and c are the inverse
elements of c and respectively,
So, (S,*) forms a group.
8.
Soln:
Composition table for G under the addition modulo r (+4) is presented below.
+4 0 1 2 3
0 0 1 2 3
1 1 2 3 0
2 2 3 0 1
3 3 0 1 2
From the table, we see that sum of any two elements of G modulo 4 is an element of
G. So, +4 satisfies closure property.
Again, 1 +4 (2 +4 3) = 1 +4 1 = 2.
And (1 +4 2) +4 3 = 3 +4 3 = 2.
This result is true for all elements of G. Hence, +4 satisfies associative property.
From the second row and second column of above table, 0 is the identity element.
Form the second row and second column of above table, 0 is the identity elemnt.
Since, 0 +4 0 = 0, 1 +4 3 = 3 +4 1 = 0
And 2 +4 2 = 0
So, the inverse elements of 0,1,2 and 3 are 0,3,2 and 1 respectively.
So, G forms a group under addition modulo 4.
9.
Soln:
Since a is an identity element, so,
a * a = a, a * b = b * a = a.
So, also a * a = a, a is the inverse of a.

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And b * b = a so that b is the inverse of b.
Now the required composite table is given below.
* A B
A A B
B B A
Exercise 3.4

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4.
Soln:
X o x = x
Or, x o x = x o e, where e is the identity element of G.
By left cancellation law, we have,
X = e.
Since, identity element of a group is unique, so, x = e is a unique solution of given
group equation.

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BijayDhobi
Chapter 2
Applications of Derivatives
Exercise 10.1

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Chapter 3
Differentials Equations
Exercise 12.1

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Chapter 4
Computational Method
Exercise 22.1

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Chapter 5
Numerical integration
Exercise 24.1

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